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class | __index_level_0__ int64 0 742k |
|---|---|---|---|---|---|---|---|---|---|
246. Prove that the equation $\frac{a^{2}}{x}+\frac{b^{2}}{x-1}=1$, where $a$ and $b$ are real numbers, not both zero, has real roots. | Given. Assuming that $x \neq 1$, after transformations we will have:
$$
x^{2}-x\left(a^{2}+b^{2}+1\right)+a^{2}=0
$$
from which we establish that $D>0$. | proof | Algebra | proof | Yes | Yes | olympiads | false | 42,427 |
247. Prove that the equation
$$
\left(a^{2}+b^{2}+c^{2}\right) x^{2}+2(a+b+c) x+3=0
$$
cannot have real roots if $a, b$, and $c$ are not equal to each other. | Indication. $D=(a+b+c)^{2}-3\left(a^{2}+b^{2}+c^{2}\right)=$ $=-\left[(a-b)^{2}+(a-c)^{2}+(b-c)^{2}\right]<0$, since
$$
\begin{gathered}
a \neq b, a \neq c \\
b \neq c
\end{gathered}
$$
Therefore, the given equation under the conditions of the problem has imaginary roots. | proof | Algebra | proof | Yes | Yes | olympiads | false | 42,428 |
248. Prove that the roots of the equation
$$
\left(a^{2}+b^{2}+c^{2}\right) x^{2}-2 x\left(a a_{1}+b b_{1}+c c_{1}\right)+a_{1}^{2}+b_{1}^{2}+c_{1}^{2}=0
$$
are imaginary, except in the case when
$$
a: a_{1}=b: b_{1}=c: c_{1}
$$ | Instruction. $D=-\left[\left(a b_{1}-a_{1} b\right)^{2}+\left(a c_{1}-a_{1} c\right)^{2}+\right.$ $\left.+\left(b c_{1}-b_{1} c\right)^{2}\right] \leqslant 0$
$249 *$. Prove that if the quadratic equation $x^{2}+p x+$ $+q=0$, where $p$ and $q$ are integers, has rational roots, then these roots can only be integers, no... | proof | Algebra | proof | Yes | Yes | olympiads | false | 42,429 |
250*. Prove that for the cubic equation $a x^{3}+b x^{2}+$ $+c x+d=0$, there exists the following relationship between the roots and the coefficients:
$$
\begin{gathered}
x_{1}+x_{2}+x_{3}=-\frac{b}{a}, x_{1} \cdot x_{2}+x_{1} \cdot x_{3}+x_{2} \cdot x_{3}=\frac{c}{a} \\
x_{1} \cdot x_{2} \cdot x_{3}=-\frac{d}{a}
\end... | Given that if $x_{1}, x_{2}, x_{3}$ are the roots of the equation
$$
a x^{3}+b x^{2}+c x+d=0
$$
or
$$
x^{3}+\frac{b}{a} x^{2}+\frac{c}{a} x+\frac{d}{a}=0
$$
then
$$
\left(x-x_{1}\right)\left(x-x_{2}\right)\left(x-x_{3}\right)=0
$$
68
or
$$
\begin{gathered}
x^{3}+x^{2}\left(-x_{1}-x_{2}-x_{3}\right)+x\left(x_{1} ... | proof | Algebra | proof | Yes | Yes | olympiads | false | 42,430 |
252. Three different numbers $x_{1}, x_{2}, x_{3}$ satisfy the relations:
\[
\begin{aligned}
& x_{1}^{3}+x_{1} p+q=0 \\
& x_{2}^{3}+x_{2} p+q=0 \\
& x_{3}^{3}+x_{3} p+q=0
\end{aligned}
\]
Prove the validity of the equality:
\[
x_{1}^{3}+x_{2}^{3}+x_{3}^{3}=3 x_{1} \cdot x_{2} \cdot x_{3}
\] | Instruction. The conditions of the problem show that $x_{1}, x_{2}, x_{3}$ are the roots of the equation $x^{3}+p x+q=0$, which means $x_{1}+x_{2}+x_{3}=0$, but under this condition
$$
x_{1}^{3}+x_{2}^{3}+x_{3}^{3}=3 x_{1} \cdot x_{2} \cdot x_{3}
$$
(see problem 113). | proof | Algebra | proof | Yes | Yes | olympiads | false | 42,432 |
253. Find the necessary and sufficient condition for the equation
$$
x^{3}+a x^{2}+b x+c=0
$$
to be reduced to the form
$$
\left(x^{2}+m\right) \cdot(x+n)=0
$$
and indicate the simplest way to obtain the roots of the equation.
