problem stringlengths 1 13.6k | solution stringlengths 0 18.5k ⌀ | answer stringlengths 0 575 ⌀ | problem_type stringclasses 8
values | question_type stringclasses 4
values | problem_is_valid stringclasses 1
value | solution_is_valid stringclasses 1
value | source stringclasses 8
values | synthetic bool 1
class | __index_level_0__ int64 0 742k |
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Promotion 1: Let $a, b, c>0$, then we have
$$\sum\left(\frac{a^{m+1}}{b^{m+1}+c^{m+1}}\right) \geq \sum\left(\frac{a^{m}}{b^{m}+c^{m}}\right)$$ | Proof: Moving the right side of the inequality to the left side, we have:
Left side $=\frac{a^{m+1}}{b^{m+1}+c^{m+1}}-\frac{a^{m}}{b^{m}+c^{m}}+\frac{b^{m+1}}{a^{m+1}+c^{m+1}}-\frac{b^{m}}{a^{m}+c^{m}}+\frac{c^{m+1}}{a^{m+1}+b^{m+1}}-\frac{c^{m}}{a^{m}+b^{m}}$
By finding a common denominator, we get: $\sum\left[\frac{a... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,210 |
Promotion 2: Given $n \geq 3, a_{i}>0(i=1,2, \mathrm{~L}, n), T=\sum_{i=1}^{n} a_{i}$, then $\sum_{i=1}^{n}\left(\frac{a_{i}}{T-a_{i}}\right) \geq \frac{n}{n-1}$. | $\begin{array}{l}\text { Prove: By Cauchy-Schwarz inequality, we have: } \sum_{i=1}^{n}\left(T-a_{i}\right) \bullet \sum_{i=1}^{n}\left(\frac{1}{T-a_{i}}\right) \geq n^{2} \text { can be transformed into: } \\ (n-1) \sum_{i=1}^{n} a_{i} \cdot \sum_{i=1}^{n}\left(\frac{1}{T-a_{i}}\right) \geq n^{2} \Leftrightarrow \sum_... | \sum_{i=1}^{n}\left(\frac{a_{i}}{T-a_{i}}\right) \geq \frac{n}{n-1} | Inequalities | proof | Yes | Yes | inequalities | false | 731,211 |
3.1 Let $a, b, c>0$, then
$$\sqrt[3]{\left(\frac{a}{b+c}\right)^{2}}+\sqrt[3]{\left(\frac{b}{a+c}\right)^{2}}+\sqrt[3]{\left(\frac{c}{a+b}\right)^{2}} \geq \frac{3}{\sqrt[3]{4}}$$ | After analysis, we can first prove the following partial inequality:
$\sqrt[3]{\left(\frac{a}{b+c}\right)^{2}} \geq \frac{3 a}{\sqrt[3]{4}(a+b+c)}$, which is equivalent to proving:
$\left(\frac{a}{b+c}\right)^{2} \geq \frac{27 a^{3}}{4(a+b+c)^{3}} \Leftrightarrow \frac{8(a+b+c)^{3}}{27} \geq 2 a(b+c)^{2}$.
$\therefore ... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,212 |
Let $a, b, c > 0, m, n \in N^{*}$, then we have
$$\sum\left(\frac{a}{b+c}\right)^{\frac{n}{n+m}} \geq \frac{n+m}{\sqrt[n+m]{n^{n} m^{m}}}$$ | Prove: $\left(\frac{n}{m} a\right)^{m}(b+c)^{n} \leq\left[\frac{n(a+b+c)}{n+m}\right]^{n+m} \cdot \Leftrightarrow(b+c)^{n} \leq \frac{n^{n} m^{m}(a+b+c)^{n+m}}{(n+m)^{n+m} a^{m}}$
$$\Leftrightarrow\left(\frac{a}{b+c}\right)^{\frac{n}{n+m}} \geq \frac{n+m}{n^{\frac{n}{n+m}} m^{\frac{m}{n+m}}} \cdot \frac{a}{a+b+c}$$
Th... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,213 |
Let $a, b, c, d \in R^{+}$, then we have
$$\sqrt[4]{\left(\frac{a}{b+c+d}\right)^{3}}+\sqrt[4]{\left(\frac{b}{a+c+d}\right)^{3}}+\sqrt[4]{\left(\frac{c}{a+b+d}\right)^{3}}+\sqrt[4]{\left(\frac{d}{a+b+c}\right)^{3}} \geq \frac{4}{\sqrt[4]{3^{3}}}$$ | $$\begin{array}{l}
\text { Prove: } 3 a(b+c+d)^{3} \leq\left[\frac{3(a+b+c+d)}{4}\right]^{4} \cdot \Leftrightarrow(b+c+d)^{3} \leq \frac{3^{3}(a+b+c+d)^{4}}{4^{4} a} \\
\Leftrightarrow\left(\frac{a}{b+c+d}\right)^{3} \geq \frac{4^{4}}{3^{3}} \cdot \frac{a^{4}}{(a+b+c+d)^{4}} \cdot \Leftrightarrow\left(\frac{a}{b+c+d}\r... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,214 |
7 If $a, b, c \in R_{+}, a+b+c=1$, then:
$$\frac{1+c}{a+b}+\frac{1+a}{b+c}+\frac{1+b}{c+a} \geq 6$$ | Proof: First, the left side of the original inequality needs to be adjusted to: " $\left(\sum \frac{2}{a+b}\right)-3$ or $3+\sum \frac{c}{a+b}$ ". Then, use the Cauchy-Schwarz inequality to prove it.
Or, add 1 to the denominator of each term, $\left(a, b, c \in R_{+}, a+b+c=1\right)$, then we have:
$$\frac{c}{a+b+1}+\... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,216 |
Similar to 8, if $a_{1}, a_{2}, \ldots, a_{n} \in R_{+}, \sum_{i=1}^{n} a_{i}=1$, then:
$$\sum_{i=1}^{n}\left(\frac{a_{i}}{1-a_{i}+\sum_{i=1}^{n} a_{i}}\right) \geq \frac{n}{2 n-1}$$ | Proof: First, the left side of the original inequality needs to be adjusted to $\sum_{i=1}^{n}\left[\frac{a_{i}}{2\left(\sum_{i=1}^{n} a_{i}\right)-a_{i}}\right]$ or $\sum_{i=1}^{n}\left(\frac{2}{1-a_{i}+\sum_{i=1}^{n} a_{i}}\right)-n$, then apply the Cauchy-Schwarz inequality. $\square$
If we add 1 to both the numera... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,217 |
If $a, b, c \in R_{+}, a+b+c=1$, then:
$$\frac{a^{2}}{b+c}+\frac{b^{2}}{c+a}+\frac{c^{2}}{a+b} \geq \frac{1}{2}$$ | Prove: $(b+c+a+c+a+b)\left(\frac{a^{2}}{b+c}+\frac{b^{2}}{a+c}+\frac{c^{2}}{b+a}\right) \geq(a+b+c)^{2}=1$. That is: $2\left(\frac{a^{2}}{b+c}+\frac{b^{2}}{a+c}+\frac{c^{2}}{b+a}\right) \geq 1$, which is to be proved. From this, we can also generalize to the following form:
Translate the text above into English, keepi... | null | Inequalities | proof | Yes | Yes | inequalities | false | 731,218 |
Similar to 12 when $a_{1}, a_{2}, \ldots, a_{n} \in R_{+}, \sum_{i=1}^{n} a_{i}=\mu, m \geq 2$ then: $\sum_{i=1}^{n}\left(\frac{a_{i}^{m}}{T-a_{i}}\right) \geq \frac{\mu^{m-1}}{(n-1) n^{m-2}}$ | Prove: $(n-1)\left(\sum_{i=1}^{n} a_{i}\right) \cdot \sum_{i=1}^{n}\left(\frac{a_{i}^{m}}{T-a_{i}}\right) \geq\left(\sum_{i=1}^{n} a_{i}^{\frac{m}{2}}\right)^{2} \geq \frac{n^{2}\left(\sum_{i=1}^{n} a_{i}\right)^{m}}{n^{m}} \geq \frac{\mu^{m}}{n^{m-2}}$ (using Cauchy and Power Mean Inequality), the original inequality ... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,219 |
13.1 Let $a, b, c, d>0$, and $a b+b d+c d+d a=1$. Prove: $\frac{a^{3}}{b+c+d}+\frac{b^{3}}{c+d+a}+\frac{c^{3}}{d+a+b}+\frac{d^{3}}{a+b+c} \geq \frac{1}{3}$. It has been found through discussion that this problem can be proved by a similar method to the one mentioned above. | Proof: $\begin{aligned} & \because \sum[a(b+c+d)] \sum \frac{a^{3}}{b+c+d} \geq\left(a^{2}+b^{2}+c^{2}+d^{2}\right)^{2}, \\ \therefore & \Leftrightarrow \sum \frac{a^{3}}{b+c+d} \geq \frac{\left(a^{2}+b^{2}+c^{2}+d^{2}\right)^{2}}{2(a b+a c+a d+b c+b d+c d)} \\ & \Leftrightarrow \sum \frac{a^{3}}{b+c+d} \geq \frac{a^{2... | \sum \frac{a^{3}}{b+c+d} \geq \frac{1}{3} | Inequalities | proof | Yes | Yes | inequalities | false | 731,220 |
Let $a, b, c>0, a b c=1$, prove: $\frac{1}{a^{2}(b+c)}+\frac{1}{b^{2}(c+a)}+\frac{1}{c^{2}(a+b)} \geq \frac{3}{2}$. | Proof: According to the problem, let \( a=\frac{y z}{x^{2}}, b=\frac{z x}{y^{2}}, c=\frac{x y}{z^{2}} \) where \( x, y, z \in \mathbb{R}^{+} \), then:
\[
\frac{1}{a^{2}(b+c)}=\frac{1}{\frac{y^{2} z^{2}}{x^{4}}\left(\frac{z x}{y^{2}}+\frac{x y}{z^{2}}\right)}=\frac{x^{3}}{y^{3}+z^{3}}
\]
Similarly, we can derive two mo... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,222 |
If $a, b, c \in R_{+}, a+b+c=1$, then:
$$\frac{a^{4}}{b^{3}+c^{3}}+\frac{b^{4}}{a^{3}+c^{3}}+\frac{c^{4}}{b^{3}+c^{3}} \geq \frac{1}{2}$$ | $\begin{array}{l}\text { Prove: } \frac{\left(c^{3}\right)^{2}}{c^{2}\left(a^{3}+b^{3}\right)}+\frac{\left(a^{3}\right)^{2}}{a^{2}\left(b^{3}+c^{3}\right)}+\frac{\left(b^{3}\right)^{2}}{b^{2}\left(c^{3}+a^{3}\right)} \\ \geq \frac{\left(a^{3}+b^{3}+c^{3}\right)^{2}}{c^{2}\left(a^{3}+b^{3}\right)+a^{2}\left(b^{3}+c^{3}\... | \frac{1}{2} | Inequalities | proof | Yes | Yes | inequalities | false | 731,223 |
Similar to 16 if $a, b, c \in R_{+}, m \geq 0, n \geq 0$, then:
$$\frac{a^{m+n}}{b^{m}+c^{m}}+\frac{b^{m+n}}{c^{m}+a^{m}}+\frac{c^{m+n}}{a^{m}+b^{m}} \geq \frac{a^{n}+b^{n}+c^{n}}{2}$$
Considering that Chebyshev's inequality can be proved by the rearrangement inequality, we then use Chebyshev's inequality:
