problem stringlengths 1 13.6k | solution stringlengths 0 18.5k ⌀ | answer stringlengths 0 575 ⌀ | problem_type stringclasses 8
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class | __index_level_0__ int64 0 742k |
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6・127 Let $n \in Z$, and the function $f: Z \rightarrow R$ satisfies
$$f(n)=\left\{\begin{array}{l}
n-10, \text { when } n>100 \text {, } \\
f(f(n+11)), \text { when } n \leqslant 100 \text { . }
\end{array}\right.$$
Prove that for any $n \leqslant 100$, we have $f(n)=91$. | [Proof] First, let $n \leqslant 100$ and $n+11>100$, i.e., $90 \leqslant n \leqslant 100$, thus
$$f(n)=f(f(n+11))=f(n+11-10)=f(n+1)$$
Therefore, $f(90)=f(91)=\cdots=f(100)=f(101)=91$.
Now let $n<90$. Take $m \in \mathbb{N}$, such that $90 \leqslant n+11 m \leqslant 100$, then we have
$$\begin{aligned}
f(n) & =f^{(2)}(... | proof | Algebra | proof | Yes | Yes | inequalities | false | 735,936 |
6. 128 Functions $f, g, h: N \rightarrow N$ satisfy the following three conditions:
(1) For different $n \in N, h(n)$ takes different values;
(2) The range of function $g(n)$ is $N$;
(3) $f(n) \equiv g(n)-h(n)+1, n \in N$.
Prove: $f(n) \equiv 1, n \in N$. | [Proof] First, prove that $g(n) \equiv h(n), n \in N$. From this and condition (3), we can obtain
$$f(n) \equiv g(n)-h(n)+1 \equiv 1, n \in N$$
Suppose there exists some $n \in N$ such that $g(n) \neq h(n)$. Since $f(n) \geqslant 1$, for any $n \in N$, we have
$$h(n)=g(n)+1-f(n) \leqslant g(n)$$
Thus, $h(n)<g(n)=k$. ... | proof | Algebra | proof | Yes | Yes | inequalities | false | 735,937 |
6. 129 Consider a non-constant function \( f(n, m) \), defined on the set of all integer pairs, taking integer values and satisfying
\[
\begin{aligned}
f(n, m)= & \frac{1}{4}(f(n-1, m)+f(n+1, m)+f(n, m-1) \\
& +f(n, m+1)), \quad n, m \in \mathbb{Z}
\end{aligned}
\]
Prove: (1) Such a function exists;
(2) For each \( k ... | [Proof] (1) For example, the function $f(n, m)=n, n, m \in Z$ satisfies all the conditions in the problem.
(2) Suppose the conclusion is not correct, i.e., for some $k \in Z$, there exists a function $f(n, m)$ that satisfies the conditions of the problem, and every value of it does not exceed $k$ (the case where it is ... | proof | Algebra | proof | Yes | Yes | inequalities | false | 735,938 |
6-131 Let $S$ be a subset of the set of all integer pairs. If the function $f: S \rightarrow S$ is invertible and for any $(n, m) \in \dot{S}$,
$$f(n, m) \in\{(n-1, m),(n+1, m),(n, m-1),(n, m+1)\}$$
then the function $f(n, m)$ is called universal. Prove: If there exists at least one universal function, then there exis... | [Proof] A point $(n, m) \in S$ is called an even point or an odd point depending on whether the sum $n+m$ is even or odd. Suppose the universal function $g(n, m)$ exists, then its inverse function $g^{-1}(n, m)$ is also universal. Define the function $f(n, m)$ as follows: when $(n, m) \in S$,
$$f(n, m)=\left\{\begin{ar... | proof | Combinatorics | proof | Yes | Yes | inequalities | false | 735,940 |
6-133 Let $N_{0}$ denote the set of non-negative integers. Find a bijection $f$ from $N_{0}$ to $N_{0}$ such that for all $m, n \in N_{0}$, we have
$$f(3 m n+m+n)=4 f(m) f(n)+f(m)+f(n)$$ | [Solution] Let $a_{1}0$ when, suppose the standard factorization of $3 n+1$ is
$$a_{i_{1}^{1}}^{\alpha} a_{i_{2}^{2}}^{\alpha} \cdots a_{i_{r}^{r}}^{\alpha} \cdot b_{j_{1}}^{\beta_{1}} b_{j_{2}^{2}}^{\beta} \cdots b_{j_{s}^{s}}^{\beta}$$
where $i_{1}0, \beta_{1}, \cdots, \beta_{s}>$ 0, and $\beta_{1}+\cdots+\beta_{s}$... | proof | Algebra | math-word-problem | Yes | Yes | inequalities | false | 735,941 |
$1 \cdot 99$ Prove: For any positive integer $n$, there exist $n$ consecutive positive integers, none of which is a prime power. | [Proof] Let $N=[(n+1)!]^{2}+1$, then the $n$ consecutive positive integers $N+1, N+2, \cdots, N+n$ satisfy the requirement.
If not, then there exist positive integers $j, m, 1 \leqslant j \leqslant n, m \geqslant 2$ and a prime $p$, such that
$$N+j=p^{m}$$
By the definition of $N$, $(1+j) \mid(N+j)$, but $(1+j)^{2} \... | proof | Number Theory | proof | Yes | Yes | inequalities | false | 735,942 |
6・134 Let $S=\{0,1,2,3, \cdots\}$ be the set of all non-negative integers. Find all functions $f$ defined on $S$ and taking values in $S$ that satisfy the following condition:
$$f(m+f(n))=f(f(m))+f(n)$$
for all $m, n \in S$. | [Solution] Let $f: S \rightarrow S$ satisfy
$$f(m+f(n))=f(f(m))+f(n)$$
for all $m, n \in S$.
Taking $m=n=0$, we get $f(f(0))=f(f(0))+f(0)$, hence
$$f(0)=0.$$
Substituting $m=0$ in (1), we get
$$f(f(n))=f(n) \quad(\text{for any } n \in S).$$
Thus, (1) becomes
$$f(m+f(n))=f(m)+f(n)$$
for all $m, n \in S$.
Let the ran... | proof | Algebra | math-word-problem | Yes | Yes | inequalities | false | 735,943 |
6. 135 Let $N$ be the set of all positive integers, and $f$ a function from $N$ to $N$ itself, such that for any $s$ and $t$ in $N$, the following holds:
$$f\left(t^{2} f(s)\right)=s(f(t))^{2}$$
Determine the smallest possible value of $f(1998)$ among all such functions $f$. | (1)
Let $t=1$, we get
$f(f(s))=s(f(1))^{2}$.
(2)
Let $s=1$, substitute into (1) to get
$f\left(t^{2}(f(1))=(f(t))^{2}\right.$.
(3)
Let $f(1)=K$, then (2) and (3) become
$f(f(s))=K^{2} s$,
(4)
$$f\left(K t^{2}\right)=(f(t))^{2}$$
(5)
Let $s=1$, substitute into (4) to get
$f(K)=K^{2}$.
Let $s=f(K)$, substitute into (1) t... | 120 | Algebra | math-word-problem | Yes | Yes | inequalities | false | 735,944 |
6-136 Let $N$ be the set of natural numbers. Let $f: N \rightarrow N$ satisfy the conditions: $f(1)=1$, and for any natural number $n$,
$$\left\{\begin{array}{l}
3 f(n) f(2 n+1)=f(2 n)(1+3 f(n)) \\
f(2 n)<6 f(n)
\end{array}\right.$$
Find all solutions to the equation
$$f(k)+f(l)=293, k<l$$ | [Solution] From the given conditions, for all $n \in N$, we have
$$1 \leqslant 3 f(n)(f(2 n+1)-f(2 n))=f(2 n)<6 f(n) .$$
Therefore,
$$\begin{array}{l}
1 \leqslant f(2 n+1)-f(2 n)<2 \\
f(2 n+1)=f(2 n)+1
\end{array}$$
Substituting (2) into (1) yields
$$f(2 n)=3 f(n) .$$
Using (2) and (3), we can prove by mathematical ... | null | Number Theory | math-word-problem | Yes | Yes | inequalities | false | 735,945 |
6- 137 Given a function $f(n)$ on the set of positive integers, satisfying the following conditions:
$f(n)=n-12$, when $n>2000$;
$f(n)=f(f(n+16))$, when $n \leqslant 2000$.
(1) Find $f(n)$.
(2) Find all solutions to the equation $f(n)=n$. | [Solution] (1) From the given information,
$$\begin{array}{l}
f(2000)=f(f(2016))=f(2004)=1992, \\
f(1999)=f(f(2015))=f(2003)=1991, \\
f(1998)=f(f(2014))=f(2002)=1990, \\
f(1997)=f(f(2013))=f(2001)=1989, \\
f(1996)=f(f(2012))=f(2000)=1992, \\
f(1995)=f(f(2011))=f(1999)=1991, \\
f(1994)=f(f(2010))=f(1998)=1990, \\
f(1993... | 1992,1991,1990,1989 | Algebra | math-word-problem | Yes | Yes | inequalities | false | 735,946 |
6-138 $S$ is the set of all non-negative integers. Find all functions $f: S \rightarrow S$, $g: S \rightarrow S, h: S \rightarrow S$ that satisfy the following two conditions:
(1) For any $m, n \in S, f(m+n)=g(m)+h(n)+2 m n$.
(2) $g(1)=h(1)=1$. | [Solution] In condition (1), let $n=0$, we have
i.e. $\square$
$$f(m)=g(m)+h(0)$$
$$\begin{array}{r}
g(m)=f(m)-h(0). \\
\text { In condition (1), let } m=0, \text { we have } \\
f(n)=g(0)+h(n),
\end{array}$$
i.e. $\square$
$$\begin{array}{l}
h(n)=f(n)-g(0). \\
\text { In condition (1), let } m=n=0, \text { we have } \... | f(n)=n^{2}-a n+2 a, \\ g(n)=h(n)=n^{2}-a n+a, \\ a \in\{0,1,2,3,4\} | Algebra | math-word-problem | Yes | Yes | inequalities | false | 735,947 |
6- 139 Find a function $f(m, n)$ that satisfies the following conditions: for every pair of non-negative integers $m, n$,
$$\begin{aligned}
(1) 2 f(m, n)= & 2+f(m+1, n-1)+f(m-1, n+1) \\
& (m \geqslant 1, n \geqslant 1)
\end{aligned}$$
$$\text { (2) } f(m, 0)=f(0, n)=0 \text {. }$$ | [Solution] From (1)
$$\begin{aligned}
& f(m, n)-f(m-1, n+1) \\
= & f(m+1, n-1)-f(m, n)+2
\end{aligned}$$
In the above equation, replace $m, n$ with $m-k, n+k$ (non-negative integer $k \leqslant m$), we get
$$\begin{aligned}
& f(m-k, n+k)-f(m-k-1, n+k+1) \\
= & f(m-k+1, n+k-1)-f(m-k, n+k)+2
\end{aligned}$$
Substitute ... | f(m, n)=m n | Algebra | math-word-problem | Yes | Yes | inequalities | false | 735,948 |
$6 \cdot 140$ Find all functions $f: Q \rightarrow Q$ (where $Q$ is the set of rational numbers) satisfying $f(1)=2$ and
$f(x y) \equiv f(x) f(y)-f(x+y)+1, x, y \in Q$. | [Solution] In the original identity, let $y=1$, then we get the identity
that is $\square$
$$f(x) \equiv f(x) f(1)-f(x+1)+1, x \in Q .$$
$$f(x+1) \equiv f(x)+1$$
Thus, for all $x \in Q, n \in Z$, we have $f(x+n)=f(x)+n$.