* See note on page 15. | According to the condition of the problem, the identity should hold:
$$
x^{3}+a x^{2}+b x+c=\left(x^{2}+m\right)(x+n)
$$
or
$$
x^{3}+a x^{2}+b x+c=x^{3}+m x+n x^{2}+m n
$$
from which
$$
m=b, n=a, m n=c
$$
or \( c=a b \) - this is the necessary condition. We will prove that this condition is also sufficient: if \(... | x_{1}=-,x_{2,3}=\\sqrt{-b} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 42,433 |
254*. Prove that if in the cubic equation
$$
x^{3}+p x^{2}+q x+n=0
$$
there is a relationship between the coefficients $q^{2}=2 n p$, then the sum of the fourth powers of the roots is equal to the square of the sum of the squares of the roots. | Proof. Let $x_{1}, x_{2}, x_{3}$ be the roots of the cubic equation, then, according to problem 250,
$$
\begin{gathered}
x_{1}+x_{2}+x_{3}=-p \\
x_{1} x_{2}+x_{1} x_{3}+x_{2} x_{3}=q \\
x_{1} \cdot x_{2} \cdot x_{3}=-n
\end{gathered}
$$
Squaring equality (1), we get:
$$
\left(x_{1}+x_{2}+x_{3}\right)^{2}=p^{2}
$$
o... | proof | Algebra | proof | Yes | Yes | olympiads | false | 42,434 |
256. Prove that if an equation of degree $n$ with integer coefficients has integer roots, then they are divisors of the constant term | Proof. Let $x_{1}$ be a root of the equation
$$
a_{0} x^{n}+a_{1} x^{n-1}+\ldots+a_{n-1} x+a_{n}=0
$$
and $x_{1}$ is an integer. Then
$$
a_{0} x_{1}^{n}+a_{1} x_{1}^{n-1}+\ldots+a_{n-1} x_{1}=-a_{n}
$$
or
$$
x_{1}\left(a_{0} x_{1}^{n-1}+a_{1} x_{1}^{n-2}+\ldots+a_{n-1}\right)=-a_{n}
$$
from which
$$
a_{0} x_{1}^... | proof | Algebra | proof | Yes | Yes | olympiads | false | 42,435 |
257. Prove that if a reciprocal equation has a root $x_{1}$, then it also has a root $\frac{1}{x_{1}}$. | Instruction. A reciprocal equation has the form:
$$
a x^{n}+b x^{n-1}+c x^{n-2}+\ldots+c x^{2}+b x+a=0
$$
i.e., the coefficients of the terms equally distant from the beginning and end of the equation are the same. Let $x_{1}$ be a root of this equation, then, substituting $\frac{1}{x_{1}}$ for $x$ in the left side o... | proof | Algebra | proof | Yes | Yes | olympiads | false | 42,436 |
258. Prove that a reciprocal equation of odd degree has $(-1)$ as a root. | Let the given reciprocal equation be of the form:
$$
a x^{2 k+1}+b x^{2 k}+\ldots+b x+a=0
$$
Then
$$
a\left(x^{2 k+1}+1\right)+b\left(x^{2 k-1}+1\right) x+\ldots+n x^{k}(x+1)=0
$$
Each term of this sum contains as a factor the sum of equal odd powers of $x$ and 1, meaning that each term and the entire sum is divisi... | proof | Algebra | proof | Yes | Yes | olympiads | false | 42,437 |
259. Show that if $|a|<b$, then $-b<a<b$. | Given the condition of the problem, $b$ is a positive number, and the number $a$ can be positive, negative, or zero. If $a=0$, then $-b<0<a<b$. If $a>0$, then $|a|=a$ and $a<b$, but $-b<a$, since any negative number is less than a positive number. Therefore, $-b<a<b$. If $a<0$, then $a<b$, since a negative number is le... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 42,438 |