$$\frac{\su... | Proof:
$$\text { Proof: } \begin{aligned}
& \frac{a^{m+n}}{b^{m}+c^{m}}+\frac{b^{m+n}}{c^{m}+a^{m}}+\frac{c^{m+n}}{a^{m}+b^{m}} \\
& \geq \frac{\sum a^{n} \cdot \sum \frac{a^{m}}{b^{m}+c^{m}}}{3} \\
& \geq \frac{\frac{3}{2} \sum a^{n}}{3}=\frac{a^{n}+b^{n}+c^{n}}{2} . \\
\text { Then we have: } & \frac{a^{n}+b^{n}+c^{n... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,224 |
Similar to 17 when $a_{1}, a_{2} \cdots \cdots \cdots a_{n} \in R_{+}, \sum_{i=1}^{n} a_{i}=1, m \in R_{+}, p \geq 1$ then: $\sum_{i=1}^{n}\left[\frac{a_{i}^{m+p}}{\left(\sum_{i}^{n} a_{i}^{m}\right)-a_{i}^{m}}\right] \geq \frac{n}{(n-1) n^{p}}$ | Prove: $\sum_{i=1}^{n}\left[\frac{a_{i}^{m+p}}{\left(\sum_{i}^{n} a_{i}^{m}\right)-a_{i}^{m}}\right] \geq \frac{\left(\sum_{i=1}^{n} a_{i}^{p}\right) \cdot\left(\sum_{i=1}^{n}\left[\frac{a_{i}^{m}}{\left(\sum_{i}^{n} a_{i}^{m}\right)-a_{i}^{m}}\right]\right)}{n} \geq\left(\sum_{i=1}^{n} a_{i}\right)^{p} \frac{n}{n^{p}(... | \frac{n}{(n-1) n^{p}} | Inequalities | proof | Yes | Yes | inequalities | false | 731,225 |
Similar to 18.1 When we study the text [6] and verify it using the MAPLE software, we find and summarize (in conjunction with the previous research content: refer to the variant) that cyclic symmetric expressions with homogeneous numerators and denominators, such as $\frac{c}{b+c}+\frac{a}{c+a}+\frac{b}{a+b} \geq \frac... | Prove: When $n \geq 2$, by the power mean inequality: $a^{\frac{n}{2}}+b^{\frac{n}{2}}+c^{\frac{n}{2}}+d^{\frac{n}{2}} \geq 4 \bullet\left(\frac{a+b+c+d}{4}\right)^{\frac{n}{2}}$ The original inequality is equivalent to
$$\begin{array}{l}
(e+f+g+h)\left(\frac{a^{n}}{e}+\frac{b^{n}}{f}+\frac{c^{n}}{g}+\frac{d^{n}}{h}\ri... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,226 |
18.2 Let $a_{1}, a_{2}, \ldots \ldots ., a_{n-1}, a_{n} \in R_{+}, b_{1}, b_{2}, \ldots \ldots, b_{n-1}, b_{n} \in R_{+}, m, n \in R_{+}$ and $m \geq 2, n \geq 1$, then: $\sum_{i=1}^{n}\left(\frac{a_{i}^{m}}{b_{i}}\right) \geq \frac{\left(\sum_{i=1}^{n} a_{i}\right)^{m}}{n^{m-2}\left(\sum_{i=1}^{n} b_{i}\right)}$. | Prove: When $m \geq 2$, by the power mean inequality, we have: $\sum_{i=1}^{n} a_{i} \frac{m}{2} \geq n\left(\frac{\sum_{i=1}^{n} a_{i}}{n}\right)^{\frac{m}{2}}$; by the Cauchy-Schwarz inequality, we get:
$$\left(\sum_{i=1}^{n} b_{i}\right) \bullet\left(\sum_{i=1}^{n} \frac{a_{i}^{m}}{b_{i}}\right)=\left[\sum_{i=1}^{n}... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,227 |
19.1 If \(a, b, c \in R_{+}\), then:
$$\sum \frac{a}{b^{2}+c^{2}} \geq \frac{4}{a+b+c}$$ | Proof: Without loss of generality, let $a \geq b \geq c$.
Such inequalities are difficult to prove using general inequalities because equality holds when $a=b$ and $c=0$ (or $b=0, a=c$ or $a=0, b=c$). Therefore, we consider using the SOS (Sum of Squares) method to prove it.
Proof: Move the right side of the inequality... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,228 |
Lemma 1 If $a, b, c$ are all real numbers, and $b+c, c+a, a+b$ are all non-zero, then
$$\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}-\frac{3}{2}=\frac{1}{2}\left[\frac{(c-a)^{2}}{(a+b)(b+c)}+\frac{(a-b)^{2}}{(b+c)(c+a)}+\frac{(b-c)^{2}}{(c+a)(a+b)}\right]$$ | $\begin{array}{l} \text { Prove: } \frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}-\frac{3}{2}=\frac{1}{2}\left[\frac{2 a}{b+c}+\frac{2 b}{c+a}+\frac{2 c}{a+b}-3\right] \\ = \frac{1}{2}\left[\left(\frac{a}{b+c}+\frac{b}{c+a}-1\right)+\left(\frac{b}{c+a}+\frac{c}{a+b}-1\right)+\left(\frac{c}{a+b}+\frac{a}{b+c}-1\right)\right]... | proof | Algebra | proof | Yes | Yes | inequalities | false | 731,229 |
Lemma 2 If $a, b, c$ are all real numbers, and $b+c, c+a, a+b$ are all non-zero, then
$$\frac{a^{2}}{b+c}+\frac{b^{2}}{c+a}+\frac{c^{2}}{a+b}-\frac{a+b+c}{2}=(a+b+c)\left(\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}-\frac{3}{2}\right)$$ | $$\begin{array}{l}
\text { Prove: } \frac{a^{2}}{b+c}+\frac{b^{2}}{c+a}+\frac{c^{2}}{a+b} \\
=\frac{a^{2}+a(b+c)}{b+c}+\frac{b^{2}+b(c+a)}{c+a}+\frac{c^{2}+c(a+b)}{a+b}-(a+b+c) \\
=\frac{a(a+b+c)}{b+c}+\frac{b(a+b+c)}{c+a}+\frac{c(a+b+c)}{a+b}-(a+b+c) \\
=(a+b+c)\left(\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}-1\right)
... | proof | Algebra | proof | Yes | Yes | inequalities | false | 731,230 |
Lemma 3 If $a, b, c$ are all real numbers, and $b+c, c+a, a+b$ are all non-zero, then
$$\frac{a^{2}}{b+c}+\frac{b^{2}}{c+a}+\frac{c^{2}}{a+b}-\frac{a+b+c}{2}=\frac{1}{4}\left[\frac{(2 a-b-c)^{2}}{b+c}+\frac{(2 b-c-a)^{2}}{c+a}+\frac{(2 c-a-b)^{2}}{a+b}\right] .$$ | Prove: $\frac{(2 a-b-c)^{2}}{b+c}=\frac{4 a^{2}}{b+c}-4 a+b+c, \frac{(2 b-c-a)^{2}}{c+a}=\frac{4 b^{2}}{c+a}-4 b+c+a$,
$$\frac{(2 c-a-b)^{2}}{a+b}=\frac{4 c^{2}}{a+b}-4 c+a+b$$
Adding the above three equations yields
$$\frac{a^{2}}{b+c}+\frac{b^{2}}{c+a}+\frac{c^{2}}{a+b}-\frac{a+b+c}{2}=\frac{1}{4}\left[\frac{(2 a-b-... | proof | Algebra | proof | Yes | Yes | inequalities | false | 731,231 |
Lemma $\mathbf{4}$ Given $a, b, c \in R$, then we have
$$a^{3}+b^{3}+c^{3}-3 a b c=\frac{1}{2}(a+b+c)\left[(a-b)^{2}+(b-c)^{2}+(c-a)^{2}\right]$$ | $\begin{array}{l}\text { Prove: } a^{3}+b^{3}+c^{3}-3 a b c \\ =(a+b)^{3}+c^{3}-3 a^{2} b-3 a b^{2}-3 a b c \\ =(a+b)^{3}+c^{3}-3 a b(a+b+c) \\ =[(a+b)+c]\left[(a+b)^{2}-(a+b) c+c^{2}\right]-3 a b(a+b+c) \\ =(a+b+c)\left[(a+b)^{2}-(a+b) c+c^{2}-3 a b\right] \\ =(a+b+c)\left(a^{2}+b^{2}+c^{2}-a b-b c-c a\right) \\ =\fra... | proof | Algebra | proof | Yes | Yes | inequalities | false | 731,232 |
Promotion 3.1: $\frac{1}{a^{3}(b+c+d)}+\frac{1}{b^{3}(c+d+a)}+\frac{1}{c^{3}(d+a+b)}+\frac{1}{d^{3}(a+b+c)} \geq \frac{4}{3}$, where $a, b, c, d>0, abcd=1$. | Proof: Let $a=\frac{y z w}{x^{3}}, b=\frac{z w x}{y^{3}}, c=\frac{w x y}{z^{3}}, d=\frac{x y z}{w^{3}}$ where $x, y, z, w \in R^{+}$, then:
$$\begin{array}{l}
\frac{1}{a^{3}(b+c+d)}=\frac{1}{\frac{y^{3} z^{3} w^{3}}{x^{9}}\left(\frac{z w x}{y^{3}}+\frac{w x y}{z^{3}}+\frac{x y z}{w^{3}}\right)}=\frac{x^{8}}{y^{4} z^{4}... | \frac{4}{3} | Inequalities | proof | Yes | Yes | inequalities | false | 731,233 |
Lemma $\mathbf{5}$ Given $a, b, c \in R$, then we have:
$$(2 a-b-c)^{2}+(2 b-a-c)^{2}+(2 c-b-a)^{2}=3\left[(a-b)^{2}+(b-c)^{2}+(c-a)^{2}\right]$$ | $\begin{array}{l}\text { Prove: }(2 a-b-c)^{2}+(2 b-c-a)^{2}+(2 c-a-b)^{2} \\ =[(a-b)+(a-c)]^{2}+[(b-a)+(b-c)]^{2}+[(c-b)+(c-a)]^{2} \\ =2\left[(a-b)^{2}+(b-c)^{2}+(c-a)^{2}\right]+2[(a-b)(a-c)+(b-a)(b-c)+(c-a)(c-b)] \\ =2\left[(a-b)^{2}+(b-c)^{2}+(c-a)^{2}+a^{2}+b^{2}+c^{2}-a b-a c-b c\right] \\ =3\left[(a-b)^{2}+(b-c... | proof | Algebra | proof | Yes | Yes | inequalities | false | 731,234 |
Strengthening 1: If $a, b, c \in R_{+}$, then
$$\left(\sum \frac{a}{b+c}\right)-\frac{3}{2} \geq \frac{1}{2} \cdot \frac{(a-b)^{2}}{a^{2}+b^{2}+c^{2}}$$ | Prove: The original inequality is equivalent to
$$\begin{array}{l}
\frac{1}{2}\left[\frac{(c-a)^{2}}{(a+b)(b+c)}+\frac{(a-b)^{2}}{(b+c)(c+a)}+\frac{(b-c)^{2}}{(c+a)(a+b)}\right] \geq \frac{1}{2} \cdot \frac{(a-b)^{2}}{a^{2}+b^{2}+c^{2}} \\
\Leftrightarrow \frac{(c-a)^{2}}{(a+b)(b+c)}+\frac{(b-c)^{2}}{(c+a)(a+b)} \geq\l... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,235 |
Strengthening 2: If $a, b, c \in R_{+}$, then
$$\frac{a^{2}}{b+c}+\frac{b^{2}}{c+a}+\frac{c^{2}}{a+b}-\frac{a+b+c}{2} \geq \mathrm{k} \cdot \frac{(2 a-b-c)^{2}}{(a+b+c)}$$
Using the MAPLE software, the maximum value of k is found to be $\frac{1}{2}$. Now, we directly prove the strongest case.