Therefore $\square$
$$f(n)=f(1)+n-1=n+1$$
Next, in the original identity, take $x=\frac{1}{n}, y... | f(x)=x+1 | Algebra | math-word-problem | Yes | Yes | inequalities | false | 735,949 |
6. 141 Let $Q^{+}$ be the set of all positive rational numbers. Try to construct a function $f: \boldsymbol{Q}^{+} \rightarrow \boldsymbol{Q}^{+}$, such that for any $x, y \in \boldsymbol{Q}^{+}$, we have $f(x f(y))=f(x) / y$. | [Solution] Let $f$ satisfy
$$f(x f(y))=\frac{f(x)}{y}, x, y \in Q^{+}$$
If $y_{1}, y_{2} \in Q^{+}$, such that $f\left(y_{1}\right)=f\left(y_{2}\right)$, then from (1) we get
$$\frac{f(x)}{y_{1}}=f\left(x f\left(y_{1}\right)\right)=f\left(x f\left(y_{2}\right)\right)=\frac{f(x)}{y_{2}},$$
Since $f(x) \neq 0$, we have... | proof | Algebra | math-word-problem | Yes | Yes | inequalities | false | 735,950 |
6-143 Let the function $f(x, y)$ be defined on the set of all pairs of rational numbers, taking only positive values and satisfying the following 3 identities:
$f(x y, z) \equiv f(x, z) f(y, z) ;$
$f(z, x y) \equiv f(z, x) f(z, y) ;$
$f(x, 1-x) \equiv 1, x, y, z \in Q$.
Then $\quad f(x, x) \equiv 1 ; f(x,-x) \equiv 1, ... | [Proof] In the first identity, take $x=y=0$ and $x=y=1$, then we get $f(0, z)=1$ and $f(1, z)=1$. Next, take $x=y=-1$, then we get
$$1=f(1, z)=f(-1, z) f(-1, \approx)=(f(-1, z))^{2}$$
Therefore, $f(-1, z)=1$. Similarly, from the second identity, we get $f(z, 0)=f(z, 1)=$ $f(z,-1)=1$, thus we have
$$f(0,0)=1, f(0, z) f... | proof | Algebra | proof | Yes | Yes | inequalities | false | 735,952 |
6. 144 Find all infinitely differentiable functions \( f: R \rightarrow R \) that satisfy
$$f(x+y) \equiv f(x)+f(y)+2 x y, x, y \in R$$ | [Solution] In the identity given in the problem, let $x=y=0$ to get $f(0)=2 f(0)$, which means $f(0)=0$. For any $x \in \mathbb{R}$, from the identity
$$f(x+y)-f(x) \equiv f(y)+2 x y, \quad y \in \mathbb{R}$$
we get
$$\begin{aligned}
f^{\prime}(x) & =\lim _{y \rightarrow 0} \frac{f(x+y)-f(x)}{y} \equiv \lim _{y \right... | f(x) = x^2 + ax | Algebra | math-word-problem | Yes | Yes | inequalities | false | 735,953 |
6-147 Find all functions $f: R \rightarrow R$ that satisfy $x f(y)+y f(x) \equiv(x+y) f(x) f(y), x, y \in R$.
untranslated text remains unchanged:
将上面的文本翻译成英文，请保留源文本的换行和格式，直接输出翻译结果。
However, the actual translation is provided above, and the note about the instruction is not part of the translated text. | [Solution] In the identity given in the problem, let $x=y=1$, then we get $2 f(1)=2(f(1))^{2}$, i.e., $f(1)=0$ or $f(1)=1$. We will examine these two cases separately as follows:
(1) If $f(1)=0$, then by setting $y=1$ in the identity, we get $f(x) \equiv 0$.
(2) If $f(1)=1$, then by still setting $y=1$, we obtain
$$f(x... | f(x) \equiv 0 \text{ or } f(x) = 1 | Algebra | math-word-problem | Yes | Yes | inequalities | false | 735,955 |
$6 \cdot 148$ Find all functions $f(x)$ and $g(x) (x \in R)$ that satisfy
$$\sin x+\cos y \equiv f(x)+f(y)+g(x)-g(y), x, y \in R$$ | [Solution] In the identity, let $x=y$ to get
$$f(x)=\frac{\sin x+\cos x}{2}.$$
Thus, $\sin x+\cos y \equiv \frac{\sin x+\cos x}{2}+\frac{\sin y+\cos y}{2}+g(x)-g(y)$
which means $g(x)-\frac{1}{2} \sin x+\frac{1}{2} \cos x \equiv g(y)-\frac{1}{2} \sin y+\frac{1}{2} \cos y$.
Let $\quad h(x)=g(x)-\frac{1}{2} \sin x+\frac... | f(x)=\frac{\sin x+\cos x}{2}, g(x)=\frac{\sin x-\cos x}{2}+c | Algebra | math-word-problem | Yes | Yes | inequalities | false | 735,956 |
6. 149 Proof: There exists a function $f: N \rightarrow N$ such that $f(f(n)) \equiv n^{2}, n \in N$. | [Proof] Let the sequence $n_{1}=2, n_{2}=3, n_{3}=5, \cdots$ be strictly increasing and cover all natural numbers that are not perfect squares, and let
$$n_{k, m}=\left(n_{k}\right)^{2^{m}} \text {, where } k \in N, m \in Z^{+} \text {. }$$
Then $n_{k, m+1}=\left(n_{k, m}\right)^{2}$, and each $n>1$ corresponds to a u... | proof | Number Theory | proof | Yes | Yes | inequalities | false | 735,957 |
6・150 $f$ and $g$ are real functions defined on $(-\infty,+\infty)$, and for all $x$ and $y$ they satisfy the equation:
$$f(x+y)+f(x-y)=2 f(x) g(y) .$$
Prove that if $f(x)$ is not identically zero, and $|f(x)| \leqslant 1$ for all $x$, then $|g(x)| \leqslant 1$ for all $y$. | $$
\begin{array}{l}
\text { [Proof] Let } M=\sup |f(x)|, \text { then by assumption } M>0 . \text { For any } \delta, 0<M-\delta \text {. At this point, for any } y, \\
2 M \geqslant|f(x+y)|+|f(x-y)| \\
\quad \geqslant|f(x+y)+f(x-y)| \\
\quad=2|f(x)||g(y)| \\
\quad \geqslant 2(M-\delta)|g(y)| .
\end{array}
$$
Thus, $|... | proof | Algebra | proof | Yes | Yes | inequalities | false | 735,958 |
6・151 Let $k$ be a positive integer, find all polynomials
$$p(x)=a_{0}+a_{1} x+\cdots \cdot+a_{n} x^{n}$$
such that $a_{i}$ are real numbers $(i=0,1,2, \cdots, n)$, and satisfy the equation
$$p(p(x))=[p(x)]^{k}$$ | ［Solution］ If $p(x)=c$ (a constant), then
$$c=c^{k}$$
Therefore, when $k=1$, $p(x)$ can be any constant;
when $k \neq 1$, $p(x)=0$ or $p(x)=1$.
If $p(x) \neq c$ (a constant), then $p(x)$ can take infinitely many values. Thus, there are infinitely many values $y$ such that
$$p(y)=y^{k} \text {. }$$
Hence, for all $x$,... | p(x)=x^{k} \quad(k \geqslant 1) | Algebra | math-word-problem | Yes | Yes | inequalities | false | 735,959 |
6. 152 Find all functions $f$ from the set of real numbers to the set of real numbers that satisfy the following conditions:
(1) $f(x)$ is strictly increasing;
(2) For all real numbers $x, f(x) + g(x) = 2x$, where $g(x)$ is the inverse function of $f(x)$. | [Solution] We prove that $f(x)-x=c$, where $c=f(0)$.
In fact, if we let $S_{c}=\{x \mid f(x)-x=c\}$, then $0 \in S_{c}$, so $S_{c}$ is non-empty.
If $a \in S_{c}$, then $f(a)=a+c$, hence $g(a+c)=a$, and $f(a+c)=2(a+c)-a=(a+c)+c$, which means $a+c \in S_{c}$.
Suppose $c' \neq c$ and $S_{c}$ is not empty, then the proof... | f(x)=x+c | Algebra | math-word-problem | Yes | Yes | inequalities | false | 735,960 |
6-153 Prove: If the function $f: R \rightarrow R$ satisfies one of the following two identities, then it must satisfy the other:
$$\begin{array}{l}
f(x+y) \equiv f(x)+f(y), x, y \in R \\
f(x y+x+y) \equiv f(x y)+f(x)+f(y), x, y \in R
\end{array}$$ | [Proof] If the function $f(x)$ satisfies the first identity, then
$$\begin{aligned}
f(x y+x+y) & \equiv f(x y)+f(x+y) \\
& \equiv f(x y)+f(x)+f(y), x, y \in R
\end{aligned}$$
i.e., $f(x)$ also satisfies the second identity. Now, assume the function $f(x)$ satisfies the second identity, and let $y=u+v+u v$, we get
$$\b... | proof | Algebra | proof | Yes | Yes | inequalities | false | 735,961 |
$1 \cdot 101$ Let $p$ be a prime number greater than 2. Prove that $\frac{2}{p}$ can be and can only be expressed in one way as
$$\frac{2}{p}=\frac{1}{x}+\frac{1}{y}$$
where $x, y$ are distinct positive integers. | [Proof] Multiply both sides of the equation
$$\frac{2}{p}=\frac{1}{x}+\frac{1}{y}$$
by $2 x y p$, and move $2 x p$ and $2 y p$ on the right side to the left, and add $p^{2}$ to both sides, equation (1) becomes
$$(2 x-p)(2 y-p)=p^{2} .$$
Since $x \neq y$, the two factors on the left side of (2) are different, and thei... | proof | Number Theory | proof | Yes | Yes | inequalities | false | 735,962 |
6. 154 Find all monotonic invertible functions \( f: R \rightarrow R \) that satisfy
$$f(x)+f^{-1}(x) \equiv 2 x, x \in R$$ | [Solution] Let $f(x)$ satisfy the conditions given in the problem. Consider the function $g(x) = f(x) - x$. First, we prove that for any $k \in \mathbb{Z}^{+}$, we have
$$f(x + k g(x)) \equiv x + (k+1) g(x), \quad x \in \mathbb{R}$$
When $k=0$, the identity $f(x) \equiv x + g(x)$ holds. Assume the identity holds for s... | f(x) \equiv x + c | Algebra | math-word-problem | Yes | Yes | inequalities | false | 735,963 |
6- 155 (1) Prove that if a continuous function $f: R \rightarrow R$ satisfies
$$f(f(f(x))) \equiv x, x \in R$$
then for any $x \in R$,
$$f(x)=x$$
(2) Provide an example of a (discontinuous) function $g: R \rightarrow R$ that satisfies $g(x) \neq x$ and $g(g(g(x))), x \in R$. | [Proof] (1) First, prove that the function $f(x)$ is an injection from $R \rightarrow R$, i.e., for any $x$, $y \in R, x \neq y$, we have $f(x) \neq f(y)$.