260. Show that if $-b<a<b$, then $|a|<b$. | Instruction. The reasoning is similar to that carried out in the previous problem, one only has to keep in mind that from the condition of the problem $(-b<b)$ it is obvious that $b$ is a positive number. | proof | Inequalities | proof | Yes | Yes | olympiads | false | 42,439 |
261. Show that $|a+b| \leqslant|a|+|b|$. | Let $a$ and $b$ be any real numbers. Then
$$
-|a| \leqslant a \leqslant|a| ;-|b| \leqslant b \leqslant|b|
$$
Adding these inequalities term by term, we will have:
$$
-(|a|+|b|) \leqslant a+b \leqslant|a|+|b|
$$
or (problem 260)
$$
|a+b| \leqslant|a|+|b|
$$ | proof | Inequalities | proof | Yes | Yes | olympiads | false | 42,440 |
263. Prove the inequality: $a^{2}+b^{2}+c^{2}+3 \geqslant 2(a+b+c)$. | Instruction. Let's form the difference between the parts of the inequality. We will have:
$$
\begin{gathered}
a^{2}+b^{2}+c^{2}+3-2 a-2 b-2 c=(a-1)^{2}+ \\
+(b-1)^{2}+(c-1)^{2} \geqslant 0
\end{gathered}
$$
since the sum of squares of real numbers is always positive or equal to 0. | proof | Inequalities | proof | Yes | Yes | olympiads | false | 42,442 |
265. Prove that if $a, b, c$ are positive numbers, then
$$
a^{2}(b+c)+a\left(b^{2}+c^{2}-b c\right)>0
$$ | Instruction. Transform the expression $b^{2}+c^{2}-b c$ into the form: $(b-c)^{2}+b c$, then
$$
\left.a^{2}(b+c)+a\left[(b-c)^{2}+b c\right)\right]
$$
will always be a positive number if $a, b, c$ are positive. | proof | Inequalities | proof | Yes | Yes | olympiads | false | 42,444 |
266. Prove that $1+2 x^{4} \geqslant x^{2}+2 x^{3}$. | Instruction. Let's form the difference between the parts of the given inequality and determine its sign, for which we will transform this difference into an obvious inequality:
$$
1+2 x^{4}-x^{2}-2 x^{3}=\left(x^{2}-x\right)^{2}+\left(x^{2}-1\right)^{2}
$$
The number is non-negative, hence
$$
1+2 x^{4} \geqslant x^{... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 42,445 |
267. Prove that for positive $a, b, c$
$$
a b(a+b)+b c(b+c)+a c(a+c) \geqslant 6 a b c
$$ | Instruction. Form the difference between the parts of the inequality to be proved, transform it to the form:
$$
b(a-c)^{2}+a(b-c)^{2}+c(b-a)^{2}
$$
Establish that the difference is non-negative, meaning the left side of the inequality is not less than the right side. Determine when they are equal. | proof | Inequalities | proof | Yes | Yes | olympiads | false | 42,446 |
268. Prove that
$$
3\left(1+a^{2}+a^{4}\right) \geqslant\left(1+a+a^{2}\right)^{2} .
$$ | Instruction. $\quad 3\left(1+a^{2}+a^{4}\right)-\left(1+a+a^{2}\right)^{2}=\left(a^{2}-a\right)^{2}+$ $+\left(a^{2}-1\right)^{2}+(a-1)^{2}-$ a non-negative number. | proof | Inequalities | proof | Yes | Yes | olympiads | false | 42,447 |
269. Prove the inequality:
$$
a^{4}+b^{4} \geqslant a^{3} b+a b^{3} ; \quad a>0 ; \quad b>0 .