$$\frac{a^{2}}{b+c}+\frac... | Prove: the original inequality is equivalent to
$$\begin{array}{l}
\frac{1}{4}\left[\frac{(2 a-b-c)^{2}}{b+c}+\frac{(2 b-c-a)^{2}}{c+a}+\frac{(2 c-a-b)^{2}}{a+b}\right] \geq \frac{(2 a-b-c)^{2}}{2(a+b+c)} \\
\Leftrightarrow \frac{(2 a-b-c)^{2}}{b+c}+\frac{(2 b-c-a)^{2}}{c+a}+\frac{(2 c-a-b)^{2}}{a+b} \geq \frac{2(2 a-b... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,236 |
Inference 2.1: If $a, b, c \in R_{+}$, then
$$\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}-\frac{3}{2} \geq \frac{(2 a-b-c)^{2}}{2(a+b+c)^{2}}$$ | $\begin{array}{l}\text { Prove simply: } \quad\left(\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}-\frac{3}{2}\right)(a+b+c) \geq \frac{(2 a-b-c)^{2}}{2(a+b+c)} \\ \Leftrightarrow \frac{a^{2}}{b+c}+\frac{b^{2}}{c+a}+\frac{c^{2}}{a+b}-\frac{a+b+c}{2} \geq \frac{(2 a-b-c)^{2}}{2(a+b+c)}\end{array}$ | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,237 |
Proposition 3.1 If $a, b, c \in R_{+}$, then
$$\sum \frac{a^{2}}{b+c}-\frac{a+b+c}{2} \geq \frac{\sum(a-b)^{2}}{a+b+c}$$ | $$\begin{array}{l}
\text { Proof: } \sum\left(\frac{2 a^{2}}{b+c}-a\right) \geq \frac{2\left[\sum(a-b)^{2}\right]}{a+b+c} \Leftrightarrow \sum\left(\frac{2 a^{2}-a b-a c}{b+c}\right) \geq \frac{2\left[\sum(a-b)^{2}\right]}{a+b+c} \\
\Leftrightarrow \sum\left(\frac{a(a-b)+a(a-c)}{b+c}\right) \geq \frac{2\left[\sum(a-b)^... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,238 |
Corollary 3.2 If \(a, b, c \in R_{+}\), then
\[
\sum \frac{a^{3}}{b+c} - \frac{\sum a^{2}}{2} \geq \frac{1}{2} \cdot \sum(a-b)^{2}
\] | $$\begin{array}{l}
\text { Proof: } \sum\left(\frac{2 a^{3}}{b+c}-a^{2}\right) \geq \sum(a-b)^{2} \Leftrightarrow \sum\left(\frac{a^{2}(a-b)+a^{2}(a-c)}{b+c}\right) \geq \sum(a-b)^{2} \\
\Leftrightarrow \sum\left[(a+b)(a-b)^{2}\left(a^{2}+a b+a c+b^{2}+b c\right)\right] \geq\left[\sum(a-b)^{2}\right](a+b)(b+c)(c+a) \\
... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,239 |
Corollary 3.3 If \(a, b, c \in R_{+}\), then
\[
\sum \frac{a^{3}}{b+c} - \frac{1}{6}\left(\sum a\right)^{2} \geq \frac{2}{3} \cdot \sum(a-b)^{2}
\] | $$\begin{array}{l}
\text { Proof: } \sum \frac{a^{2}(a-b)+a^{2}(a-c)+a(a-b)(a+b)+a(a-c)(a+c)+2 a\left(a^{2}-b c\right)}{b+c} \geq 4 \sum(a-b)^{2} \\
\Leftrightarrow \sum(a-b)^{2}(a+b)\left(2 a^{2}+2 c a+3 a b+2 b c+2 b^{2}\right)+\sum 2 a\left(a^{2}-b c\right)(a+b)(c+a) \\
\geq 4\left(\sum(a-b)^{2}\right)(a+b)(b+c)(c+a... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,240 |
Strengthening 4: If $a, b, c \in R_{+}$, then we have $\sum\left(\frac{a}{b+c}\right)-\frac{3}{2} \geq k \cdot \frac{(a-b)^{2}+(b-c)^{2}+(c-a)^{2}}{a^{2}+b^{2}+c^{2}}$.
Ref. [11] only proved that $k=\frac{1}{3}$. Using the MAPLE software, we found the maximum value of $k$ to be $\frac{\sqrt{3}-1}{2}$, although $k$ is i... | $$\begin{array}{l}
: \sum\left(\frac{a}{b+c}\right)-\frac{3}{2}-\frac{\sqrt{3}-1}{2} \cdot \frac{(a-b)^{2}+(b-c)^{2}+(c-a)^{2}}{a^{2}+b^{2}+c^{2}} \geq 0 \\
\Leftrightarrow \sum \frac{(a-b)+(a-c)}{b+c} \geq \frac{(\sqrt{3}-1)\left(\sum(a-b)^{2}\right)}{a^{2}+b^{2}+c^{2}} \\
\Leftrightarrow \sum(a-b)^{2}(a+b) \geq(\sqrt... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,242 |
Strengthening 5: $\sum \frac{a}{b+c} \geq \frac{3}{2}+\frac{17}{6} \cdot \frac{(a-b)(b-c)(c-a)}{(a+b)(b+c)(c+a)} \quad(a, b, c \geq 0)$ | Proof: Without loss of generality, let $\max (a, b, c) = a$.
Move to the left: $\sum \frac{a}{b+c} - \frac{3}{2} - \frac{17}{6} \cdot \frac{(a-b)(b-c)(c-a)}{(a+b)(b+c)(c+a)} \geq 0$
Left side $= \frac{1}{3} \cdot \frac{3a^3 + 3b^3 + 3c^3 + 7b^2c - 10ab^2 - 10bc^2 + 7ac^2 + 7a^2b - 10a^2c}{(b+c)(c+a)(a+b)}$
When $a \geq... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,252 |
Strengthening 7 We seek intrinsic connections similar to Nesbitt's inequality, obtained through the MAPLE software, if \(a, b, c \in R_{+}\), then (the coefficient 2 is the strongest case):
$$2 \cdot \sum \frac{a}{b+c} \leq (a+b+c) \sum \frac{a}{b^{2}+c^{2}}$$ | $$\begin{array}{l}
\text { Proof: To prove the original inequality, we need to show: } \\
\sum \frac{a(a+b+c)}{b^{2}+c^{2}}-\frac{2 a}{b+c} \geq 0 \Leftrightarrow \sum\left[\frac{a b(a-b)+(a-c) a c}{\left(b^{2}+c^{2}\right)(b+c)}\right]+\sum \frac{2 a b c}{\left(b^{2}+c^{2}\right)(b+c)} \geq 0 \\
\Leftrightarrow \sum \... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,253 |
Strengthen $\mathbf{8}$ When $a, b, c \in R_{+}$, there is the following form:
$$k \cdot \frac{\left(\sum a^{3}\right)-\left(\sum a^{2}\right)\left(\sum a\right)}{\sum a^{2}} \leq\left(\sum \frac{a}{b+c}-\frac{3}{2}\right)(a+b+c)$$
Through MAPLE software, $k=1.5$ is the maximum value
$$\frac{3 \cdot\left(\sum a^{3}\ri... | To prove the original inequality, it is necessary to prove:
$$\begin{array}{l}
(a+b)(b+c)(c+a) \cdot \frac{2 \cdot \sum a^{3}-\sum a \cdot \sum a^{2}}{\sum a^{2}} \leq\left[\sum(a-b)^{2}(a+b)\right](a+b+c) \\
(a+b)(b+c)(c+a) \frac{2 \sum a^{3}-\sum\left(a^{2} b+a^{2} c\right)}{\sum a^{2}} \leq\left[\sum(a-b)^{2}(a+b)\r... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,254 |
Example 1. (2009 5th Northern Mathematical Olympiad) If $x, y, z > 0$ and $x^{2} + y^{2} + z^{2} = 3$, prove:
$$\frac{x^{2009} - 2008(x - 1)}{y + z} + \frac{y^{2009} - 2008(y - 1)}{z + x} + \frac{z^{2009} - 2008(z - 1)}{x + y} \geq \frac{1}{2}(x + y + z)$$ | Proof: By the AM-GM inequality,
$$x^{2009}-2008(x-1)=x^{2009}+1+\ldots \ldots .+1_{(2008 \uparrow 1)}-2008 x \geq 2009 x-2008 x=x,$$
Therefore, $\frac{x^{2009}-2008(x-1)}{y+z} \geq \frac{x}{y+z}$,
Similarly, $\frac{y^{2009}-2008(y-1)}{z+x} \geq \frac{y}{z+x}, \frac{z^{2009}-2008(z-1)}{x+y} \geq \frac{z}{x+y}$,
By the ... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,256 |
Example 2. Let $a, b, c>0$, prove that: $\frac{1}{a(1+b)}+\frac{1}{b(1+c)}+\frac{1}{c(1+a)} \geq \frac{3}{\sqrt[3]{a b c}(1+\sqrt[3]{a b c})}$ | Proof: Let $a b c=k^{3}>0, a=k \cdot \frac{a_{2}}{a_{1}}, b=k \cdot \frac{a_{3}}{a_{2}}, c=k \cdot \frac{a_{1}}{a_{3}}$, where $a_{1}, a_{2}, a_{3}>0$.