Indeed, suppose for $x, y \in R, u=f(x)=f(y)$, then
$$x=f^{3}(x)=f^{2}(u)=f^{3}(y)=y .$$
Next, prove that $f(x)$ is strictly monotonic. Otherwise, there exist $x_{... | proof | Algebra | proof | Yes | Yes | inequalities | false | 735,964 |
6. 156 Find all continuous functions \( f:(1,+\infty) \rightarrow \mathbb{R} \) that satisfy
$$f(x y) \equiv x f(y)+y f(x), \quad x, y>1$$ | [Solution] First, we prove that for any $k>0$, we have
$$f\left(x^{k}\right) \equiv k x^{k-1} f(x) \text {, where } x>1 \text {. }$$
The proof is divided into three steps:
(1) Let $k \in \mathbb{N}$, if $k=1$, then $f\left(x^{1}\right)=1 \cdot x^{0} \cdot f(x)$. Assume the identity holds for some $k \in \mathbb{N}$. B... | f(x) = c x \ln x | Algebra | math-word-problem | Yes | Yes | inequalities | false | 735,965 |
6・158 Let the function $f: R \rightarrow R$ be defined as follows: if $x$ is irrational, then $f(x)=0$; if $p \in Z, q \in Z$, and the fraction $\frac{p}{q}$ is in its simplest form, then
$$f\left(\frac{p}{q}\right)=\frac{1}{q^{3}} .$$
Prove: This function is differentiable at each point $x=\sqrt{k}$, where $k$ is a n... | [Proof] The following proof shows that if $k \in \mathbb{N}$ is not a perfect square, then $f'(\sqrt{k})=0$, because $\sqrt{k} \notin \mathbb{Q}$, hence $f(\sqrt{k})=0$. The remaining part is to prove the limit
$$\lim _{x \rightarrow \sqrt{k}} \frac{f(x)-f(\sqrt{k})}{x-\sqrt{k}}$$
exists and equals 0. Take any $\varep... | proof | Calculus | proof | Yes | Yes | inequalities | false | 735,966 |
6・159 Functions $f, g: R \rightarrow R$ are not constant and satisfy
$$\begin{array}{l}
f(x+y) \equiv f(x) g(y)+g(x) f(y), \\
g(x+y) \equiv g(x) g(y)-f(x) f(y), \\
x, y \in R \text {. }
\end{array}$$
Find all possible values of $f(0)$ and $g(0)$. | [Solution] In the two identities, let $x=y=0$, we get
$$f(0)=2 f(0) g(0) \text { and } g(0)=(g(0))^{2}-(f(0))^{2}.$$
Since $\quad g(0) \neq \frac{1}{2}$,
otherwise from the second equation we get
$$(f(0))^{2}=\frac{1}{4}-\frac{1}{2}<0.$$
Therefore, from the first equation, we get $f(0)=0$, and from the second equatio... | f(0)=0, g(0)=1 | Algebra | math-word-problem | Yes | Yes | inequalities | false | 735,967 |
6・160 (1) Let the function $f(x)$ be defined for all $x>0$, and satisfy the following conditions:
(1) The function $f(x)$ is strictly increasing on $(0,+\infty)$;
(2) For all $x>0$, $f(x)>-\frac{1}{x}$;
(3) For all $x>0$, $f(x) f\left(f(x)+\frac{1}{x}\right)=1$. Find the function value $f(1)$.
(2) Provide a function th... | [Solution] (1) Let $f(1)=a$, then by condition (3), when $x=1$, we have $a f(a+$
1) $=1$. That is, $f(a+1)=\frac{1}{a}$.
Now let $x=a+1$, then by condition (3) we get
$$f(a+1) f\left(f(a+1)+\frac{1}{a+1}\right)=1$$
That is, $f\left(\frac{1}{a}+\frac{1}{a+1}\right)=a$,
Therefore, $\quad f\left(\frac{1}{a}+\frac{1}{a+1... | a=\frac{1+\sqrt{5}}{2} | Algebra | math-word-problem | Yes | Yes | inequalities | false | 735,968 |
6-161 Let $R$ be the set of all real numbers. Find all functions $f: R \rightarrow R$ such that for all $x$ and $y$ in $R$, we have
$$f\left(x^{2}+f(y)\right)=y+(f(x))^{2}$$ | [Solution] Let $t=f^{2}(0), f^{(2)}(x)=f[f(x)]$. In the functional equation
$$f\left(x^{2}+f(y)\right)=y+f^{2}(x)$$
let $x=0$, we get
$$f^{(2)}(y)=y+t$$
Thus, from (2) and (1),
$$\begin{aligned}
x^{2}+f^{(2)}(y)+t & =f^{(2)}\left[x^{2}+f^{(2)}(y)\right] \\
& =f\left[f(y)+(f(x))^{2}\right] \\
& =f\left[f^{2}(x)+f(y)\r... | f(x)=x | Algebra | math-word-problem | Yes | Yes | inequalities | false | 735,969 |
6・163 Let $I=[0,1], G=\{(x, y) \mid x \in I, y \in I\}$, find all mappings $f$ from $G$ to $I$ such that for any $x, y, z \in I$, we have
(1) $f(f(x, y), z)=f(x, f(y, z))$;
(2) $f(x, 1)=x, f(1, y)=y$;
(3) $f(z x, z y)=z^{k} f(x, y)$, where $k$ is a positive number independent of $x, y, z$. | [Solution] From conditions (3) and (2), we have
$$\begin{array}{l}
f(x, y)=y^{k} f\left(\frac{x}{y}, 1\right)=y^{k-1} x, 0 \leqslant x \leqslant y, 00 ; \\
y, \text { when } 0 \leqslant y \leqslant x, x>0 .
\end{array}\right. \\
f_{2}(x, y)=x y, \text { when }(x, y) \neq(0,0) .
\end{array}$$
When $x=y=0$, from conditi... | f(x, y) = \begin{cases}
y^{k-1} x, & 0 \leqslant x \leqslant y, 0 < y \leq 1; \\
x y, & 0 \leqslant y \leqslant x, 0 < x \leq 1; \\
0, & x = y = 0.
\end{cases} | Algebra | math-word-problem | Yes | Yes | inequalities | false | 735,970 |
$1 \cdot 102$ (1) The product of $n$ integers equals $n$, and their sum equals 0, prove that the number $n$ is divisible by 4.
(2) Let $n$ be a natural number divisible by 4, prove that it is possible to find $n$ integers whose product is $n$, and whose sum is 0. | [Proof] (1) If $n$ is odd, then from the product equal to $n$, it follows that all $n$ integers are odd, and their sum is not equal to 0, which contradicts the problem statement, hence $n$ must be even. If $n$ is not divisible by 4, then from the product equal to $n$, it follows that there is only one even factor, so t... | proof | Number Theory | proof | Yes | Yes | inequalities | false | 735,971 |
6- 164 Find all functions $f$ defined on the set of positive real numbers, taking positive real values, and satisfying the conditions:
(1) For all positive numbers $x, y, f(x f(y))=y f(x)$;
(2) As $x \rightarrow \infty$, $f(x) \rightarrow 0$. | [Solution] In (1), take $y=x$, we get
$$f(x f(x))=x f(x)$$
For any $x$, let $x f(x)=a$,
(1)
then (1) becomes $f(a)=a$
(2)
(3)
In (1), take $x=y=a$, we get
$$f(a f(a))=a f(a)$$
Substitute (3) into it, we get $f\left(a^{2}\right)=a^{2}$;
By induction, for all positive integers $n$,
$$f\left(a^{n}\right)=a^{n}$$
In (1... | f(x)=1/x | Algebra | math-word-problem | Yes | Yes | inequalities | false | 735,972 |
6- 166 Let $f$ be a function defined on the set of non-negative real numbers and taking values in the same set. Find all functions $f$ that satisfy the following conditions:
(i) $f(x f(y)) \cdot f(y)=f(x+y)$;
(ii) $f(2)=0$;
(iii) $f(x) \neq 0$, when $0 \leqslant x<2$. | [Solution] Let $f$ satisfy the given conditions.
When $x \geqslant 2$, we have
$$\begin{aligned}
f(x) & =f((x-2)+2) \\
& =f((x-2) f(2)) \cdot f(2) \\
& =0
\end{aligned}$$
Thus, $f(x)=0$;
If $0 \leqslant x<2$, we set $y \geqslant 2-x$. Then
$$\begin{array}{l}
x+y \geqslant 2 \\
f(x+y)=0 \\
f(y f(x)) \cdot f(x)=f(y+x)=f... | f(x)=\left\{\begin{array}{lc}
\frac{2}{2-x}, & 0 \leqslant x<2 \text { when; } \\
0, & x \geqslant 2 \text { when. }
\end{array}\right.} | Algebra | math-word-problem | Yes | Yes | inequalities | false | 735,974 |
6・167 $f$ is a function defined on $(1,+\infty)$ and takes values in $(1,+\infty)$, satisfying the condition: for any $x, y>1$ and $u, v>0$, we have
$$f\left(x^{u} y^{v}\right) \leqslant f(x)^{\frac{1}{4 u}} f(y)^{\frac{1}{4 v}}$$
Determine all such functions $f$. | ［Solution］ Suppose $f$ satisfies the conditions given in the problem, then we have
$$f\left(x^{u} y^{v}\right) \leqslant f(x)^{\frac{1}{4 u}} f(y)^{\frac{1}{4 v}} .$$
Taking $u=\frac{1}{2}$, we get
$$f\left(x^{\frac{1}{2}} y^{v}\right) \leqslant f(x)^{\frac{1}{2}} f(y)^{\frac{1}{4 v}}$$
Choose $v$ such that
$$y^{v}=x... | f(x)=C^{\frac{1}{\ln x}} \quad(C>0) | Inequalities | math-word-problem | Yes | Yes | inequalities | false | 735,975 |
6-168 Find all $d \in(0,1]$ with the following property: if $f(x)$ is any continuous function defined on $[0,1]$ and $f(0)=f(1)$, then there exists $x_{0} \in[0,1-d]$ such that
$$f\left(x_{0}\right)=f\left(x_{0}+d\right)$$ | [Solution] First, we show that for any $k \in \mathbb{N}, d=\frac{1}{k}$ satisfies the conditions of the problem. Let $f(x)$ be any continuous function that satisfies the conditions of the problem. Since $f(0)=f(1)$, $d$ satisfies the condition. Consider natural numbers $k>1$, and define the function
$$g(x)=f\left(x+\f... | proof | Calculus | math-word-problem | Yes | Yes | inequalities | false | 735,976 |
6. 169 For a given increasing function $f: R \rightarrow R$, define the function $g(x, y)$ as follows:
$$g(x, y)=\frac{f(x+y)-f(x)}{f(x)-f(x-y)}, x \in R, y>0 .$$
Assume that when $x=0$ for all $y>0$ and when $x \neq 0$ for all $y \in(0,|x|]$, we have
$$2^{-1}<g(x, y)<2$$
Prove that for all $x \in R$ and $y>0$, we ha... | [Proof] Let $h(x, y)=f(y)-f(x)$, then the function $h(x, y)$ is increasing in variable $y$ and decreasing in variable $x$, and for all $y>x$, we have $h(x, y)>0$, and
$$g(x, y) \equiv \frac{h(x, x+y)}{h(x-y, x)}$$
First, prove that for each $x \in \mathbb{R}, y>0$, we have $g(x, y)>0$. Since for all $y>0$ we have $h(0... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 735,977 |
$6 \cdot 170$ Find all functions $f: R \rightarrow R$, for all real numbers $x$, satisfying: $f(\sqrt{2} x)+f((4+3 \sqrt{2}) x)=2 f((2+\sqrt{2}) x)$. | [Solution] Let $x=0$, substituting into the original equation yields an identity, therefore, $f(0)$ can be any real number. Denote
$$f(0)=a, a \text { is any real number. }$$
Let
$$t=(2+\sqrt{2}) x,$$
Notice that
$$4+3 \sqrt{2}=(2+\sqrt{2})(\sqrt{2}+1)$$
Thus, the original equation can be transformed into
$$f\left(\... | proof | Algebra | math-word-problem | Yes | Yes | inequalities | false | 735,978 |
$6 \cdot 171$ Find all real-coefficient polynomials $f(x)$ of degree greater than or equal to 1 that satisfy the equation
$$f\left(x^{2}\right)=f(x) f(x-1),$$
and prove your conclusion. | [Solution] From the given equation, we know that if $x$ is a root of the polynomial $f(x)$, then $x^{2}$ is also a root of the polynomial $f(x)$. If the modulus of a non-zero root $x_{0}$ of $f(x)$ is not equal to 1, then the moduli of $x_{0}^{2}, x_{0}^{4}, x_{0}^{8}, x_{0}^{16}, \cdots$ are all different, and they ar... | f(x)=(x^2+x+1)^t, t \in \mathbb{N} | Algebra | proof | Yes | Yes | inequalities | false | 735,979 |
6・172 Determine all real-coefficient polynomials $p(x)$ such that $t p(t-1)=(t-2) p(t)$ for all real numbers $t$.