$$ | Instruction. $a^{4}+b^{4}-a^{3} b-a b^{3}=a^{3}(a-b)-b^{3}(a-b)=$ $=(a-b)^{2}\left(a^{2}+a b+b^{2}\right)-$ the product is non-negative. Therefore, $a^{4}+b^{4} \geqslant a^{3} b+a b^{3}$, and the equality sign holds when $a=b$. | proof | Inequalities | proof | Yes | Yes | olympiads | false | 42,448 |
270. Prove that for $n$, greater than $-1$, $n^{3}+1$ is greater than $n^{2}+n$; for $n$, less than $-1$, $n^{3}+1$ is less than $n^{2}+n$. | Instruction. $\left(n^{3}+1\right)-\left(n^{2}+n\right)=(n+1)(n-1)^{2}$. For $n$, less than (-1), the given expression is negative, meaning $n^{3}+1$ is less than $n^{2}+n$. For $n$, greater than (-1), the obtained difference is positive, i.e., $n^{3}+1$ is greater than $n^{2}+n$. | proof | Inequalities | proof | Yes | Yes | olympiads | false | 42,449 |
271. Prove the inequality:
$$
a^{4}+2 a^{3} b+2 a b^{3}+b^{4} \geqslant 6 a^{2} b^{2}
$$
where $a$ and $b$ have the same sign. | Instruction. $\quad\left(a^{4}+2 a^{3} b+2 a b^{3}+b^{4}\right)-6 a^{2} b^{2}=\left(a^{4}-2 a^{2} b^{2}+\right.$ $\left.+b^{4}\right)+2 a b\left(a^{2}-2 a b+b^{2}\right)=(a-b)^{2}\left(a^{2}+b^{2}+4 a b\right)$, under the conditions of the problem - a non-negative quantity. | proof | Inequalities | proof | Yes | Yes | olympiads | false | 42,450 |
272. Prove that if $a<b<c$, then
$$
a^{2} b+b^{2} c+c^{2} a<a^{2} c+b^{2} a+c^{2} b
$$ | Instruction. Let's form the difference
$$
a^{2} b+b^{2} c+c^{2} a-a^{2} c-b^{2} a-c^{2} b
$$
To transform it into a product, we add and subtract $a b c$ :
$$
\begin{gathered}
a^{2}(b-c)-a c(b-c)-a b(b-c)+b c(b-c)= \\
=(b-c)(a-b)(a-c)
\end{gathered}
$$
We will get a negative value, since $a<b<c$. | proof | Inequalities | proof | Yes | Yes | olympiads | false | 42,451 |
273. Prove that if $a+b$ is a positive number, then $\frac{a}{b^{2}}+\frac{b}{a^{2}}$ is greater than $\frac{1}{a}+\frac{1}{b}$. | Instruction. $\left(\frac{a}{b^{2}}+\frac{b}{a^{2}}\right)-\left(\frac{1}{a}+\frac{1}{b}\right)=\frac{a^{3}+b^{3}-a b^{2}-a^{2} b}{a^{2} b b^{2}}=$ $=\frac{(a+b)(a-b)^{2}}{a^{2} b^{2}}$. The obtained expression is non-negative when $a+b>0$. | proof | Inequalities | proof | Yes | Yes | olympiads | false | 42,452 |
274. Prove that if $a$ and $b$ are positive numbers, then
$$
\frac{a^{3}+b^{3}}{2} \geqslant\left(\frac{a+b}{2}\right)^{3}
$$ | Instruction. Transform the difference between the parts of the inequality to the form:
$$
\frac{3(a+b) \cdot(a-b)^{2}}{8}
$$ | proof | Inequalities | proof | Yes | Yes | olympiads | false | 42,453 |
275. Prove that for positive $a$ and $b$ and $a \geqslant b$
$$
\frac{a^{4}+b^{4}}{2} \geqslant\left(\frac{a+b}{2}\right)^{4}
$$ | Instruction. Transform the difference between the left and right sides of the inequality to be proved into the form:
$$
\frac{(a-b)^{2}\left[3(a+b)^{2}+4\left(a^{2}+a b+b^{2}\right)\right]}{16}
$$
Under the condition of the problem, this expression is non-negative. | proof | Inequalities | proof | Yes | Yes | olympiads | false | 42,454 |