Then $\frac{1}{a(1+b)}+\frac{1}{b(1+c)}+\frac{1}{c(1+a)} \geq \frac{3}{\sqrt[3]{a b c}(1+\sqrt[3]{a b c})}$
$\Leftrightarrow \frac{1}{k \frac{a_{2}}{a_{1}}+k^{2} \frac{... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,257 |
Example 3. $a, b, c \in R^{+}$, prove: $\frac{a^{2}+b c}{a(b+c)}+\frac{b^{2}+c a}{b(c+a)}+\frac{c^{2}+a b}{c(a+b)} \geq 3$. | Prove $\sum \frac{a^{2}+b c}{a(b+c)}=\sum \frac{c}{a+b}+\sum \frac{c a}{a b+b c}$, then use $\sum \frac{c}{a+b} \geq \frac{3}{2}, \sum \frac{c a}{a b+b c} \geq \frac{3}{2}$ and add them together. | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,258 |
Example 4. Given that $x, y, z$ are real numbers greater than -1, prove: $\frac{1+x^{2}}{1+y+z^{2}}+\frac{1+y^{2}}{1+z+x^{2}}+\frac{1+z^{2}}{1+x+y^{2}} \geq 2$. Analysis: This problem involves proving a fractional inequality, and we can try using the Cauchy-Schwarz inequality to eliminate the denominators. | Prove: $2 x \leq 1+x^{2}=a, 2 y \leq 1+y^{2}=b, 2 z \leq 1+z^{2}=c$, then
$$\begin{array}{l}
\frac{1+x^{2}}{1+y+z^{2}}+\frac{1+y^{2}}{1+z+x^{2}}+\frac{1+z^{2}}{1+x+y^{2}} \\
=\frac{a}{y+c}+\frac{b}{z+a^{2}}+\frac{c}{x+b} \geq \frac{a}{\frac{b}{2}+c}+\frac{b}{\frac{c}{2}+a}+\frac{c}{\frac{a}{2}+b} \\
=2\left(\frac{a}{b+... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,259 |
Example 5. (36th IMO Problem) Let $a, b, c$ be positive real numbers, $abc=1$, prove that: $\sum \frac{1}{a^{3}(b+c)} \geq \frac{3}{2}$. | Prove: Let $x=\frac{1}{a}, y=\frac{1}{b}, z=\frac{1}{c}$, then $x y z=1$. The original inequality can be transformed into: $\sum \frac{x^{2}}{y+z} \geq \frac{3}{2}$. This inequality can be proven using the AM-GM inequality, Cauchy-Schwarz inequality, and rearrangement inequality (combined with similar 11). | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,260 |
Example 6. Let positive numbers $a, b, c$ satisfy $abc=1$, prove that: $\frac{a}{b^{2}(c+1)}+\frac{b}{c^{2}(a+1)}+\frac{c}{a^{2}(b+1)} \geq \frac{3}{2}$. | Proof: Let $a=\frac{x}{y}, b=\frac{y}{z}, c=\frac{z}{x}, x, y, z>0$, we need to prove that $\frac{x^{2} z^{2}}{y^{3}(x+z)}+\frac{x^{2} y^{2}}{z^{3}(y+x)}+\frac{z^{2} y^{2}}{x^{3}(z+y)} \geq \frac{3}{2}$, which is equivalent to $\frac{z x}{y^{2}} \cdot \frac{z x}{x y+y z}+\frac{x y}{z^{2}} \cdot \frac{x y}{y z+z x}+\fra... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,261 |
Example 7. (Problem 556 from Issue 2, 2002 of "Mathematics Teaching") In an acute triangle, prove that $\frac{\cos A}{\cos (B-C)}+\frac{\cos B}{\cos (C-A)}+\frac{\cos C}{\cos (A-B)} \geq \frac{3}{2}$. | $\begin{array}{l}\text { Prove: } \frac{\cos A}{\cos (B-C)}=\frac{\sin B \sin C-\cos B \cos C}{\cos B \cos C+\sin B \sin C}=\frac{\tan B \tan C-1}{1+\tan B \tan C} \\ =\frac{\tan A \tan B \tan C-\tan A}{\tan A+\tan A \tan B \tan C}=\frac{\tan B+\tan C}{2 \tan A+\tan B+\tan C}, \\ \text { Similarly, } \frac{\cos B}{\cos... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,262 |
Example 8. $0<A, B, C<\frac{\pi}{2}, A+B+C=\pi$, prove:
$$\sqrt{1-\sin A \sin B}+\sqrt{1-\sin B \sin C}+\sqrt{1-\sin C \sin A} \geq \frac{3}{2}$$ | Prove: Let $A=\frac{\pi}{2}-\frac{A_{1}}{2}, B=\frac{\pi}{2}-\frac{B_{1}}{2}, C=\frac{\pi}{2}-\frac{C_{1}}{2}$, then the problem is equivalent to proving in any triangle $\Delta A_{1} B_{1} C_{1}$: $\sqrt{1-\sin A_{1} \sin B_{1}}+\sqrt{1-\sin B_{1} \sin C_{1}}+\sqrt{1-\sin C_{1} \sin A_{1}} \geq \frac{3}{2}$.
$$\begin{... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,263 |
Example 9. (2000 Korean Mathematical Olympiad) Let $a>b>c>0, x>y>z>0$, prove: $\frac{a^{2} x^{2}}{(b y+c z)(b z+c y)}+\frac{b^{2} y^{2}}{(c z+a x)(c x+a z)}+\frac{c^{2} z^{2}}{(a x+b y)(a y+b x)} \geq \frac{3}{4}$. | Proof: By the AM-GM inequality,
$$(b y+c z)(b z+c y) \leq\left[\frac{(b y+c z)+(b z+c y)}{2}\right]^{2}=\frac{1}{4}(b+c)^{2}(y+z)^{2}$$
Thus, it suffices to prove
$$\frac{a^{2} x^{2}}{(b+c)^{2}(y+z)^{2}}+\frac{b^{2} y^{2}}{(c+a)^{2}(z+x)^{2}}+\frac{c^{2} z^{2}}{(a+b)^{2}(x+y)^{2}} \geq \frac{3}{16}$$
By the Cauchy-Sc... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,264 |
Conjecture 1: With the help of Professor Yang Lu's Bottema2009 software, Strengthening 5 can still be further strengthened to: For $a, b, c > 0$, then:
$$\sum \frac{a}{b+c}-\frac{3}{2} \geq \frac{\sqrt{13+16 \sqrt{2}}}{2} \cdot \frac{|(a-b)(b-c)(c-a)|}{(a+b)(b+c)(c+a)}$$
The coefficient of this inequality is the root ... | $$\begin{array}{l}
\text { Prove: By moving the inequality to the left side, combining terms and eliminating the denominator, we get: } \\
\sqrt{13+16 \sqrt{2}} a^{2} b+\sqrt{13+16 \sqrt{2}} a c^{2}-\sqrt{13+16 \sqrt{2}} a^{2} c+\sqrt{13+16 \sqrt{2}} b^{2} c-\sqrt{13+16 \sqrt{2}} b^{2} a-\sqrt{13+16 \sqrt{2}} b c^{2} \... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,265 |
Promotion 5: $\alpha_{i} \geq 0, a_{i} \geq 0\left(1 \leq i, k \leq n ; i, k, n \in N_{+}\right), \lambda 、 \alpha 、 \beta \in \mathrm{R}_{+}, \mu \geq 1$ when, we have:
$$\left[\frac{\lambda\left(a_{1}+a_{2}+\ldots \ldots+a_{k}\right)}{\beta_{1} a_{k+1}+\ldots \ldots+\beta_{n-k} a_{n}}\right]^{\mu}+\left[\frac{\lambda... | Proof: Step 1: Apply the AM-GM inequality to raise to the power; Step 2: Use the Cauchy-Schwarz inequality to eliminate the denominator (refer to the above proof and similar proof 2), and the original inequality can be proved. | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,266 |
1.1 Let $a, b, c>0$, then
$$\sum \frac{a^{3}}{b^{2}+c^{2}} \geq \sum \frac{a^{2}}{b+c}$$ | Prove: Moving the right side of the inequality to the left side, we have:
Left side $=\sum\left(\frac{a^{3}}{b^{2}+c^{2}}-\frac{a^{2}}{b+c}\right)=\sum\left(\frac{a^{2}\left(a b+a c-b^{2}-c^{2}\right)}{\left(b^{2}+c^{2}\right)(b+c)}\right)$
Transforming it, we get: $\sum\left(\frac{a^{2} c(a-c)}{\left(b^{2}+c^{2}\right... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,267 |
If $1.3 a_{i} \geq 0\left(1 \leq i \leq n, i \in R_{+}\right)$, then we have: $\sum_{j=1}^{n}\left[\frac{a_{j}^{2}}{\left(\sum_{i=1}^{n} a_{i}^{2}\right)-a_{j}^{2}}\right] \geq \sum_{j=1}^{n}\left[\frac{a_{j}}{\left(\sum_{i=1}^{n} a_{i}\right)-a_{j}}\right]$ | $\begin{array}{l}\text { Prove: } \sum_{j=1}^{n}\left[\frac{a_{j}^{2}}{\left(\sum_{i=1}^{n} a_{i}^{2}\right)-a_{j}^{2}}\right]-\sum_{j=1}^{n}\left[\frac{a_{j}}{\left(\sum_{i=1}^{n} a_{i}\right)-a_{j}}\right]=\sum_{j=1}^{n}\left[\frac{a_{j}{ }^{2}}{\left(\sum_{i=1}^{n} a_{i}^{2}\right)-a_{j}^{2}}-\frac{a_{j}}{\left(\sum... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,268 |
Similar to 2.2 and when $\mu \geq 1$, we have: $\sum_{i=1}^{n}\left[\frac{a_{i}}{\left(\sum_{i=1}^{n} a_{i}\right)-a_{i}}\right]^{\mu} \geq \frac{n}{(n-1)^{\mu}}$ | Prove: According to the power mean inequality: when $\alpha>\beta$, we have:
$$\begin{aligned}
& \left(\frac{\sum_{i=1}^{n} a_{i}^{\alpha}}{n}\right)^{\frac{1}{\alpha}} \geq\left(\frac{\sum_{i=1}^{n} a_{i}^{\beta}}{n}\right)^{\frac{1}{\beta}} \\
\text { Then }: & \sum_{i=1}^{n}\left[\frac{a_{i}}{\left(\sum_{i=1}^{n} a_... | \frac{n}{(n-1)^{\mu}} | Inequalities | proof | Yes | Yes | inequalities | false | 731,269 |
Proposition If $h_{a} 、 h_{b} 、 h_{c} 、 r$ are the lengths of the altitudes and the inradius of a triangle, respectively, and $0<\lambda \leqslant 2$, then
$$\frac{1}{h_{a}-\lambda r}+\frac{1}{h_{b}-\lambda r}+\frac{1}{h_{c}-\lambda r} \geqslant \frac{3}{(3-\lambda) r},$$
with equality if and only if the triangle is e... | Proof: $\because \frac{1}{h_{a}}+\frac{1}{h_{b}}+\frac{1}{h_{c}}=\frac{1}{r}$,
$$\therefore \frac{h_{a}-\lambda r}{h_{a}}+\frac{h_{b}-\lambda r}{h_{b}}+\frac{h_{c}-\lambda r}{h_{c}}=3-\lambda .$$
Thus, $\frac{h_{a}}{h_{a}-\lambda r}+\frac{h_{b}}{h_{b}-\lambda r}+\frac{h_{c}}{h_{c}-\lambda r} \geqslant \frac{9}{3-\lamb... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,270 |
Given
$$\begin{array}{l}
a_{11} x_{1}^{a_{1}}+a_{12} x_{21}^{a_{1}}+\cdots+a_{1 n} x_{n}^{a_{1}}=y_{1}^{a_{1}}, \\
a_{21} x_{1}^{a_{2}}+a_{22} x_{2}^{a_{2}}+\cdots+a_{2 n} x_{n}^{2}=y_{2}^{a_{2}} \\
\cdots \cdots . \\
a_{n 1} x_{1}^{a_{1}}+a_{n 2} x_{12}^{a_{1}}+\cdots+a_{n 1} x_{n}^{a_{n}}=y_{n^{\prime \prime}},
\end{... | Proof: By the weighted arithmetic mean-geometric mean inequality
$$\begin{array}{l}
\frac{q_{1} a_{1}+q_{2} a_{2}+\cdots+q_{n} a_{n}}{q_{1}+q_{2}+\cdots+q_{n}} \\
\geqslant\left(a_{1}^{q_{1}} a_{2}^{q_{2}} \cdots a_{n}^{q_{n}}\right)^{\frac{1}{q_{1}+q_{2}+\cdots+q_{n}}},
\end{array}$$
Let $q_{i}=a_{1 i}, a_{i}=x_{i}^{... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,271 |
Lemma Given $a$ as a positive real number, $b, c$ as non-negative real numbers, then the cubic algebraic equation
$$\lambda^{3}-\frac{1}{4}\left(a^{2}+b^{2}+c^{2}\right) \lambda-\frac{a b c}{4}=0$$
has a unique positive root $\lambda=\sqrt{\frac{a^{2}+b^{2}+c^{2}}{3}} \cos \frac{\theta}{3}$, where $\theta=$ $\arccos \... | Prove that given $p=-\frac{1}{4}\left(a^{2}+b^{2}+c^{2}\right), q=-\frac{a b c}{4}$, using the discriminant of a cubic algebraic equation, we have:
$$\Delta=\left(\frac{q}{2}\right)^{2}+\left(\frac{p}{3}\right)^{3}=\left(\frac{a b c}{8}\right)^{2}-\left(\frac{a^{2}+b^{2}+c^{2}}{12}\right)^{3} \leqslant 0$$
Then the eq... | proof | Algebra | proof | Yes | Yes | inequalities | false | 731,273 |
【Example 1】Let $x, y, z$ be positive real numbers, and satisfy $x+y+z=1$. Prove that: $\frac{x}{x+y z}+\frac{y}{y+z x}+\frac{z}{z+x y} \leqslant \frac{9}{4}$. | Proof: Without loss of generality, let $x \geqslant y \geqslant z$, from $x+y+z=1$ we get $x \geqslant \frac{1}{3}, z = 1-x-y$.