The text has been translated while preserving the original line breaks and format. | [Solution] Let $t=2$, we get $2 p(1)=0$, so $p(1)=0$. Thus, 1 is a root of the polynomial $p(x)$. Let $t=1$, we get $p(0)=-p(1)=0$, so 0 is also a root of the polynomial $p(x)$. Let $p(x)=x(x-1) q(x)$, then
$$\begin{array}{l}
t(t-1)(t-2) q(t-1)=(t-2) \cdot t(t-1) q(t) \\
q(t-1)=q(t), t \neq 0,1,2
\end{array}$$
By sett... | p(x)=c x(x-1), c \in \mathbb{R} | Algebra | math-word-problem | Yes | Yes | inequalities | false | 735,980 |
6-173 Prove: There does not exist a function $f(x)$ that satisfies
$$f[f(x)]=x^{2}-1996$$
for all real numbers $x$.
untranslated part:
将上面的文本翻译成英文，请保留源文本的换行和格式，直接输出翻译结果。
Since the last part is a note about the translation instruction, it is not part of the content to be translated. Here is the translation of the co... | [Proof] If there exists a function $f(x)$ that satisfies the conditions, we let
$$g(x)=f[f(x)]=x^{2}-1996.$$
Let $a, b$ be the two real roots of $x=x^{2}-1996$, then $a, b$ are fixed points of the function $g(x)$.
Let $f(a)=p$, then
$$f[f(p)]=f[f(f(a))]=f(a)=p,$$
thus $p$ is also a fixed point of $g(x)$, $p \in\{a, ... | proof | Algebra | proof | Yes | Yes | inequalities | false | 735,981 |
1-103 Proof: Among any $2 n+1$ different integers whose absolute values do not exceed $2 n-1$, there exist three numbers whose sum equals 0. | [Proof] When $n=1$, the proposition becomes: among any 3 different integers with absolute values not exceeding 1, there are 3 numbers whose sum is 0, which is obviously true.
Assume that the proposition holds for $n=k$, i.e., among any $2k+1$ different integers with absolute values not exceeding $2k-1$, there must be ... | proof | Combinatorics | proof | Yes | Yes | inequalities | false | 735,982 |
6-174 Let $R$ be the set of real numbers. Does there exist a function $f: R \rightarrow R$ that simultaneously satisfies the following three conditions:
(a) There exists a positive number $M$ such that for all $x$, $-M \leqslant f(x) \leqslant M$;
(b) The value of $f(1)$ is 1;
(c) If $x \neq 0$, then
$$f\left(x+\frac{1... | ［Solution］ It does not exist.
In fact, if there exists $f: R \rightarrow R$ satisfying all the conditions, then by $(a)$, there exists an integer multiple of $\frac{1}{4}$, $c$, such that
$$\left\{\begin{array}{l}
f(x)2, c-\frac{1}{4}>0$.
By the definition of $c$ and $x_{0}$ and $(c)$, we have
$$\begin{aligned}
c & >f\... | proof | Algebra | math-word-problem | Yes | Yes | inequalities | false | 735,983 |
6-175 Let $f$ be a function from $R \rightarrow R$, and
(1) For any $x, y \in R$,
$$f(x)+f(y)+1 \geqslant f(x+y) \geqslant f(x)+f(y)$$
(2) For any $x \in[0,1)$, $f(0) \geqslant f(x)$;
(3) $-f(-1)=f(1)=1$.
Find all functions $f$ that satisfy the conditions. | ［Solution］ From (1) and (2) we get
$$2 f(0)+1 \geqslant 2 f\left(\frac{1}{2}\right)+1 \geqslant f(1) \geqslant f(1)+f(0)$$
But from (3) we know $f(1)=1$, so the above equation can be simplified to
$$2 f(0) \geqslant 0 \geqslant f(0)$$
Thus, we have $\quad f(0) \geqslant 0 \geqslant f(0)$,
$$f(0)=0 .$$
From (2), for ... | f(x)=[x] | Algebra | math-word-problem | Yes | Yes | inequalities | false | 735,984 |
6・176 Let the function $f: R \rightarrow R$ satisfy the condition
$$\begin{array}{c}
f\left(x^{3}+y^{3}\right)=(x+y)\left((f(x))^{2}-f(x) \cdot f(y)+(f(y))^{2}\right), \\
x, y \in R
\end{array}$$
Prove that for all $x \in R$, we have
$$f(1996 x)=1996 f(x)$$ | [Proof] In (1), let $x=y=0$, we get $f(0)=0$.
In (1), let $y=0$, we get
$$f\left(x^{3}\right)=x \cdot(f(x))^{2}, x \in \mathbb{R}$$
Rewrite the above as
$$f(x)=x^{\frac{1}{3}}\left(f\left(x^{\frac{1}{3}}\right)\right)^{2}, x \in \mathbb{R}$$
It is evident that when $x \geqslant 0$, $f(x) \geqslant 0$; when $x \leqsla... | proof | Algebra | proof | Yes | Yes | inequalities | false | 735,985 |
6・177 (a) Do there exist functions $f: R \rightarrow R, g: R \rightarrow R$, such that for all $x$ $\in R$, we have
$$f(g(x))=x^{2}, g(f(x))=x^{3} ?$$
(b) Do there exist functions $f: R \rightarrow R, g: R \rightarrow R$, such that for all $x \in R$, we have $f(g(x))=x^{2}, g(f(x))=x^{4} ?$ | [Solution] (a) If such functions $f, g$ exist, from
$$g(f(x))=x^{3}$$
we know that when $x_{1} \neq x_{2}$, $f\left(x_{1}\right) \neq f\left(x_{2}\right)$.
Therefore, $f(0), f(1), f(-1)$ are three distinct real numbers.
On the other hand, by the given conditions, we have
$$(f(x))^{2}=f(g(f(x)))=f\left(x^{3}\right)$$
... | proof | Algebra | math-word-problem | Yes | Yes | inequalities | false | 735,986 |
6-178 Let $R$ be the set of all real numbers, and $R^{+}$ be the subset of all positive real numbers. $\alpha, \beta$ are given real numbers. Find all functions $f: R^{+} \longrightarrow R$ such that for all $x$ and $y$ in $R^{+}$, we have
$$f(x) f(y)=y^{\alpha} f\left(\frac{x}{2}\right)+x^{\beta} f\left(\frac{y}{2}\ri... | [Solution] In the given functional equation, let $y=x$, we get
$$f\left(\frac{x}{2}\right)=\frac{(f(x))^{2}}{x^{\alpha}+x^{\beta}}$$
Rewrite $x$ as $y$ in (1), we get
$$f\left(\frac{y}{2}\right)=\frac{(f(y))^{2}}{y^{\alpha}+y^{\beta}}$$
Substitute (1) and (2) into the original functional equation, we get
$$f(x) \cdot... | f(x)=0 \text{ when } \alpha \neq \beta; \text{ and } f(x)=0 \text{ or } f(x)=2^{1-\alpha} x^{\alpha} \text{ when } \alpha=\beta | Algebra | math-word-problem | Yes | Yes | inequalities | false | 735,987 |
$6 \cdot 179$ Find all functions $f: R \rightarrow R$, such that $x f(x)-y f(y)=(x-y) f(x+y)$. | [Solution] If $f$ is a real constant function, i.e., $f(x)$ is always equal to some real constant $b$, then it clearly satisfies the given functional equation.
Assume $f$ is not a real constant function.
Let the line passing through the two points $(x, f(x))$ and $(y, f(y))$ be denoted as $l_{x, y}$. The equation of t... | f(x) = ax + b, \quad a, b \in \mathbb{R} | Algebra | math-word-problem | Yes | Yes | inequalities | false | 735,988 |
6. 180 Determine all functions \( f: \mathbb{R} \rightarrow \mathbb{R} \), where \(\mathbb{R}\) is the set of real numbers, such that for all \( x, y \in \mathbb{R} \), the following holds:
\[ f(x - f(y)) = f(f(y)) + x f(y) + f(x) - 1 \]
establish. | ［Solution］Let
$$A=\operatorname{Im}f$$
be the image set of the function $f$, and let
$$f(0)=c$$
In the given functional equation, setting $x=y=0$, we have
$$f(-c)=f(c)+c-1$$
Thus,
$$c \neq 0$$
In the given functional equation, setting $x=f(y)$, we have
$$f(0)=f(x)+x^{2}+f(x)-1$$
which implies
$$f(x)=\frac{c+1}{2}-... | f(x)=1-\frac{x^{2}}{2} | Algebra | math-word-problem | Yes | Yes | inequalities | false | 735,989 |
7・ 1 There are 9 chairs around a round table, and 4 people sit down randomly. What is the probability that no two people are sitting next to each other? | [Solution] The first person can sit in any position, and the remaining three people have a total of $8 \cdot 7 \cdot 6$ ways to sit in a circle.