277. Prove that for positive $a, b, c$, the following inequality holds:
$$
(a+b+c)^{3}-\left(a^{3}+b^{3}+c^{3}\right)>(a+b)(b+c)(a+c)
$$ | Instruction. Factorize the left side of the inequality (see problem 107), form the difference between the left and right sides of the inequality. The obtained expression
$$
2(a+b)(b+c)(a+c)
$$
under the conditions of the problem is a positive number. | proof | Inequalities | proof | Yes | Yes | olympiads | false | 42,456 |
278*. Prove that for any positive values of $x$ and $y$
$$
x^{5}+y^{5} \geqslant x^{4} y+x y^{4}
$$ | Instruction. Transform the difference between the right and left sides of the inequality to be proved into the form:
$$
(x-y)^{2}\left(x^{2}+y^{2}\right)(x+y)
$$
76 | proof | Inequalities | proof | Yes | Yes | olympiads | false | 42,457 |
280. Prove that if $a, b, c$ are positive integers, then
$$
a b + b c + a c \leqslant 3 a b c
$$ | Instruction. Given the conditions of the inequality $a b c \geqslant a b$, $a b c \geqslant b c$, $a b c \geqslant a c$, they are obvious. Adding them term by term, we obtain the required inequality. | proof | Inequalities | proof | Yes | Yes | olympiads | false | 42,459 |
281. Prove that the sum of the squares of two numbers is not less than (greater than or equal to) their double product. | Instruction. Taking the obvious inequality $(a-b)^{2} \geqslant 0$, from which
$$
a^{2}-2 a b+b^{2} \geqslant 0 \text { and } a^{2}+b^{2} \geqslant 2 a b
$$
The equality sign holds when $a=b$. | ^{2}+b^{2}\geqslant2 | Inequalities | proof | Yes | Yes | olympiads | false | 42,460 |
282. Prove that for $a$ positive $\frac{2 a}{1+a^{2}} \leqslant 1$. | Instruction. $(1-a)^{2} \geqslant 0$, hence, $1+a^{2} \geqslant 2 a$. | proof | Inequalities | proof | Yes | Yes | olympiads | false | 42,461 |
283. Prove that if $a: b=b: c$, then $a^{2}+c^{2} \geqslant 2 b^{2}$. | From the condition of the problem, it follows that $b^{2}=a c$, therefore, $2 b^{2}=2 a c$ and, taking $(a-c)^{2} \geqslant 0$, we get:
$$
a^{2}+c^{2} \geqslant 2 b^{2}
$$ | proof | Inequalities | proof | Yes | Yes | olympiads | false | 42,462 |
284. Prove that if $k>0$, then $k+\frac{1}{k} \geqslant 2$. | Instruction. Consider the obvious inequality $(k-1)^{2} \geqslant 0 . * *$[^7] | proof | Inequalities | proof | Yes | Yes | olympiads | false | 42,463 |
285. Prove that the sum of a given fraction with positive terms and its reciprocal is not less than two. | Instruction. Consider the inequality $a^{2}+b^{2} \geqslant 2 a b$, divide both of its sides by the positive number $a b$:
$$
\frac{a}{b}+\frac{b}{a} \geqslant 2
$$
This derivation significantly simplifies the proof of inequalities in problems $286-288$. | \frac{}{b}+\frac{b}{}\geqslant2 | Inequalities | proof | Yes | Yes | olympiads | false | 42,464 |
286. Prove that if $a, b, c$ are positive numbers, then
$$
\frac{a b(a+b)+a c(a+c)+b c(b+c)}{a b c} \geqslant 6