The original inequality is equivalent to $\frac{x}{x+y(1-x-y)}+\frac{y}{y+(1-x-y) x} +\frac{1-x-y}{1-x-y+x y} \leqslant \frac{9}{4}$
Rearranging gives $\frac{x}{(x+y)(1-y)}+\frac{y}{(x+y)(1-... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,277 |
【Example 2】Let $x, y, z$ be positive real numbers, and satisfy $x+y+z=1$, prove:
$$\frac{x}{2 x^{2}+y^{2}+z^{2}}+\frac{y}{x^{2}+2 y^{2}+z^{2}}+\frac{z}{x^{2}+y^{2}+2 z^{2}} \leqslant \frac{9}{4} .$$ | Prove: Since $2 x^{2}+y^{2}+z^{2}=2 x^{2}+y^{2}+(1-x-y)^{2}=$
$$\begin{array}{l}
3 x^{2}+2 y^{2}+2 x y-2 x-2 y+1, \\
\quad \text { and } x+y z=x+y(1-x-y)=x+y-x y-y^{2}, \\
\quad \text { hence }\left(2 x^{2}+y^{2}+z^{2}\right)-(x+y z)=3 y^{2}+(3 x-3) y+ \\
\left(3 x^{2}-3 x+1\right),
\end{array}$$
i.e., $\left(2 x^{2}+... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,278 |
Let $x, y, z \in(0,+\infty)$, and $x^{2}+y^{2}+$ $z^{2}=1$. Find the range of the function $x+y+z-x y z$ | First, prove $f>1$: Clearly, $0<y^{2}, z>z^{2}$. Thus,
$$\begin{aligned}
f & =x+y+z-x y z \\
& \geqslant x+y+z-x \cdot \frac{y^{2}+z^{2}}{2} \\
& =x+y+z-x \cdot \frac{1-x^{2}}{2}\left(\text { because } x^{2}+y^{2}+z^{2}\right. \\
& =1) \\
& =x \cdot \frac{1+x^{2}}{2}+y+z \\
& \geqslant x \cdot x+y+z \\
& >x^{2}+y^{2}+z... | \left[1, \frac{8 \sqrt{3}}{9}\right] | Algebra | math-word-problem | Yes | Yes | inequalities | false | 731,280 |
Question: Let $x, y, z \in(0,+\infty)$ and $x^{2}+y^{2}+$ $z^{2}=1$, find the range of the function $f=x+y+z-x y z$.
Find the range of the function $f=x+y+z-x y z$ given that $x, y, z \in(0,+\infty)$ and $x^{2}+y^{2}+z^{2}=1$. | Given $x^{2}+y^{2}+z^{2}=1$, so $y^{2}+z^{2}=1-x^{2}$. Therefore, we can set $y=\sqrt{1-x^{2}} \cos \theta, z=\sqrt{1-x^{2}} \sin \theta, \theta \in\left(0, \frac{\pi}{2}\right)$, thus $f=x+\sqrt{1-x^{2}}(\sin \theta+\cos \theta) - x(1-x^{2}) \sin \theta \cos \theta$.
Let $t=\sin \theta+\cos \theta$, then $\sin \theta... | 1 | Algebra | math-word-problem | Yes | Yes | inequalities | false | 731,281 |
Question 1 Find the minimum value of the function $y=2 \sqrt{(x-1)^{2}+4}+$ $\sqrt{(x-8)^{2}+9}$. | Let $a, b$ be undetermined constants, and $a, b$ are not both zero.
Then by the Cauchy-Schwarz inequality for two variables $\left(p^{2}+q^{2}\right)\left(x^{2}+y^{2}\right) \geqslant (p x+q y)^{2}$ (equality holds if and only if $q x=p y$), we have
$$\begin{aligned}
\sqrt{a^{2}+b^{2}} y= & 2 \sqrt{\left(a^{2}+b^{2}\ri... | 5 \sqrt{5} | Algebra | math-word-problem | Yes | Yes | inequalities | false | 731,283 |
Question 2 (Text [2] Question 13) Let $a, b, c$ be positive real numbers, and $\sqrt{a^{2}+b^{2}}+c=1$, prove that: $a b+2 a c \leqslant \frac{1}{\sqrt{3}}$. | Prove that if $y=a b+2 a c, \alpha$ is a positive constant to be determined, then
$$\begin{aligned}
\alpha y & =\alpha a(b+2 c) \leqslant\left[\frac{\alpha a+(b+2 c)}{2}\right]^{2} \\
& =\left[\frac{1}{2}(\alpha a+b)+c\right]^{2} \\
& =\left[\frac{1}{2} \sqrt{(\alpha a+1 \cdot b)^{2}}+c\right]^{2} \\
& \leqslant\left[\... | a b+2 a c \leqslant \frac{1}{\sqrt{3}} | Inequalities | proof | Yes | Yes | inequalities | false | 731,284 |
Question 3 (2008 Shaanxi Province Mathematics Exam for Science, Last Question) Given the sequence $\left\{a_{n}\right\}$ with the first term $a_{1}=\frac{3}{5}, a_{n+1}=\frac{3 a_{n}}{2 a_{n}+1}, n=1,2$,
( I ) Find the general term formula for the sequence $\left\{a_{n}\right\}$;
(II) Prove that for any $x>0, a_{n} \ge... | (I) $a_{n}=\frac{3^{n}}{3^{n}+2}, n=1,2, \cdots$ (Process omitted);
(II) Proof omitted;
(III) Proof: If following the "hint" in (II), it is easy to prove (III). If this college entrance examination problem did not have (I) and (III) and asked candidates to freely develop, how would one prove (III)? ——Try the method of ... | a_{1}+a_{2}+\cdots+a_{n}>\frac{n^{2}}{n+1}, n=1,2, \cdots | Algebra | proof | Yes | Yes | inequalities | false | 731,285 |
Example 1 Given $a, b, c \geqslant 0$, and $a+b+c=1$, prove:
$$2\left(a^{2}+b^{2}+c^{2}\right)+9 a b c \geqslant 1$$ | Assume without loss of generality that $a \leqslant b \leqslant c$, then $0 \leqslant a \leqslant \frac{1}{3}$, which implies $9 a-4<0$. Therefore,
$$\begin{array}{l}
2\left(a^{2}+b^{2}+c^{2}\right)+9 a b c \\
=2 a^{2}+2(b+c)^{2}+b c(9 a-4) \\
\geqslant 2 a^{2}+2(1-a)^{2}+\left(\frac{b+c}{2}\right)^{2}(9 a-4) \\
=2 a^{... | 1 | Inequalities | proof | Yes | Yes | inequalities | false | 731,286 |
Example 2 Let $a, b, c \in R^{+}$, prove that:
$$\frac{a^{2}}{a+b}+\frac{b^{2}}{b+c}+\frac{c^{2}}{c+a} \geqslant \frac{a+b+c}{2} .$$ | Prove that if $a=b+\alpha, b=c+\beta, c=a+\gamma$, then $\alpha +\beta+\gamma=0$, thus
$$\begin{array}{l}
4\left(\frac{a^{2}}{a+b}+\frac{b^{2}}{b+c}+\frac{c^{2}}{c+a}\right) \\
=\frac{(a+b+\alpha)^{2}}{a+b}+\frac{(b+c+\beta)^{2}}{b+c}+\frac{(c+a+\gamma)^{2}}{c+a} \\
=2(a+b+c)+2(\alpha+\beta+\gamma) \\
\quad+\frac{\alph... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,287 |
Example 11 (Adapted from Problem 15 of the First Round of the 2009 National High School Mathematics Competition) Prove:
$$3 \sqrt{3}+\sqrt{13} \leqslant \sqrt{27+x}+\sqrt{13-x}+\sqrt{x} \leqslant 11$$ | Prove that for the function $y=f(x)=\sqrt{27+x}+\sqrt{13-x}+\sqrt{x}$, the derivative is
$$\begin{array}{l}
y^{\prime}=\frac{1}{2 \sqrt{27+x}}-\frac{1}{2 \sqrt{13-x}}+\frac{1}{2 \sqrt{x}} . \\
\text { Let } y^{\prime}=0 \text {, we get } \frac{1}{\sqrt{27+x}}-\frac{1}{\sqrt{13-x}}+\frac{1}{\sqrt{x}}=0
\end{array}$$
Si... | 3 \sqrt{3}+\sqrt{13} \leqslant \sqrt{27+x}+\sqrt{13-x}+\sqrt{x} \leqslant 11 | Inequalities | proof | Yes | Yes | inequalities | false | 731,288 |
Example 13 For any real numbers $x, y, z$, we have $\left(x^{2}+1\right)\left(y^{2}+1\right)\left(z^{2}+1\right) \geqslant \frac{3}{4}(x+y+z)^{2}$.
The equality holds if and only if $x=y=z= \pm \frac{1}{\sqrt{2}}$. | To prove the original inequality, it suffices to prove
$$3(x+y+z)^{2}-4\left(x^{2}+1\right)\left(y^{2}+1\right)\left(z^{2}+1\right) \leqslant 0$$
Thus, construct the quadratic function
$$f(t)=\left(x^{2}+1\right) t^{2}-\sqrt{3}(x+y+z) t+\left(y^{2}+1\right)\left(z^{2}+1\right)$$.