On the other hand, if no two people are to sit next to each other, the first person can sit in any position, and the remaining 3 people can sit in alternating seats in 6 ways... | \frac{1}{14} | Combinatorics | math-word-problem | Yes | Yes | inequalities | false | 735,990 |
7.2 In the set $\left\{-3,-\frac{5}{4},-\frac{1}{2}, 0, \frac{1}{3}, 1, \frac{4}{5}, 2\right\}$, two numbers are drawn without replacement. Find the probability that the two numbers are the slopes of a pair of perpendicular lines. | [Solution] Two numbers represent the slopes of a pair of perpendicular lines, meaning the product of the two numbers is -1. Pairs of numbers that satisfy this condition are: -3 and $\frac{1}{3}$, $-\frac{5}{4}$ and $\frac{4}{5}$, $-\frac{1}{2}$ and 2, totaling 3 pairs. Choosing 2 numbers from 8, there are $C_{8}^{2}=\f... | \frac{3}{28} | Combinatorics | math-word-problem | Yes | Yes | inequalities | false | 735,991 |
$7 \cdot 3$ Tossing a coin, each time heads appears you get 1 point, and tails appears you get 2 points, prove that the probability of getting exactly $n$ points is $\frac{1}{3}\left[2+\left(-\frac{1}{2}\right)^{n}\right]$. | Let $p_{n}$ denote the probability of getting exactly $n$ points.
The only case of not getting exactly $n$ points is getting $n-1$ points and then tossing a tail. Since the probability of the event "not getting exactly $n$ points" is $1-p_{n}$, the probability of "getting exactly $n-1$ points" is $p_{n-1}$. The probabi... | p_{n}=\frac{1}{3}\left[2+\left(-\frac{1}{2}\right)^{n}\right] | Combinatorics | proof | Yes | Yes | inequalities | false | 735,992 |
Theorem 1 Let $a, b, c$ be distinct real numbers, and $\lambda$ be a real number, then
$$\begin{array}{l}
\left(\frac{a}{a-b}+\lambda\right)^{2}+\left(\frac{b}{b-c}+\lambda\right)^{2}+\left(\frac{c}{c-a}+\lambda\right)^{2} \\
\geqslant 2 \lambda^{2}+2 \lambda+1
\end{array}$$ | Prove that for $x=\frac{a}{a-b}, y=\frac{b}{b-c}, z=\frac{c}{c-a}$, we have
$$\begin{array}{l}
(x-1)(y-1)(z-1)=\frac{b}{a-b} \cdot \frac{c}{b-c} \cdot \frac{a}{c-a} \\
=\frac{a}{a-b} \cdot \frac{b}{b-c} \cdot \frac{c}{c-a}=x y z
\end{array}$$
That is, $(x-1)(y-1)(z-1)=x y z$,
which can be transformed to $x+y+z=x y+y z... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 735,993 |
Theorem 2 Let $a, b, c$ be distinct real numbers, and $\lambda$ be a real number, then
$$\begin{array}{l}
\left(\frac{a+b}{a-b}+\lambda\right)^{2}+\left(\frac{b+c}{b-c}+\lambda\right)^{2}+\left(\frac{c+a}{c-a}+\lambda\right)^{2} \\
\geqslant 2\left(\lambda^{2}+1\right)
\end{array}$$ | Prove that if $x=\frac{a+b}{a-b}, y=\frac{b+c}{b-c}, z=\frac{c+a}{c-a}$, then
$$\begin{array}{l}
(x-1)(y-1)(z-1)=\frac{8 a b c}{(a-b)(b-c)(c-a)} \\
=(x+1)(y+1)(z+1),
\end{array}$$
which implies $x y+y z+z x=-1$. Consequently, we have
$$\begin{aligned}
& \left(\frac{a+b}{a-b}+\lambda\right)^{2}+\left(\frac{b+c}{b-c}+\l... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 735,998 |
Theorem 3 Let $a, b, c$ be distinct real numbers, and $\lambda$ be a real number, then
$$\begin{array}{l}
\left(\frac{c}{a-b}+\lambda\right)^{2}+\left(\frac{a}{b-c}+\lambda\right)^{2}+\left(\frac{b}{c-a}+\lambda\right)^{2} \\
\geqslant 2\left(\lambda^{2}+1\right)
\end{array}$$ | Prove that for $x=\frac{c}{a-b}, y=\frac{a}{b-c}, z=\frac{b}{c-a}$, we have
$$\begin{array}{l}
(x-1)(y-1)(z-1) \\
=\frac{(b+c-a)(c+a-b)(a+b-c)}{(a-b)(b-c)(c-a)} \\
=(x+1)(y+1)(z+1)
\end{array}$$
Thus, $x y+y z+z x=-1$, and we get
$$\begin{aligned}
& \left(\frac{c}{a-b}+\lambda\right)^{2}+\left(\frac{a}{b-c}+\lambda\ri... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,000 |
In 2008, the 12th question of the first round of the National High School Mathematics Competition (A paper) is: A small ball with a radius of 1 can move freely in all directions inside a regular tetrahedron container with an internal edge length of $4 \sqrt{6}$. Then the area of the container's inner wall that the smal... | Solution 1: Let the radius of the small sphere $O$ be 1, and the edge length of the inner tetrahedron $ABCD$ be $4\sqrt{6}$. The points of contact between the small sphere $O$ and the faces of the tetrahedron $ABCD$ form four congruent small equilateral triangles (as shown in Figure 1, two small equilateral triangles a... | not found | Geometry | math-word-problem | Yes | Yes | inequalities | false | 736,001 |
Example 1 Let $a, b, c \in \mathbf{R}_{+}$, and $abc=1$. Prove:
$$\frac{1}{\sqrt{1+a}}+\frac{1}{\sqrt{1+b}}+\frac{1}{\sqrt{1+c}} \leqslant \frac{3 \sqrt{2}}{2} .$$ | Explanation: Directly using the condition $abc=1$ to reduce it to a one-variable problem is quite difficult. However, we can first treat $ab$ as a whole to reduce the three-variable problem to a two-variable one, and then use inequalities to further reduce it to a one-variable problem.
Assume $a \leqslant b \leqslant ... | \frac{3 \sqrt{2}}{2} | Inequalities | proof | Yes | Yes | inequalities | false | 736,003 |
Example 2 Let $a, b, c \in \mathbf{R}_{+}$, and $abc=1$. Prove:
$$\frac{ab}{a^{5}+b^{5}+ab}+\frac{bc}{b^{5}+c^{5}+bc}+\frac{ca}{c^{5}+a^{5}+ca} \leqslant 1$$ | Notice that
$$\begin{array}{l}
a^{5}+b^{5}-a^{2} b^{2}(a+b) \\
=\left(a^{2}-b^{2}\right)\left(a^{3}-b^{3}\right) \geqslant 0
\end{array}$$
Then $\frac{a b}{a^{5}+b^{5}+a b}=\frac{a b \cdot a b c}{a^{5}+b^{5}+a b \cdot a b c}$
$$\begin{array}{l}
=\frac{a^{2} b^{2} c}{a^{5}+b^{5}+a^{2} b^{2} c} \leqslant \frac{a^{2} b^{... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,004 |
Example $\mathbf{3}$ Let $a, b, c \in \mathbf{R}_{+}$, and $abc=1$. Prove:
$$\frac{1}{a^{3}(b+c)}+\frac{1}{b^{3}(a+c)}+\frac{1}{c^{3}(a+b)} \geqslant \frac{3}{2}^{[2]}$$ | Explanation: The left side of the inequality to be proven is somewhat "top-light and bottom-heavy" compared to the right side. We can use the reciprocal substitution $a=\frac{1}{x}, b=\frac{1}{y}$, $c=\frac{1}{z}$ to make its form more "harmonious".
Let $a=\frac{1}{x}, b=\frac{1}{y}, c=\frac{1}{z}$. Then $x y z=1$.
It... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,005 |
Example 4 Let $a, b, c \in \mathbf{R}_{+}$, and $abc=1$. Prove that:
$$\frac{1}{1+2a}+\frac{1}{1+2b}+\frac{1}{1+2c} \geqslant 1^{[2]} \text{.}$$ | Let $a=\frac{x}{y}, b=\frac{y}{z}, c=\frac{z}{x}\left(x, y, z \in \mathbf{R}_{+}\right)$.
$$\begin{array}{l}
\text { Then } \frac{1}{1+2 a}+\frac{1}{1+2 b}+\frac{1}{1+2 c} \\
=\frac{y}{y+2 x}+\frac{z}{z+2 y}+\frac{x}{x+2 z} \text {. }
\end{array}$$
By the Cauchy-Schwarz inequality, we have
$$\begin{array}{l}
{[y(y+2 x... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,006 |
Example 5 Let $a, b, c \in \mathbf{R}_{+}$, and $abc=1$. Prove that:
$$\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{3}{a+b+c} \geqslant 4 \text {. }$$ | Let $a=\frac{x}{y}, b=\frac{y}{z}, c=\frac{z}{x}\left(x, y, z \in \mathbf{R}_{+}\right)$.
Assume without loss of generality that $x \geqslant y \geqslant z$. Then
$$\begin{array}{c}
y^{2} z+z^{2} x+x^{2} y-\left(x^{2} z+y^{2} x+z^{2} y\right) \\
=(x-y)(y-z)(x-z) \geqslant 0 .
\end{array}$$
Therefore, $\frac{1}{a}+\fra... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,007 |
Example 6 Let $a, b, c \in \mathbf{R}_{+}$, and $abc=1$. Prove that:
$$\frac{1}{1+a+b}+\frac{1}{1+b+c}+\frac{1}{1+c+a} \leqslant 1 .$$ | Let $a=x^{3}, b=y^{3}, c=z^{3}(x, y, z \in \mathbf{R}_{+})$. Then $x y z=1$.
Notice that $x^{3}+y^{3} \geqslant x^{2} y+y^{2} x$. Then
$$\begin{aligned}
& \frac{1}{1+a+b}=\frac{1}{1+x^{3}+y^{3}} \\
& \leqslant \frac{1}{1+x^{2} y+y^{2} x}=\frac{1}{x y z+x^{2} y+y^{2} x} \\
= & \frac{1}{x y(x+y+z)}=\frac{z}{x+y+z}
\end{... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,008 |
Example 7 Let real numbers $x, y, z$ all be not equal to 1, and satisfy $xyz=1$. Prove:
$$\frac{x^{2}}{(x-1)^{2}}+\frac{y^{2}}{(y-1)^{2}}+\frac{z^{2}}{(z-1)^{2}} \geqslant 1^{[3]}$$ | Let $\frac{x}{x-1}=a, \frac{y}{y-1}=b, \frac{z}{z-1}=c$. Then $x=\frac{a}{a-1}, y=\frac{b}{b-1}, z=\frac{c}{c-1}$.
From $x y z=1$, we get
$$a b c=(a-1)(b-1)(c-1)$$
which simplifies to $a+b+c-1=a b+b c+c a$.