$$ | Instruction. Transform the left side of the inequality to be proven:
$$
\begin{aligned}
& \frac{a b(a+b)+a c(a+c)+b c(b+c)}{a b c}=\frac{a}{c}+\frac{b}{c}+\frac{a}{b}+\frac{c}{b}+\frac{b}{a}+\frac{c}{a}= \\
& =\left(\frac{a}{c}+\frac{c}{a}\right)+\left(\frac{c}{b}+\frac{b}{c}\right)+\left(\frac{a}{b}+\frac{b}{a}\right... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 42,465 |
287. Prove that if $a, b, c$ are positive and unequal numbers, then $(a+b+c)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)>9$. | $$
\begin{gathered}
\frac{(a+b+c) \cdot(b c+a c+a b)}{a b c}=\frac{3 a b c+a^{2} b+a^{2} c+b^{2} c+b c^{2}+a c^{2}+a b^{2}}{a b c}= \\
=3+\left(\frac{c}{a}+\frac{a}{c}\right)+\left(\frac{a}{b}+\frac{b}{a}\right)+\left(\frac{c}{b}+\frac{b}{c}\right)>9
\end{gathered}
$$
Instruction. Transform the left side of the inequa... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 42,466 |
288. Prove that if $a, b, c$ are positive and unequal numbers, then
$$
\frac{b c}{a}+\frac{a c}{b}+\frac{a b}{c}>a+b+c
$$ | Consider the expression $2\left(\frac{b c}{a}+\frac{a c}{b}+\frac{a b}{c}\right)=$
$$
=b\left(\frac{c}{a}+\frac{a}{c}\right)+c\left(\frac{b}{a}+\frac{a}{b}\right)+a\left(\frac{c}{b}+\frac{b}{c}\right)>2(a+b+c)
$$
from which
$$
\frac{b c}{a}+\frac{a c}{b}+\frac{a b}{c}>a+b+c
$$ | proof | Inequalities | proof | Yes | Yes | olympiads | false | 42,467 |
289. Prove that for real $a, b, c$, the inequality holds:
$$
a^{2}+b^{2}+c^{2} \geqslant a b+a c+b c
$$
78 | Instruction. Take the obvious inequalities:
$$
a^{2}+b^{2} \geqslant 2 a b, a^{2}+c^{2} \geqslant 2 a c, b^{2}+c^{2} \geqslant 2 b c
$$
add them term by term. | proof | Inequalities | proof | Yes | Yes | olympiads | false | 42,468 |
290. Prove that
$(a+b-c)^{2}+(a+c-b)^{2}+(b+c-a)^{2} \geqslant a b+b c+a c$. | Instruction. Transform the left side of the inequality:
$$
\begin{gathered}
(a+b-c)^{2}+(a+c-b)^{2}+(b+c-a)^{2}= \\
=3\left(a^{2}+b^{2}+c^{2}\right)-2(a b+b c+a c) \geqslant a b+b c+a c,
\end{gathered}
$$
since according to problem 289
$$
a^{2}+b^{2}+c^{2} \geqslant a b+b c+a c
$$ | proof | Inequalities | proof | Yes | Yes | olympiads | false | 42,469 |
291. Prove the inequality:
$$
\left(\frac{a+b+c}{3}\right)^{2} \geqslant \frac{a b+a c+b c}{3}
$$ | Indication. We will use the inequality from problem 289
$$
a^{2}+b^{2}+c^{2} \geqslant a b+a c+b c
$$
Add to both sides of the taken inequality
$$
2(a b+a c+b c)
$$
we get:
$$
(a+b+c)^{2} \geqslant 3(a b+a c+b c)
$$
from which
$$
\left(\frac{a+b+c}{3}\right)^{2} \geqslant \frac{a b+a c+b c}{3}
$$ | proof | Inequalities | proof | Yes | Yes | olympiads | false | 42,470 |
292. Prove that for positive $\boldsymbol{x}, \boldsymbol{y}, \boldsymbol{z}$, satisfying the condition: $x+y+z=1$, the following inequality holds:
$$
x^{2}+y^{2}+z^{2} \geqslant \frac{1}{3}
$$ | Indication. We will use the inequality from problem 289 and the condition of this problem: $x+y+z=1$.