Since \(x^{2}+1>0\),
$$\begin{array}{... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,290 |
Example 14 In $\triangle A B C$, prove that: $\sin ^{2} \frac{A}{2}+\sin ^{2} \frac{B}{2}+\sin ^{2} \frac{C}{2} \geqslant \frac{3}{4}$. | To prove the inequality is equivalent to
$$\begin{aligned}
\frac{\sin ^{2} \frac{A}{2}}{\sin ^{2} \frac{A}{2}+\cos ^{2} \frac{A}{2}}+ & \frac{\sin ^{2} \frac{B}{2}}{\sin ^{2} \frac{B}{2}+\cos ^{2} \frac{B}{2}} \\
& +\frac{\sin ^{2} \frac{C}{2}}{\sin ^{2} \frac{C}{2}+\cos ^{2} \frac{C}{2}} \geqslant \frac{3}{4}
\end{ali... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,291 |
Example 15 (2010 Shandong Province High School Mathematics Preliminary Competition Question) Let non-negative real numbers $a, b, c$ satisfy $a+b+c=1$, prove: homogenize $9 a b c \leqslant a b+b c+c a \leqslant \frac{1}{4}(1+9 a b c)$. | To prove the left inequality, we apply the 3-variable mean inequality, obtaining
$$\begin{array}{l}
a b + b c + c a \\
= (a b + b c + c a)(a + b + c) \\
\geqslant 3 \sqrt[3]{a b \cdot b c \cdot c a} \cdot 3 \sqrt[3]{a b c} \\
= 9 a b c
\end{array}$$
Thus, \(a b + b c + c a \geqslant 9 a b c\).
For the right inequalit... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,292 |
Example 16 Let $a, b, c \in \mathbf{R}^{+}$, and $a b c \geqslant 1$, positive integers $m, n$ satisfy $(n+1)^{2}>m$, prove:
$\left(a+\frac{m}{a+n}\right)\left(b+\frac{m}{b+n}\right)\left(c+\frac{m}{c+n}\right)$
$\geqslant\left(\frac{m+n+1}{n+1}\right)^{3}$. I didn't understand the background of the problem. | Prove that from the 2-variable mean inequality, we get
$\frac{a+n}{(n+1)^{2}}+\frac{1}{a+n} \geqslant \frac{2}{n+1}$,
we obtain $\frac{m(a+n)}{(n+1)^{2}}+\frac{m}{a+n} \geqslant \frac{2 m}{n+1}$,
transforming, we have
$a+\frac{m}{a+n} \geqslant \frac{2 m}{n+1}+a-\frac{m a}{(n+1)^{2}}-\frac{m n}{(n+1)^{2}}$,
which means... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,293 |
Example 1 Given any 7 real numbers, prove that: there exist at least two numbers $x, y$ such that $0 \leqslant \frac{x-y}{1+x y}<\frac{\sqrt{3}}{3}$. | Analysis If we start from 7 real numbers, the difficulty is high, but the inequality to be proved resembles the trigonometric formula $\tan (\alpha-\beta)=\frac{\tan \alpha-\tan \beta}{1+\tan \alpha \tan \beta}$, which leads us to think of solving it using a trigonometric model.
Proof Let the given 7 real numbers be $... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,294 |
Example 3 Let $x, y, z \geqslant 0, x+y+z=1$, prove that:
$$x^{2} y^{2}+y^{2} z^{2}+z^{2} x^{2}+x^{2} y^{2} z^{2} \leqslant \frac{1}{16}$$ | Assume $x \geqslant y \geqslant z$, then $y^{2} \leqslant x^{2}, z^{2} \leqslant y z, y^{2} z^{2} \leqslant y z$, thus $x^{2} y^{2}+y^{2} z^{2}+z^{2} x^{2}+x^{2} y^{2} z^{2}$
$\leqslant x^{2} y^{2}+x^{2} z^{2}+x^{2} y z+x^{2} y z$ $=x^{2}(y+z)^{2}$, (here it is converted to a homogeneous form!)
and $x(y+z) \leqslant\le... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,296 |
Example 4 Given positive numbers $x, y, z$ satisfying $x+y+z=3$, prove:
$$\frac{\sqrt{x}}{2 x+3}+\frac{\sqrt{y}}{2 y+3}+\frac{\sqrt{z}}{2 z+3} \leqslant \frac{3}{5} .$$ | Prove that the tangent line equation of $f(x)=\frac{\sqrt{x}}{2 x+3}$ at $x=1$ is $y=\frac{1}{50} x+\frac{9}{50}$, and the following proof:
$$\frac{\sqrt{x}}{2 x+3} \leqslant \frac{1}{50} x+\frac{9}{50}, x \in(0,3)$$
Using the method of analysis, it suffices to prove $50 \sqrt{x} \leqslant 2 x^{2}+21 x+27$, $x \in(0,3... | \frac{3}{5} | Inequalities | proof | Yes | Yes | inequalities | false | 731,297 |
Example 5 Given that $a, b, c$ are non-negative numbers, and $a+b+c=1$, prove: $a b+b c+c a-3 a b c \leqslant \frac{1}{4}$. This... let's consolidate it again. | Proof: Let $S=ab+bc+ca-3abc$.
Using the local adjustment method to find the maximum value of $S$. Without loss of generality, assume $a \geqslant b \geqslant c$, then $c \leqslant \frac{1}{3}$.
Fix $c$, i.e., $a+b$ is constant,
$S=ab+bc+ca-3abc=ab(1-3c)+c(a+b)$
$1-3c \geqslant 0, ab$ reaches its maximum value when $a=... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,298 |
Example 6 (Mathematics Bulletin, Issue 9, 2010, Problem 1872) Prove: Among any 13 real numbers, there must exist two real numbers $x, y$, such that $y>\frac{x-0.3}{1+0.3 x}$ | Prove that for any 13 real numbers \(a_{1}, a_{2}, \cdots, a_{13}\), there exist 13 real numbers \(\theta_{1}, \theta_{2}, \cdots, \theta_{13}\) in \(\left(-\frac{\pi}{2}\right., \left.\frac{\pi}{2}\right)\) such that \(\tan \theta_{k} = a_{k} (k=1,2, \cdots, 13)\).
Divide the interval \(\left(-\frac{\pi}{2}, \frac{\p... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,299 |
Example 7 Given $a, b, c \in \mathbf{R}, a, b$ are not both zero, prove:
$$\frac{\left(a^{2}+b^{2}+a c\right)^{2}+\left(a^{2}+b^{2}+b c\right)^{2}}{a^{2}+b^{2}} \geqslant(a+b+c)^{2} .$$ | Prove that the inequality to be proved is equivalent to
$$\begin{array}{l}
\sqrt{\frac{\left(a^{2}+b^{2}+a c\right)^{2}+\left(a^{2}+b^{2}+b c\right)^{2}}{\left(a^{2}+b^{2}\right)^{2}}} \geqslant \frac{|a+b+c|}{\sqrt{a^{2}+b^{2}}} \\
\text { i.e., } \sqrt{\left(1+\frac{a c}{a^{2}+b^{2}}\right)^{2}+\left(1+\frac{b c}{a^{... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,300 |
Example 8 (Math Problem $1613,2006,5$ ) Let $a, b, c \in \boldsymbol{R}^{+}$, $\lambda \geqslant 0$, prove:
$$\begin{array}{l}
\frac{\sqrt{a^{2}+\lambda b^{2}}}{a}+\frac{\sqrt{b^{2}+\lambda c^{2}}}{c}+\frac{\sqrt{c^{2}+\lambda a^{2}}}{c} \\
\geqslant 3 \sqrt{1+\lambda}
\end{array}$$ | $$\begin{array}{l}
\frac{\sqrt{a^{2}+\lambda b^{2}}}{a}+\frac{\sqrt{b^{2}+\lambda c^{2}}}{c}+\frac{\sqrt{c^{2}+\lambda a^{2}}}{c} \\
=\sqrt{1+\frac{\lambda b^{2}}{a^{2}}}+\sqrt{1+\frac{\lambda c^{2}}{b^{2}}}+\sqrt{1+\frac{\lambda a^{2}}{c^{2}}} \\
=\left|1+\frac{\sqrt{\lambda} b}{a} i\right|+\left|1+\frac{\sqrt{\lambda... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,301 |
Example 9 Given positive numbers $a, b, c$ satisfying $a+b+c=1$, prove:
$$2 \sqrt{3} \leqslant \sqrt{3 a^{2}+1}+\sqrt{3 b^{2}+1}+\sqrt{3 c^{2}+1}<4$$ | Prove that on one hand, construct vectors $\overrightarrow{O A}=(\sqrt{3} a, 1), \overrightarrow{O B}=$ $(\sqrt{3} b, 1), \overrightarrow{O C}=(\sqrt{3} c, 1)$, then
$$\begin{aligned}
& \sqrt{3 a^{2}+1}+\sqrt{3 b^{2}+1}+\sqrt{3 c^{2}+1} \\
= & |\overrightarrow{O A}|+|\overrightarrow{O B}|+|\overrightarrow{O C}| \\
\geq... | 2 \sqrt{3} \leqslant \sqrt{3 a^{2}+1}+\sqrt{3 b^{2}+1}+\sqrt{3 c^{2}+1}<4 | Inequalities | proof | Yes | Yes | inequalities | false | 731,302 |
Example 10 Given the sequence $\left\{a_{n}\right\}$, where the first term $a_{1}=\frac{3}{2}$, and for any $n>1, n \in \mathbf{N}^{*}$, we have $a_{n+1}=a_{n}+1+\frac{\sqrt{1+4 a_{n}}}{2}$, prove that: $\frac{(n+1)^{2}}{4}<a_{n}<3 \times 2^{n-1}-\frac{3}{2}$. | Prove the right inequality first.
Since $a_{n+1}=a_{n}+1+\frac{\sqrt{1+4 a_{n}}}{2}$
$$a_{n}+1+\frac{\sqrt{4 a_{n}}}{2} \text {. }$$
Therefore, $a_{n+1}>a_{n}+1+\sqrt{a_{n}}$, hence
$$a_{n+1}>\left(\sqrt{a_{n}}+\frac{1}{2}\right)^{2}+\frac{3}{4}>\left(\sqrt{a_{n}}+\frac{1}{2}\right)^{2}$$
So $\sqrt{a_{n+1}}-\sqrt{a_{... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,303 |
Theorem Let $a_{i} \in \mathbf{R}^{+}(i=1,2, \cdots, n), k, l$ and $m \in \mathbf{N}$, and $m \geqslant 2 l$. Then
$$\begin{array}{l}
\sum_{i=0}^{n-1} \frac{a_{t+1}^{m}}{\sum_{j=0}^{k-1} a_{t+c+j}^{l}} \geqslant \frac{1}{k} \sum_{i=1}^{n} a_{i}^{m-1} . \\
\left(a_{0}=a_{n}, a_{n+i}=a_{i}\right)
\end{array}$$ | Proof: By Cauchy's inequality, we get
$$\begin{array}{l}
\sum_{t=0}^{n-1} \frac{a_{t+1}^{m}}{\sum_{j=0}^{k-1} a_{i+t+1}^{l}} \cdot \sum_{t=0}^{n-1} a_{t+1}^{m-2 l} \cdot \sum_{j=0}^{k-1} a_{i+t+1}^{l} \\
\geqslant\left(\sum_{i=1}^{n} a_{i}^{m-1}\right)^{2} .