Thus, $a^{2}+b^{2}+c^{2}$
$$\begin{array}{l}
=(a+b+c)^{2}-2(a b+b c+c a) \\
=(a+b+c)^{2}-2(a+b+c-1) \\
=(a+b+c-1... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,009 |
Given $a, b, c \in \mathbf{R}_{+}, abc=1$ Prove:
$$\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{3}{a+b+c} \geqslant 4^{[1]} \text {. }$$ | Proof: Let $x=\frac{1}{a}, y=\frac{1}{b}, z=\frac{1}{c}$. Then $x y z=1$.
Equation (1) becomes $x+y+z+\frac{3}{y z+z x+x y} \geqslant 4$
$$\begin{array}{l}
\text { By }(x+y+z)^{2}=x^{2}+y^{2}+z^{2}+2(y z+z x+x y) \\
\geqslant 3(y z+z x+x y),
\end{array}$$
Thus, $x+y+z+\frac{3}{y z+z x+x y}$
$$\geqslant x+y+z+\frac{9}{... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,011 |
Strengthened Proposition Let $a, b, c \in \mathbf{R}_{+}, abc=1$ then
$$\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{6}{a+b+c} \geqslant 5 \text {. }$$ | Proof: Since equation (4) is symmetric with respect to $a, k, c$, without loss of generality, assume $a$ is the largest among them. Then $a \geqslant 1, bc \leqslant 1$.
Let $f(a, b, c) = \frac{1}{a} + \frac{1}{b} + \frac{1}{c} + \frac{6}{a+b+c}$. Then
\[ f(a, b, c) - f(a, \sqrt{bc}, \sqrt{bc}) \]
\[ = \frac{1}{b} + \... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,012 |
Proposition Let $x_{1}, x_{2}, \cdots, x_{n} \in R(n \geq 2), m, p \in$ $N$ and have the same parity, then
$$\begin{aligned}
& \frac{x_{1}^{m}+x_{2}^{m}+\cdots+x_{n}^{m}}{n} \cdot \frac{x_{1}^{p}+x_{2}^{p}+\cdots+x_{n}^{p}}{n} \\
\leq & \frac{x_{1}^{m+p}+x_{2}^{m+p}+\cdots+x_{n}^{m+p}}{n}
\end{aligned}$$
Equality hold... | To prove (2), that is to prove
$$\sum_{j=1}^{n} \sum_{i=1}^{n} x_{1}^{\alpha} \cdot x_{j}^{\delta} \geq \sum_{j=1}^{n} \sum_{i=1}^{n} x_{i}^{\beta} \cdot x_{j}^{\gamma}$$
If $i=j$, then $x_{i}^{\alpha} \cdot x_{j}^{\delta}=x_{i}^{\beta} \cdot x_{j}^{\gamma}$. If $i \neq j$, then the left side of (3) has terms $x_{i}^{... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,013 |
Inference 3 Let $m>0, p>0, x_{1}, x_{2}, \cdots, x_{n} \in$ $R^{+}$. Then
$$\frac{\sum_{i=1}^{n} x_{i}^{m+p}}{n} \geq \frac{\sum_{i=1}^{n} x_{i}^{m}}{n} \cdot \frac{\sum_{i=1}^{n} x_{i}^{p}}{n}$$
Equality holds if and only if $x_{1}=x_{2}=\cdots=x_{n}$. | Proof: By the above theorem, let $\delta=0$, take $\beta=m, \gamma=p$. Then $\alpha=m+p$. This completes the proof. | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,016 |
Example: Let $x_{1}, x_{2}, \cdots, x_{n} \in R^{+}$, and $\sum_{i=1}^{n} x_{i}=1$. Prove:
$$\left(x_{1}+\frac{1}{x_{1}}\right)^{4}+\left(x_{2}+\frac{1}{x_{2}}\right)^{4}+\cdots+\left(x_{n}+\frac{1}{x_{n}}\right)^{4} \geq n\left(n+\frac{1}{n}\right)^{4}$$ | $$\begin{aligned}
& \frac{\left(x_{1}+\frac{1}{x_{1}}\right)^{4}+\left(x_{2}+\frac{1}{x_{2}}\right)^{4}+\cdots+\left(x_{n}+\frac{1}{x_{n}}\right)^{4}}{n} \\
\geq & {\left[\frac{\left(x_{1}+\frac{1}{x_{1}}\right)^{2}+\left(x_{2}+\frac{1}{x_{2}}\right)^{2}+\cdots+\left(x_{n}+\frac{1}{x_{n}}\right)^{2}}{n}\right]^{2} } \\... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,019 |
Example 1 (1996 Polish Mathematical Competition Question) Given $a, b, c \geqslant -\frac{3}{4}$, and $a+b+c=1$, prove: $\frac{a}{a^{2}+1}+\frac{b}{b^{2}+1}+$ $\frac{c}{c^{2}+1} \leqslant \frac{9}{10}$. | Proof: Let $f(x)=\frac{x}{x^{2}+1}, x \geqslant-\frac{3}{4}$, then $f\left(\frac{1}{3}\right)=$ $\frac{3}{10}, f^{\prime}(x)=\frac{1-x^{2}}{\left(x^{2}+1\right)^{2}}, f^{\prime}\left(\frac{1}{3}\right)=\frac{18}{25}, f(x)$ at $x=\frac{1}{3}$ the equation of the tangent line is: $g(x)=f^{\prime}\left(\frac{1}{3}\right)\... | \frac{9}{10} | Inequalities | proof | Yes | Yes | inequalities | false | 736,022 |
Example 2 Given $a, b, c>0$, prove: $\frac{a}{(b+c)^{2}}+$
$$\frac{b}{(c+a)^{2}}+\frac{c}{(a+b)^{2}} \geqslant \frac{9}{4(a+b+c)} .$$ | Analysis: Multiplying both sides of the inequality by $a+b+c$, we get the homogeneous inequality: $\sum \frac{a(a+b+c)}{(b+c)^{2}} \geqslant \frac{9}{4}, \sum$ represents the cyclic sum. Dividing the numerator and denominator of each term by $(a+b+c)^{2}$, we get the equivalent inequality:
$$\begin{array}{l}
\sum \frac... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,023 |
Example 3 (1997 Japan Mathematical Olympiad) Given $a$, $b, c>0$, prove that: $\frac{(b+c-a)^{2}}{(b+c)^{2}+a^{2}}+\frac{(c+a-b)^{2}}{(c+a)^{2}+b^{2}}+$ $\frac{(a+b-c)^{2}}{(a+b)^{2}+c^{2}} \geqslant \frac{3}{5}$. | Proof: Since the inequality is homogeneous, we can assume $a+b+c=1$. The original inequality then becomes: $\frac{(1-2 a)^{2}}{(1-a)^{2}+a^{2}}+\frac{(1-2 b)^{2}}{(1-b)^{2}+b^{2}}$ $+\frac{(1-2 c)^{2}}{(1-c)^{2}+c^{2}} \geqslant \frac{3}{5}$. Let $f(x)=\frac{(1-2 x)^{2}}{(1-x)^{2}+x^{2}}$, $0<x<1$, then $f\left(\frac{1... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,024 |
Example 4 Given that $a, b, c$ are the three sides of a triangle, prove: $\sum$
$$\frac{a}{\sqrt{2 b^{2}+2 c^{2}-a^{2}}} \geqslant \sqrt{3} \text {. }$$ | Proof: Let $a^{2}=x, b^{2}=y, c^{2}=z$, then the inequality becomes: $\sum \sqrt{\frac{x}{2 y+2 z-x}} \geqslant \sqrt{3}$. Since the fraction inside the square root is homogeneous, we can assume $x+y+z=1$, leading to the equivalent inequality: $\sum$ $\sqrt{\frac{x}{2(1-x)-x}} \geqslant \sqrt{3}$, which simplifies to $... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,025 |
Example 5 (1999 Belarus Mathematical Olympiad) Given $a$, $b, c>0$, and $a^{2}+b^{2}+c^{2}=3$. It's a bit hard to figure out at first.
Prove: $\frac{1}{1+a b}+\frac{1}{1+b c}+\frac{1}{1+c a} \geqslant \frac{3}{2}$. | Prove: $a b+b c+c a \leqslant a^{2}+b^{2}+c^{2}=3$. Let $a b=x$, $b c=y, c a=z$, then $x+y+z \leqslant 3$, the original inequality becomes: $\frac{1}{1+x}$ $+\frac{1}{1+y}+\frac{1}{1+z} \geqslant \frac{3}{2}$. Let $f(x)=\frac{1}{1+x}, 0<x<3$, then $f(1)=\frac{1}{2}, f^{\prime}(x)=-\frac{1}{(1+x)^{2}}, f^{\prime}(1)=-\f... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,026 |
Example 6 (8th Hong Kong Mathematical Olympiad) Given $a, b, c, d > 0$, and $a+b+c+d=1$, prove that $6\left(a^{3}+b^{3}+c^{3}+d^{3}\right) \geqslant \left(a^{2}+b^{2}+c^{2}+d^{2}\right) + 1 / 8$. | Prove: The inequality is equivalent to: $6 a^{3}-a^{2}+6 b^{3}-b^{2}+6 c^{3}-$ $c^{2}+6 d^{3}-d^{2} \geqslant \frac{1}{8}$. Let $f(x)=6 x^{3}-x^{2}, 0<x<1$, then $f\left(\frac{1}{4}\right)=\frac{1}{32}, f^{\prime}(x)=18 x^{2}-2 x, f^{\prime}\left(\frac{1}{4}\right)=\frac{5}{8}, f(x)$ is the tangent line equation at $x=... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,027 |
Problem 1: (Mathematical Bulletin 2001. 12 Mathematical Problem Solution Column 1324) In $\triangle A B C$, $a, b, c$ are the lengths of its sides, respectively. Prove that: $\frac{a+b}{b+c-a}+\frac{b+c}{c+a-b}+\frac{c+a}{a+b-c} \geqslant 6$.
| Proof: From (*) we know that
\[
\frac{a+b}{b+c-a}+\frac{b+c}{c+a-b}+
\]
\[
\begin{array}{l}
\frac{c+a}{a+b-c}=\frac{x+2 y+z}{2 z}+\frac{x+y+2 z}{2 x}+ \\
\frac{2 x+y+z}{2 y}=\frac{3}{2}+\left(\frac{y}{z}+\frac{z}{x}+\frac{x}{y}\right)+\frac{1}{2}\left(\frac{x}{z}+\frac{y}{x}\right. \\
\left.+\frac{z}{y}\right) \geqslan... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,028 |
Question 2: (Mathematics Teaching 2001.5 - 6 Mathematics Problem Solving Column 543) In $\triangle A B C$, $a$, $b$, and $c$ are the lengths of its sides. Prove: $\sqrt{\frac{a}{b+c-a}}+\sqrt{\frac{b}{c+a-b}}+$ $\sqrt{\frac{c}{a+b}-c} \geqslant 3$.
Note: The original problem has a typo. The correct expression should b... | $$\begin{array}{l}
\text { Proof: From (*) we know } \sqrt{\frac{a}{b+c-a}}+ \\
\sqrt{\frac{b}{c+a-b}}+\sqrt{\frac{c}{a+b-c}}=\sqrt{\frac{x+y}{2 z}}+ \\
\sqrt{\frac{y+z}{2 x}}+\sqrt{\frac{z+x}{2 y}} \geqslant \sqrt{\frac{\sqrt{x y}}{z}}+\sqrt{\frac{\sqrt{y z}}{x}}+ \\
\sqrt{\frac{\sqrt{x z}}{y}} \geqslant 3 .
\end{arra... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,029 |
Question 3: (Mathematics Teaching 2001.5 - 6 Mathematics Problem Solving Column 548) In $\triangle A B C$, $a$, $b$, and $c$ are the lengths of its sides. Prove that: $\sqrt{\frac{b+c-a}{a}}+\sqrt{\frac{a-b+c}{b}}+$ $\sqrt{\frac{a+b}{c}-\bar{c}}>2 \sqrt{2}$. | Proof: From (*) we know $\sqrt{\frac{b+c-a}{a}}+$
$$\begin{array}{l}
\sqrt{\frac{a-b+c}{b}}+\sqrt{\frac{a+b-c}{c}}=\sqrt{\frac{2 z}{x+y}}+ \\
\sqrt{\frac{2 x}{y+z}}+\sqrt{\frac{2 y}{z+x}}=\sqrt{2}\left[\frac{z}{\sqrt{z} \sqrt{x+y}}+\frac{x}{\sqrt{x} \sqrt{y+z}}\right. \\
\left.+\frac{y}{\sqrt{y} \sqrt{z+x}}\right]>\sqr... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,030 |
Problem 4: (Mathematics for Middle School 1996.2 Mathematical Olympiad Problem 40) Let $x, y, z \in R^{+}$, prove that: $\frac{x}{2 x+y+z}+$ $\frac{y}{x+2 y+z}+\frac{z}{x+y+2 z} \leqslant \frac{3}{4}$. | $$\begin{array}{l}
\text { Prove: From (*) } \frac{x}{2 x+y+z}+\frac{y}{x+2 y+z}+ \\
\frac{z}{x+y+2 z}=\frac{1}{2}\left[\frac{a-b+c}{a+c}+\frac{a+b-c}{a+b}+\right. \\
\left.\frac{-a+b+c}{b+c}\right]=\frac{3}{2}-\frac{1}{2}\left[\frac{b}{a+c}+\frac{c}{a+b}+\right. \\
\left.\frac{a}{b+c}\right] \text {. }
\end{array}$$
... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,031 |
Problem 5: (Mathematical Bulletin 1996.1 Mathematical Problem and Solution Column 991) In $\triangle A B C$, $a$, $b$, $c$ are the side lengths, $s$ is the semi-perimeter, and $\Delta$ is the area. Prove: $(s-a)^{4}+(s-$ $b)^{4}+(s-c)^{4} \geqslant \Delta^{2}$ | To prove: From (*), the desired conclusion is equivalent to $x^{4}+y^{4}+z^{4} \geqslant x y z(x+y+z)$,
This follows from the well-known inequality $x^{2}+y^{2}+z^{2} \geqslant x y+y z+z x$:
$$\begin{array}{c}
x^{4}+y^{4}+z^{4} \geqslant x^{2} y^{2}+y^{2} z^{2}+z^{2} x^{2} \geqslant \\
(x y)(y z)+(y z)(z x)+(z x)(x y)... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,032 |
Problem 6: (《Mathematics Bulletin》 +2001.10 P-44) In $\triangle ABC$, $a, b, c$ are the side lengths, $s$ is the semi-perimeter, $r_{a}, r_{b}, r_{c}$ are the radii of the excircles, respectively. Prove that: $\sqrt{\frac{a}{r_{a}}}+\sqrt{\frac{b}{r_{b}}}+\sqrt{\frac{c}{r_{c}}} \geqslant \sqrt{\frac{2 s}{r}}$. $r$ is t... | Proof: From $(*)$ and $r_{a}=\frac{r s}{s-a}$, etc., we know that the original inequality is equivalent to
$$
\sqrt{x(y+z)}+\sqrt{y(z+x)}+\sqrt{z(x+y)}
$$
$$
\leqslant \sqrt{2}(x+y+z) \Leftrightarrow \sqrt{2 x(y+z)}+\sqrt{2 y(z+x)} \\
+\sqrt{2 z(x+y)} \leqslant 2(x+y+z).
$$
According to the two-variable arithmetic-geo... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,033 |
Practice problem: Given the sequence $\left\{\frac{1}{n(n+1)}\right\}$, find $S_{n}$ | Analysis: $\because(n+1)-n=1, \therefore a_{n}=$ $\frac{(n+1)-n}{n(n+1)}=\frac{1}{n}-\frac{1}{n+1}$
Solution: $\because a_{n}=\frac{1}{n}-\frac{1}{n+1}$
$$\begin{array}{c}
\therefore S_{n}=a_{1}+a_{2}+\cdots+a_{n}=\left(1-\frac{1}{2}\right)+\left(\frac{1}{2}\right. \\
\left.-\frac{1}{3}\right)+\cdots+\left(\frac{1}{n}... | \frac{n}{n+1} | Algebra | math-word-problem | Yes | Yes | inequalities | false | 736,034 |
Example 1 (2006 China Northern Olympiad Problem) Let $a$, $b$, $c$ be positive real numbers, and $a+b+c=3$. Prove:
$$\begin{array}{l}
\frac{a^{2}+9}{2 a^{2}+(b+c)^{2}}+\frac{b^{2}+9}{2 b^{2}+(c+a)^{2}} \\
+\frac{c^{2}+9}{2 c^{2}+(a+b)^{2}} \leqslant 5
\end{array}$$ | 【Analysis】For the sake of explanation, the author will abbreviate the proposition as follows:
Given $a, b, c \in \mathbf{R}^{+}, \sum a=3$, prove:
$\sum \frac{a^{2}+9}{2 a^{2}+(b+c)^{2}} \leqslant 5$ (where $\sum$ denotes cyclic summation, the same applies below).
From the condition, it is necessary to prove $\sum \fr... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,036 |
Example 2 Let $a, b, c > 0, a + b + c \leqslant abc$, prove: $\frac{1}{\sqrt{1 + ab}} + \frac{1}{\sqrt{1 + bc}} + \frac{1}{\sqrt{1 + ca}} \leqslant \frac{3}{2}$. | 【Analysis】From $a+b+c \leqslant a b c$ we get $\frac{1}{a b}+\frac{1}{b c}+\frac{1}{c a} \leqslant 1$. Let $x=\frac{1}{a b}, y=\frac{1}{b c}, z=\frac{1}{c a}$, then the original proposition is transformed into: Given $x, y, z > 0, \sum x \leqslant 1$, prove: $\sum \frac{1}{\sqrt{1+\frac{1}{x}}} \leqslant \frac{3}{2}$. ... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,037 |
Example 3 (2004 Singapore Mathematical Olympiad) Let $0<a, b, c<1$ and $ab+bc+ac=1$, prove that: $\frac{a}{1-a^{2}}+\frac{b}{1-b^{2}}+\frac{c}{1-c^{2}} \geqslant \frac{3}{2} \sqrt{3}$. | 【Analysis】By the basic inequality, we know $(a+b+c)^{2} \geqslant 3(ab + bc + ca) = 3$, thus $a+b+c \geqslant \sqrt{3}$. Therefore, the proposition is transformed into:
Given $0 < a, b, c < 1$, we need to prove $\sum \frac{a}{1-a^{2}} \geqslant \frac{3}{2} \sqrt{3}$.
Consider the function $h(x) = \frac{x}{1-x^2} - 3... | \frac{3}{2} \sqrt{3} | Inequalities | proof | Yes | Yes | inequalities | false | 736,038 |
Example 4 Given that $a, b, c$ are the lengths of the three sides of a triangle, prove:
$$\sum \frac{1}{\sqrt{b}+\sqrt{c}-\sqrt{a}} \geqslant \frac{3(\sqrt{a}+\sqrt{b}+\sqrt{c})}{a+b+c} .$$ | 【Analysis】The inequality to be proved can be transformed into
$$\begin{array}{r}
\sum \frac{a+b+c}{(\sqrt{b}+\sqrt{c}-\sqrt{a})(\sqrt{b}+\sqrt{c}+\sqrt{a})} \geqslant 3, \\
\text { i.e., } \sum \frac{a+b+c}{b+c+2 \sqrt{b c}-a} \geqslant 3 . \ldots \ldots . . . . . . . . . . . . .(4)
\end{array}$$
Notice that (4) is a ... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,039 |
Example 1 Given $a, b, c, d > 0$, and $a + b + c + d = 1$. Prove: $\sum \frac{1}{1 + a^3} \leq \frac{256}{65}$ (Note " $\sum$ " denotes cyclic summation, the same applies hereinafter $)^{[1]}$ . | Proof: Let $f(x)=\frac{1}{1+x^{3}}(0<x<1)$. Then
$$\begin{array}{l}
f\left(\frac{1}{4}\right)=\frac{64}{65}, f^{\prime}(x)=-\frac{3 x^{2}}{\left(1+x^{3}\right)^{2}} \\
f^{\prime}\left(\frac{1}{4}\right)=-\frac{768}{4225}
\end{array}$$
Thus, the equation of the tangent line to $f(x)$ at $x=\frac{1}{4}$ is
$$\begin{arra... | \frac{256}{65} | Inequalities | proof | Yes | Yes | inequalities | false | 736,040 |
Example 2 Given $x_{i}>0(i=1,2,3,4,5)$, and $\sum_{i=1}^{5} \frac{1}{1+x_{i}}=1$. Prove: $\sum_{i=1}^{5} \frac{x_{i}}{4+x_{i}^{2}} \leqslant 1$.
(2003, China Western Mathematical Olympiad)
Analysis: The condition does not meet the requirements.
Let $\frac{1}{1+x_{i}}=a_{i}(i=1,2,3,4,5)$. Then the original condition bec... | Prove: Let $\frac{1}{1+x_{i}}=a_{i}(i=1,2,3,4,5)$. Then
$$\begin{array}{l}
a_{i}>0, \sum_{i=1}^{5} a_{i}=1 \\
\frac{x_{i}}{4+x_{i}^{2}}=\frac{\frac{1}{a_{i}}-1}{4+\left(\frac{1}{a_{i}}-1\right)^{2}}=\frac{a_{i}-a_{i}^{2}}{5 a_{i}^{2}-2 a_{i}+1}
\end{array}$$
Thus, the original inequality is equivalent to
$$\sum_{i=1}^... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,041 |
Example 11 Let $a, b, c$ be positive numbers, and $a+b+c=1$. Prove:
$$\left(\frac{1}{b+c}-a\right)\left(\frac{1}{c+a}-b\right)\left(\frac{1}{a+b}-c\right) \geqslant\left(\frac{7}{6}\right)^{3[2]}$$
Analysis: The left side of the inequality is not a sum. To transform the product into a sum, we can think of the AM-GM in... | Prove: The inequality is equivalent to
$$\sqrt[3]{\prod \frac{1-a+a^{2}}{1-a}} \geqslant \frac{7}{6}$$
where, " $\Pi$ " denotes the cyclic product. Since the geometric mean is greater than or equal to the harmonic mean, we have
$$\sqrt[3]{\Pi \frac{1-a+a^{2}}{1-a}} \geqslant \frac{3}{\sum \frac{1-a}{1-a+a^{2}}}$$
Thu... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,042 |
Example 3 Given $a, b, c > 0$, and $a^{4} + b^{4} + c^{4} = 3$. Prove: $\sum \frac{1}{4 - ab} \leqslant 1$. | Prove: Notice
$$\begin{array}{l}
3=a^{4}+b^{4}+c^{4}=\left(a^{2}\right)^{2}+\left(b^{2}\right)^{2}+\left(c^{2}\right)^{2} \\
\geqslant a^{2} b^{2}+b^{2} c^{2}+c^{2} a^{2} . \\
\text { Let } a^{2} b^{2}=x, b^{2} c^{2}=y, c^{2} a^{2}=z \text {. Then } \\
x+y+z \leqslant 3,0<x, y, z<3, \\
0<\sqrt{x}, \sqrt{y}, \sqrt{z}<2 ... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,043 |
Example 4 Let $x, y, z > 0, x^{4} + y^{4} + z^{4} = 1$. Find the minimum value of $\sum \frac{x^{3}}{1 - x^{8}}$.