\end{array}$$
By \( m \geqslant 2 l, m, l \in \mathbf{N} \) ... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,304 |
Given $x, y \in \mathbf{R}^{+}$, and $x \neq y$, prove:
$$\left|\frac{1}{1+x^{3}}-\frac{1}{1+y^{3}}\right|<|x-y| \text {. }$$ | Proof: By imitating the above proof, first transform the left side to get
$$\begin{aligned}
& \left\lvert\, \frac{1}{1+x^{3}}-\frac{1}{1+y^{3}}\right. \\
= & |x-y| \cdot \frac{x^{2}+x y+y^{2}}{\left(1+x^{3}\right)\left(1+y^{3}\right)} \\
< & |x-y| \cdot \frac{x^{2}+x y+y^{2}}{1+x^{3}+y^{3}}
\end{aligned}$$
Thus, it su... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,307 |
If $a, b, c$ are positive numbers satisfying $a+b+c=1$, then $\sqrt{a+\frac{1}{b}}+\sqrt{b+\frac{1}{c}}+\sqrt{c+\frac{1}{a}} \geqslant \sqrt{30}$. | According to the extreme principle, the condition for the inequality to hold with equality is $a=b=c=\frac{1}{3}$. Therefore, multiplying each term on the left side of the inequality by the constant $\sqrt{\frac{1}{3}+3}$ and combining with the Cauchy-Schwarz inequality, we get:
$$\begin{array}{l}
\sqrt{\frac{1}{3}+3} ... | \sqrt{30} | Inequalities | proof | Yes | Yes | inequalities | false | 731,310 |
Given: $a^{3}+b^{3}=2$. Prove: $a+b \leqslant 2$. | Prove: From $a^{3}+1+1 \geqslant 3 \sqrt[3]{a^{3} \cdot 1 \cdot 1}=3 a$, we know $b^{3}+1+1 \geqslant 3 b$, adding the two inequalities yields $a^{3}+b^{3}+4 \geqslant 3(a+b)$.
From $a^{3}+b^{3}=2$, we get
$$a+b \leqslant \frac{1}{3}\left(a^{3}+b^{3}+4\right)=\frac{1}{3} \times 6=2 \text {. }$$ | a+b \leqslant 2 | Inequalities | proof | Yes | Yes | inequalities | false | 731,311 |
Text [1] and [2] introduce the following inequality:
If $a, b, c>0$, and $a+b+c=1$, then
$$\frac{1}{1+a^{2}}+\frac{1}{1+b^{2}}+\frac{1}{1+c^{2}} \leqslant \frac{27}{10}$$
Text [3] provides an elementary proof of (1). Text [4] generalizes
(1) to:
If $a, b, c, d>0$, and $a+b+c+d=1$, then
$$\frac{1}{1+a^{3}}+\frac{1}{1+... | Lemma If $f(x)=\frac{1}{\left(1+x^{2}\right)^{2}}, x \in(0,1)$, then
$$f(x) \leqslant-\frac{4096}{4913} x+\frac{5376}{4913}$$
Proof It is obvious that inequality (3) is equivalent to
$$4913 \leqslant(-4096 x+5376)\left(1+x^{2}\right)^{2}$$
Transforming, we get $4096 x^{5}-5376 x^{4}+8192 x^{3}-10752 x^{2}$
$$+4096 x-... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,315 |
Theorem If $a, b, c, d \in \mathbf{R}^{+}, a+b+c+d=1$ then
$$\begin{array}{l}
\frac{1}{\left(1+a^{2}\right)^{2}}+\frac{1}{\left(1+b^{2}\right)^{2}}+\frac{1}{\left(1+c^{2}\right)^{2}}+\frac{1}{\left(1+d^{2}\right)^{2}} \\
\leqslant \frac{824}{289}
\end{array}$$ | Notice that, for $a, b, c, d \in(0,1)$, from the inequality $(*)$ we have
$$\begin{array}{l}
\frac{1}{\left(1+a^{2}\right)^{2}} \leqslant \frac{4096}{4913} a+\frac{5376}{4913} \\
\frac{1}{\left(1+b^{2}\right)^{2}} \leqslant \frac{4096}{4913} b+\frac{5376}{4913} \\
\frac{1}{\left(1+c^{2}\right)^{2}} \leqslant \frac{4096... | \frac{824}{289} | Inequalities | proof | Yes | Yes | inequalities | false | 731,316 |
For positive numbers $x$, $y$, and $z$ satisfying $x^{2}+y^{2}+z^{2}=1$, find
$$\frac{x}{1-x^{2}}+\frac{y}{1-y^{2}}+\frac{z}{1-z^{2}}$$
the minimum value. | Solution: Clearly, $x \cdot y, z \in (0,1)$,
By the AM-GM inequality for three variables,
$2 x^{2}\left(1-x^{2}\right)\left(1-x^{2}\right) \leqslant\left(\frac{2}{3}\right)^{3}=\frac{8}{27}$,
with equality if and only if $2 x^{2}=1-x^{2}$, i.e., $x^{2}=\frac{1}{3}$.
$$\begin{array}{l}
2 x^{2}\left(1-x^{2}\right)\left(1... | \frac{3 \sqrt{3}}{2} | Algebra | math-word-problem | Yes | Yes | inequalities | false | 731,317 |
Question 3 Given that $x, y$ are positive real numbers, and $n \geqslant 2(n \in \mathrm{N})$, prove:
$$\sqrt[n]{\frac{x}{x+\left(2^{n}-1\right) y}}+\sqrt[n]{\frac{y}{y+\left(2^{n}-1\right) x}} \geqslant 1 .$$ | Prove that for positive numbers $a, b$, noting the combinatorial identity $C_{n}^{1}+C_{n}^{2}+\cdots+C_{n}^{r}+\cdots+C_{n}^{n}=2^{n}-1$, and using the binomial theorem and the $(2^{n}-1)$-element mean inequality, we get
$$\begin{aligned}
(a+b)^{n}= & a^{n}+C_{n}^{1} a^{n-1} b+C_{n}^{2} a^{n-2} b^{2}+\cdots+C_{n}^{r} ... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,318 |
Example 1 Let $a, b, c \in \mathbf{R}^{+}$. Prove:
$$\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b} \geqslant \frac{3}{2} .$$ | Prove: Let $a+b+c=s$. Then
$$\begin{array}{l}
\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b} \\
= \frac{a}{s-a}+\frac{b}{s-b}+\frac{c}{s-c} \\
= \frac{a-s+s}{s-a}+\frac{b-s+s}{s-b}+\frac{c-s+s}{s-c} \\
=-3+\left(1-\frac{a}{s}\right)^{-1}+\left(1-\frac{b}{s}\right)^{-1} \\
+\left(1-\frac{c}{s}\right)^{-1} \\
\geqslant-3+3\l... | \frac{3}{2} | Inequalities | proof | Yes | Yes | inequalities | false | 731,319 |
Example 2 Let $a, b, c \in \mathbf{R}^{+}$. Prove:
$$\frac{a^{2}}{b+c}+\frac{b^{2}}{c+a}+\frac{c^{2}}{a+b} \geqslant \frac{a+b+c}{2} .$$ | $$\begin{array}{l}
\frac{a^{2}}{b+c}+\frac{b^{2}}{c+a}+\frac{c^{2}}{a+b} \\
= \frac{a^{2}}{s-a}+\frac{b^{2}}{s-b}+\frac{c^{2}}{s-c} \\
= \frac{a^{2}-s^{2}+s^{2}}{s-a}+\frac{b^{2}-s^{2}+s^{2}}{s-b} \\
+\frac{c^{2}-s^{2}+s^{2}}{s-c} \\
=-(a+s)+\frac{s^{2}}{s-a}-(b+s)+\frac{s^{2}}{s-b} \\
-(c+s)+\frac{s^{2}}{s-c} \\
=-4 ... | \frac{a+b+c}{2} | Inequalities | proof | Yes | Yes | inequalities | false | 731,320 |
Example 4 If $a_{i}$ have the same sign, $a=\sum a_{i} \neq 0, n \geqslant 2$, $n \in \mathbf{N}$. Then
$$\sum \frac{a_{i}}{2 a-a_{i}} \geqslant \frac{n}{2 n-1}$$ | $\begin{array}{l}\text { Prove: } \sum \frac{a_{i}}{2 a-a_{i}}=\sum \frac{a_{1}-2 a+2 a}{2 a-a_{i}} \\ =-n+\sum\left(\frac{2 a-a_{i}}{2 a}\right)^{-1} \\ =-n+\sum\left(1-\frac{a_{i}}{2 a}\right)^{-1} \\ \geqslant-n+n \cdot\left(\frac{\sum\left(1-a_{i} / 2 a\right)}{n}\right)^{-1} \\ =-n+n \cdot \frac{n}{n-\sum a_{i} / ... | \frac{n}{2 n-1} | Inequalities | proof | Yes | Yes | inequalities | false | 731,321 |
Example 5 Given $0<a_{1}, a_{2}, \cdots, a_{n}<1, \sum a_{i}$ $=a$. Prove:
$$\sum \frac{a_{i}}{1-a_{i}} \geqslant \frac{n a}{n-a}$$
(shapiro) | $\begin{array}{l}\text { Prove: } \sum \frac{a_{i}}{1-a_{i}}=\sum \frac{a_{i}-1+1}{1-a_{i}} \\ =-n+\sum\left(1-a_{i}\right)^{-1} \\ \geqslant-n+n \cdot\left(\frac{\sum\left(1-a_{i}\right)}{n}\right)^{-1} \\ =-n+n \cdot \frac{n}{n-\sum a_{i}} \\ =-n+\frac{n^{2}}{n-a}=\frac{n a}{n-a} .\end{array}$ | \frac{n a}{n-a} | Inequalities | proof | Yes | Yes | inequalities | false | 731,322 |
Example 6 Let $x_{i} \in \mathbf{R}^{+}$, and $\sum a_{i}=a, n \in \mathbf{N}$, and $n \geqslant 2$. Prove:
$$\sum \frac{x_{i}}{a-x_{i}} \geqslant \frac{n}{n-1}$$ | $\begin{array}{l}\text { Prove: } \sum \frac{x_{i}}{a-x_{i}}=\sum \frac{x_{i}-a+a}{a-x_{i}} \\ =-n+\sum\left(1-\frac{x_{i}}{a}\right)^{-1} \\ \geqslant-n+n \cdot\left(\frac{\sum\left(1-x_{i} / a\right)}{n}\right)^{-1} \\ =-n+n \cdot \frac{n}{n-\sum x_{i} / a} \\ =-n+\frac{n^{2}}{n-1}=\frac{n}{n-1} .\end{array}$ | \frac{n}{n-1} | Inequalities | proof | Yes | Yes | inequalities | false | 731,323 |
Example 7 Let $a_{i} \in \mathbf{R}^{+}$, and all are less than $c, \sum a_{i}=$ $s$. Then
$$\sum \frac{a_{i}}{c-a} \geqslant \frac{n s}{n c-s} .(n \geqslant 2)$$ | $\begin{array}{l}\text { Prove: } \sum \frac{a_{i}}{c-a_{i}}=\sum \frac{a_{i}-c+c}{c-a_{i}} \\ =-n+\sum\left(1-\frac{a_{i}}{c}\right)^{-1} \\ \geqslant-n+n \cdot\left(\frac{\sum\left(1-a_{i} / c\right)}{n}\right)^{-1} \\ =-n+\frac{n^{2}}{n-\sum a_{i} / c} \\ =-n+\frac{n^{2}}{n-s / c} \\ =\frac{n s}{n c-s} .\end{array}$ | \frac{n s}{n c-s} | Inequalities | proof | Yes | Yes | inequalities | false | 731,324 |
Example 8 Let $\alpha_{i}(i=1,2, \cdots, n)$ all be acute angles, and satisfy $\sum \cos ^{2} \alpha_{i}=s$. Then