(2000, Jiangsu Province High School Mathematics Competition)
Analysis: Let $x^{4} = a, y^{4} = b, z^{4} = c$, the expression $\sum \frac{x^{3}}{1 - x^{8}} = \sum \frac{\left(x^{4}\right)^{\frac{3}{4}}}{1 - ... | Let $x^{4}=a, y^{4}=b, z^{4}=c$. Then
$a$,
$$\begin{array}{l}
\cdot c>0, a+b+c=1, \\
\sum \frac{x^{3}}{1-x^{8}}=\sum \frac{a^{\frac{3}{4}}}{1-a^{2}} .
\end{array}$$
Let $f(x)=\frac{x^{\frac{3}{4}}}{1-x^{2}}$.
It is easy to know that the tangent line equation of $f(x)$ at $x=\frac{1}{3}$ is
$$g(x)=\frac{9 \sqrt[4]{3}}{... | \frac{9 \sqrt[4]{3}}{8} | Algebra | math-word-problem | Yes | Yes | inequalities | false | 736,044 |
Example 5 Given $x, y, z > 0$. Prove:
$$\frac{x^{3}}{y(x+y)^{2}}+\frac{y^{3}}{z(y+z)^{2}}+\frac{z^{3}}{x(z+x)^{2}} \geqslant \frac{3}{4} \text {. }$$
Analysis: Since each term on the left side of the inequality can only contain one variable, the left side must be transformed, simultaneously producing the conditions re... | Prove: The original inequality can be transformed into
$$\frac{\left(\frac{x}{y}\right)^{3}}{\left(\frac{x}{y}+1\right)^{2}}+\frac{\left(\frac{y}{z}\right)^{3}}{\left(\frac{y}{z}+1\right)^{2}}+\frac{\left(\frac{z}{x}\right)^{3}}{\left(\frac{z}{x}+1\right)^{2}} \geqslant \frac{3}{4}$$
Let $\frac{x}{y}=a, \frac{y}{z}=b,... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,045 |
Example 6 Given $a, b, c, d > 0$. Prove:
$$\sum \frac{a^{3}}{(a+b)(a+c)(a+d)} \geqslant \frac{1}{2} .$$
Analysis: From the product in the denominator
$$(a+b)(a+c)(a+d)$$
we think of the AM-GM inequality
$$\sqrt[3]{x y z} \leqslant \frac{x+y+z}{3} .$$ | Proof: Since the inequality is a homogeneous inequality, we can assume $a+b+c+d=1$. Notice that
$$\begin{array}{l}
(a+b)(a+c)(a+d) \\
\leqslant\left[\frac{(a+b)+(a+c)+(a+d)}{3}\right]^{3} \\
=\left(a+\frac{b+c+d}{3}\right)^{3}=\left(a+\frac{1-a}{3}\right)^{3} \\
=\frac{1}{27}(2 a+1)^{3} .
\end{array}$$
Therefore, $\fr... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,046 |
Example 7 Given $a, b, c > 0$. Prove:
$$\sum \frac{a}{2 \sqrt{b^{2}-b c+c^{2}}+3 a} \leqslant \frac{3}{5}.$$
Analysis: When $b=c$,
$$\sqrt{b^{2}-b c+c^{2}}=\sqrt{b^{2}}=b=\frac{b+c}{2},$$
This leads to the conjecture that $2 \sqrt{b^{2}-b c+c^{2}} \geqslant b+c$. | Proof: Notice
$$\begin{array}{l}
2 \sqrt{b^{2}-b c+c^{2}}=\sqrt{4\left(b^{2}+c^{2}\right)-4 b c} \\
\geqslant \sqrt{\left(b^{2}+c^{2}\right)+3 \times 2 b c-4 b c}=\sqrt{(b+c)^{2}}
\end{array}$$
Then $2 \sqrt{b^{2}-b c+c^{2}} \geqslant b+c$
$$\Rightarrow \sum \frac{a}{2 \sqrt{b^{2}-b c+c^{2}}+3 a} \leqslant \sum \frac{... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,047 |
Example 8 Given $x, y, z > -\frac{1}{3}$, and $x+y+z = 3$. Prove: $\sum \frac{1}{\sqrt{3 x+1}+\sqrt{3 y+1}} \geqslant \frac{3}{4}$.
Analysis: $x, y$ are respectively in the two radicals $\sqrt{3 x+1}$, $\sqrt{3 y+1}$, which leads to the inequality
$$\sqrt{a}+\sqrt{b} \leqslant \sqrt{2(a+b)}$$
This can bring $x, y$ in... | Proof: By Cauchy-Schwarz inequality, we have
$$\begin{array}{l}
\sqrt{3 x+1}+\sqrt{3 y+1} \\
\leqslant \sqrt{\left(1^{2}+1^{2}\right)(3 x+1+3 y+1)} \\
=\sqrt{2} \cdot \sqrt{3(x+y)+2} \\
=\sqrt{2} \cdot \sqrt{3(3-z)+2} \\
=\sqrt{2} \cdot \sqrt{11-3 z} .
\end{array}$$
Thus,
$$\sum \frac{1}{\sqrt{3 x+1}+\sqrt{3 y+1}} \g... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,048 |
Example 9 Let positive numbers $x, y, z$ satisfy $xyz \geqslant 1$. Prove:
$$\sum \frac{x^{5}-x^{2}}{x^{5}+y^{2}+z^{2}} \geqslant 0 \text {. }$$
(46th IMO)
Analysis: The degrees of $x, y, z$ are different. According to the condition $xyz \geqslant 1$, we can multiply $x^{2}, y^{2}, z^{2}$ by $xyz$ to make the degrees o... | Proof: $\frac{x^{5}-x^{2}}{x^{5}+y^{2}+z^{2}} \geqslant \frac{x^{5}-x^{2} \cdot x y z}{x^{5}+\left(y^{2}+z^{2}\right) \cdot x y z}$
$$\begin{array}{l}
=\frac{x^{4}-x^{2} y z}{x^{4}+\left(y^{2}+z^{2}\right) y z}=\frac{2 x^{4}-x^{2} \cdot 2 y z}{2 x^{4}+\left(y^{2}+z^{2}\right) \cdot 2 y z} \\
\geqslant \frac{2 x^{4}-x^{... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,049 |
Example 10 Given that $a, b, c$ are the lengths of the three sides of a triangle. Prove:
$$\sum \frac{1}{\sqrt{b}+\sqrt{c}-\sqrt{a}} \geqslant \frac{3(\sqrt{a}+\sqrt{b}+\sqrt{c})}{a+b+c}$$
Analysis: The right side of the inequality is not a constant, so both sides must be multiplied by $\frac{a+b+c}{\sqrt{a}+\sqrt{b}+... | Prove: The inequality can be transformed into
$$\sum \frac{a+b+c}{(\sqrt{b}+\sqrt{c}-\sqrt{a})(\sqrt{b}+\sqrt{c}+\sqrt{a})} \geqslant 3 .$$
This is a homogeneous inequality, so we can set \(a+b+c=1\). Thus, the inequality is equivalent to
$$\begin{array}{l}
\sum \frac{1}{(\sqrt{b}+\sqrt{c})^{2}-(\sqrt{a})^{2}} \geqsla... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,050 |
Example 1 Let $x, y, z$ be positive real numbers, and $x+y+z=1$. Find the minimum value of the function $f(x, y, z)=\frac{3 x^{2}-x}{1+x^{2}}+\frac{3 y^{2}-y}{1+y^{2}}+\frac{3 z^{2}-z}{1+z^{2}}$, and provide a proof. (2003 Hunan Province Mathematics Competition Question) | Solution: For $0<x<1$, we first prove that $\frac{3 x^{2}-x}{1+x^{2}} \geqslant$ $\frac{9 x-3}{10}(1)$, this inequality is equivalent to $9 x^{3}-33 x^{2}+19 x-3 \leqslant 0 \Leftrightarrow(3 x$ $-1)^{2}(x-3) \leqslant 0$, which is obviously true. Similarly, we have $\frac{3 y^{2}-y}{1+y^{2}} \geqslant$ $\frac{9 y-3}{1... | 0 | Algebra | proof | Yes | Yes | inequalities | false | 736,051 |
Example 2 Let $a, b, c, d>0$ and $a+b+c+d=1$, prove: $6\left(a^{3}+b^{3}+c^{3}+d^{3}\right) \geqslant a^{2}+b^{2}+c^{2}+d^{2}+\frac{1}{8}$ (2005 8th Hong Kong Mathematical Olympiad) | Proof: Let $f(x)=6 x^{3}-x^{2}, 00$ and $a+b+c+d=1 \cdot f(x)=6 x^{3}-x^{2}$ has a tangent line at $x=\frac{1}{4}$ which is $y=\frac{5}{8} x-\frac{1}{8}$. We will now prove that $f(x) \geqslant \frac{5}{8} x- \frac{1}{8}$, i.e., $6 x^{3}-x^{2} \geqslant \frac{5}{8} x-\frac{1}{8}$.
This inequality is equivalent to $(4 ... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,052 |
Example 3 Given positive numbers $a, b, c$, satisfying $a+b+c=3$, prove that: $\quad \frac{a^{2}+9}{2 a^{2}+(b+c)^{2}}+\frac{b^{2}+9}{2 b^{2}+(a+c)^{2}}+$ $\frac{c^{2}+9}{2 c^{2}+(b+a)^{2}} \leqslant 5$ (2006 2nd Northern Mathematical Olympiad) | Proof: Let $f(x)=\frac{x^{2}+9}{2 x^{2}+(3-x)^{2}}, 0<x<3$, i.e., $f(x)=\frac{x^{2}+9}{3 x^{2}-6 x+9}$. The tangent line of $f(x)$ at $x=1$ is $g(x)=\frac{1}{3} x+\frac{4}{3}$. We need to prove that when $0<x<3$, $f(x) \leqslant g(x)$, i.e., $\frac{x^{2}+9}{3 x^{2}-6 x+9} \leqslant \frac{1}{3} x+\frac{4}{3}$. This ineq... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,053 |
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