$$\sum \cot ^{2} \alpha_{i} \geqslant \frac{n s}{n-s}$$ | $\begin{array}{l}\text { Prove: } \sum \cot ^{2} \alpha_{i}=\sum \frac{\cos ^{2} \alpha_{i}}{1-\cos ^{2} \alpha_{i}} \\ =\sum \frac{\cos ^{2} \alpha_{i}-1+1}{1-\cos ^{2} \alpha_{i}} \\ =-n+\sum\left(1-\cos ^{2} \alpha_{i}\right)^{-1} \\ \geqslant-n+n \cdot\left(\frac{\sum\left(1-\cos ^{2} \alpha_{i}\right)}{n}\right)^{... | \frac{n s}{n-s} | Inequalities | proof | Yes | Yes | inequalities | false | 731,325 |
1 Inequality of sequences derived from the functional inequality $\ln (1+x)>x-\frac{x^2}{2}(x>0)$ | Prove that if $f(x)=\ln (1+x)-x+\frac{x^{2}}{2}(x>0)$, then $f^{\prime}(x)=\frac{1}{1+x}-1+x=\frac{x^{2}}{1+x}>0$,
thus when $x>0$, we have $f(x)>f(0)=0$, i.e., $\ln (1+$ $x)>x-\frac{x^{2}}{2}$ | \ln (1+x)>x-\frac{x^2}{2} | Inequalities | proof | Yes | Yes | inequalities | false | 731,326 |
Example 1 (2009 National High School Mathematics League Additional Problem Adapted) If $n \in \mathbf{N}^{*}$, prove:
$$\frac{1}{1^{2}+1}+\frac{2}{2^{2}+1}+\frac{3}{3^{2}+1}+\cdots+\frac{n}{n^{2}+1} \leqslant \frac{1}{2}+\ln n .$$ | Proof: Let the general term of the sequence $\left\{a_{n}\right\}$ be $a_{n}=\frac{1}{1^{2}+1}+\frac{2}{2^{2}+1}+\frac{3}{3^{2}+1}+\cdots+\frac{n}{n^{2}+1}-\frac{1}{2}-\ln n$, then
$$\begin{array}{l}
a_{n+1}=\frac{1}{1^{2}+1}+\frac{2}{2^{2}+1}+\frac{3}{3^{2}+1}+\cdots+\frac{n}{n^{2}+1}+ \\
\frac{n+1}{(n+1)^{2}+1}-\frac... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,327 |
Example 4 If $n \in \mathbf{N}^{*}$, prove: $1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\cdots+$ $\frac{1}{n} \geqslant \ln \frac{\mathrm{e}^{n}}{n!}$. | Proof: Let the general term of the sequence $\left\{a_{n}\right\}$ be $a_{n}=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\ldots+\frac{1}{n}-\ln \frac{\mathrm{e}^{n}}{n!}$, then
$$a_{n+1}=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\ldots+\frac{1}{n}+\frac{1}{n+1}-\ln$$
$$\frac{\mathrm{e}^{n+1}}{(n+1)!}$$
Thus, $a_{n+1}-a_{n}=\fra... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,332 |
Example 5 If $n \in \mathbf{N}^{*}$ and $n \geqslant 2$, prove: $\frac{\ln 2}{2^{2}}+\frac{\ln 3}{3^{2}}+\frac{\ln 4}{4^{2}}$ $+\cdots+\frac{\ln n}{n^{2}}<\frac{2 n-3}{4}$ | Proof: Let the general term of the sequence $\{a_n\}$ be $a_n = \frac{\ln 2}{2^2} + \frac{\ln 3}{3^2} + \frac{\ln 4}{4^2} + \cdots + \frac{\ln n}{n^2} - \frac{2n-3}{4}$. Then,
$$a_{n+1} = \frac{\ln 2}{2^2} + \frac{\ln 3}{3^2} + \frac{\ln 4}{4^2} + \cdots + \frac{\ln n}{n^2} + \frac{\ln (n+1)}{(n+1)^2} -$$
$\frac{2(n+1)... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,333 |
Example 6 If $n \in \mathbf{N}^{*}$ and $n \geqslant 2$, prove: $\frac{\mathrm{e}^{2^{2}}}{2^{2}}+\frac{\mathrm{e}^{3^{2}}}{3^{2}}+\frac{\mathrm{e}^{4^{2}}}{4^{2}}$ $+\cdots+\frac{\mathrm{e}^{n^{2}}}{n^{2}}>\frac{2 n^{2}+n-3}{2(n+1)}$ | Proof: Let the general term of the sequence $\{a_n\}$ be
$$a_n = \frac{e^{2^2}}{2^2} + \frac{e^{3^2}}{3^2} + \frac{e^{4^2}}{4^2} + \ldots + \frac{e^{n^2}}{n^2} - \frac{2n^2 + n - 3}{2(n+1)}, \text{ then }$$
$$a_{n+1} = \frac{e^{2^2}}{2^2} + \frac{e^{3^2}}{3^2} + \frac{e^{4^2}}{4^2} + \ldots + \frac{e^{n^2}}{n^2} + \fr... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,335 |
Question 2-1 Given the sequence $\left\{a_{n}\right\}$ with the first term $a_{1}=\frac{3}{5}, a_{n+1}$
$$=\frac{3 a_{n}}{2 a_{n}+1}, n=1,2,3, \cdots$$
(1) Find the general term formula for $\left\{a_{n}\right\}$;
(2) Prove: For any $x>0, a_{n} \geqslant \frac{1}{1+x}-$
$$\frac{1}{(1+x)^{2}}\left(\frac{2}{3^{n}}-x\righ... | Prove by mathematical induction.
(1) When $n=1$, we have $a_{1}=\frac{3}{5}>\frac{1}{2}$, so the inequality holds at this point.
(2) Assume that when $n=k$, $S_{k}=a_{1}+a_{2}+\cdots+a_{k}>\frac{k^{2}}{k+1}$ holds.
Then, when $n=k+1$,
$$\begin{aligned}
S_{k+1} & =a_{1}+a_{2}+\cdots+a_{k}+a_{k+1} \\
& >\frac{k^{2}}{k+1... | proof | Algebra | proof | Yes | Yes | inequalities | false | 731,337 |
Question 2-2 Given the function $f(x)=\frac{1}{3^{x}+2}$, for any positive integer $n$, prove:
$$f(1)+f(2)+\cdots+f(n)<\frac{1}{2}\left(1-\frac{1}{n+1}\right)$$ | To prove the inequality is equivalent to
$$f(1)+f(2)+\cdots+f(n) < \frac{1}{2}\left(1-\frac{1}{n+1}\right)$$
we need to show that
$$f(k) < \frac{1}{2}\left(\frac{k}{k+1} - \frac{k-1}{k}\right) \quad \text{for} \quad k \geq 3$$
This is equivalent to proving
$$3^k + 2 > 2k^2 + 2k$$
We can prove this by mathematical indu... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,338 |
Question 3-2 Given $x, y>0, n \in \mathrm{N}^{*}$, prove:
$$\frac{x^{n}}{1+x^{2}}+\frac{y^{n}}{1+y^{2}} \leqslant \frac{x^{n}+y^{n}}{1+x y}$$ | $$\begin{array}{l}
(1+x^{2})(1+y^{2})=1+x^{2}+y^{2}+x^{2} y^{2} \\
\geqslant 1+2 x y+x^{2} y^{2}=(1+x y)^{2}, \\
\text { i.e., }(1+x^{2})(1+y^{2}) \geqslant(1+x y)^{2} . \\
x^{n-1} y+x y^{n-1} \leqslant \frac{1}{n}\left[(n-1) x^{n}+y^{n}\right]+\frac{1}{n}\left[x^{n}+\right. \\
\left.(n-1) y^{n}\right]=x^{n}+y^{n},
\en... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,339 |
Question 4-2 If $a, b, c, d \in\left[\frac{1}{\sqrt{2}}, \sqrt{2}\right]$, prove: $a b + b c + c d + d a \geqslant \frac{4}{5}\left(a^{2}+b^{2}+c^{2}+d^{2}\right)$. | It is easy to prove that the function $y=x+\frac{1}{x}$ is a decreasing function on $\left[\frac{1}{2}, 1\right]$ and an increasing function on $[1,2]$. Therefore, when $x=$ $\frac{1}{2}$ or $x=2$, the maximum value of the function $y=x+\frac{1}{x}$ on $\left[\frac{1}{2}, 2\right]$ is $y_{\max }=\frac{5}{2}$, i.e.,
$$x... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,340 |
For any real numbers $x, y, z$, we have
$$\left(x^{2}+1\right)\left(y^{2}+1\right)\left(z^{2}+1\right) \geqslant \frac{3}{4}(x+y+z)^{2}$$
Equality holds in inequality (4) if and only if $x=y=z= \pm \frac{1}{\sqrt{2}}$.
This inequality is concise and elegant, and by constructing a quadratic function, using the techniq... | To prove the inequality (4), it suffices to prove
$$3(x+y+z)^{2}-4\left(x^{2}+1\right)\left(y^{2}+1\right)\left(z^{2}+1\right) \leqslant 0$$
Construct the quadratic function
$$f(t)=\left(x^{2}+1\right) t^{2}-\sqrt{3}(x+y+z) t+\left(y^{2}+1\right)\left(z^{2}+1\right)$$.
Since \(x^{2}+1>0\),
$$\begin{aligned}
f(t)= & \... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,341 |
Given $x, y, z \in(0,+\infty)$, and $x^{2}+y^{2}+z^{2}$ $=1$, find the range of the function $f=x+y+z-x y z$. | Solution: First, find the optimal upper bound of $f$.
Let $x=\frac{\sqrt{3 \lambda}}{3}+\alpha, y=\frac{\sqrt{3 \lambda}}{3}+\beta, z=\frac{\sqrt{3 \lambda}}{3}+\gamma$,
then $\left(\frac{\sqrt{3 \lambda}}{3}+\alpha\right)^{2}+\left(\frac{\sqrt{3 \lambda}}{3}+\beta\right)^{2}+\left(\frac{\sqrt{3 \lambda}}{3}+\gamma\rig... | \left(1, \frac{8 \sqrt{3}}{9}\right] | Algebra | math-word-problem | Yes | Yes | inequalities | false | 731,342 |
Example: Points $P$, $Q$, $M$, and $N$ are all on the ellipse $x^{2}+\frac{y^{2}}{2}=1$, and $F$ is the focus of the ellipse on the positive $y$-axis. It is known that $\overrightarrow{P F}$ is collinear with $\overrightarrow{P Q}$, $\overrightarrow{M F}$ is collinear with $\overrightarrow{F N}$, and $\overrightarrow{P... | Analysis: To find the extremum of the area, it is only necessary to first derive the analytical expression of the area and then solve it using functional relationships or inequalities.
Solution: Given that $\overrightarrow{P F}$ and $\overrightarrow{P Q}$, $\overrightarrow{M F}$ and $\overrightarrow{F N}$ are collinea... | 2 | Geometry | math-word-problem | Yes | Yes | inequalities | false | 731,343 |
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