$$
(x+y+z)^{2}=x^{2}+y^{2}+z^{2}+2(x y+y z+x z)=1
$$
or
$$
x^{2}+y^{2}+z^{2}=1-2(x y+y z+x z) \geqslant 1-2\left(x^{2}+y^{2}+z^{2}\right)
$$
from which
$$
3\left(x^{2}+y^{2}+z^{2}\right) \geqslant 1 \text { or } x... | x^{2}+y^{2}+z^{2}\geqslant\frac{1}{3} | Inequalities | proof | Yes | Yes | olympiads | false | 42,471 |
293*. Prove that for positive $a, b, c$ the inequality holds: $a b c \geqslant(b+c-a)(a+c-b)(a+b-c)$. | Instruction. Let's take the obvious inequalities:
$$
a^{2} \geqslant a^{2}-(b-c)^{2}, b^{2} \geqslant b^{2}-(a-c)^{2}, c^{2} \geqslant c^{2}-(a-b)^{2}
$$
Multiplying these inequalities term by term, we will have:
$$
\begin{gathered}
a^{2} b^{2} c^{2} \geqslant(a-b+c)(a+b-c)(b+a-c)(b-a+c)(c+a-b) \times \\
\times(c-a+... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 42,472 |
294*. Prove the inequality:
$$
\left(a_{1}+a_{2}+\ldots+a_{n}\right)^{2} \leqslant n\left(a_{1}^{2}+a_{2}^{2}+\ldots+a_{n}^{2}\right)
$$ | Instruction. Let us use the obvious inequality:
$$
\begin{gathered}
\left(a_{1}-a_{2}\right)^{2}+\left(a_{1}-a_{3}\right)^{2}+\ldots+\left(a_{1}-a_{n}\right)^{2}+\ldots+ \\
+\left(a_{n-1}-a_{n}\right)^{2} \geqslant 0
\end{gathered}
$$
transforming which, we get:
$$
(n-1)\left(a_{1}^{2}+a_{2}^{2}+\ldots+a_{n}^{2}\rig... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 42,473 |
295*. Prove that for any positive $a$, greater than 1, the following inequality holds:
$$
\frac{1+a^{2}+a^{4}+\ldots+a^{2 n}}{a+a^{3}+\ldots+a^{2 n-1}}>\frac{n+1}{n}
$$ | Instruction. Let's take the obvious inequalities:
$$
\begin{aligned}
& a^{2 n}(a-1)>a-1 \\
& a^{2 n-1}\left(a^{2}-1\right)>a^{2}-1 \\
& \cdot \cdot \cdot \cdot \cdot \cdot \cdot \\
& a^{n+1}\left(a^{n}-1\right)>a^{n}-1
\end{aligned}
$$
Adding the taken inequalities term by term, we get:
$$
n a^{2 n+1}-\left(a^{2 n}+... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 42,474 |
296. Prove that the arithmetic mean of two non-negative numbers is not less than their geometric mean. | Instruction. Let's take the obvious inequality: $(a-b)^{2} \geqslant 0$, from which
$$
a^{2}+b^{2} \geqslant 2 a b, \text { or } a^{2}+b^{2}+2 a b \geqslant 4 a b,
$$
or
$$
(a+b)^{2} \geqslant 4 a b, \text { or } a+b \geqslant 2 \sqrt{a b}
$$
or
$$
\frac{a+b}{2} \geqslant \sqrt{a b}
$$
Note. For the proof, the di... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 42,475 |
297. Prove the inequality:
$$
a+b+c \geqslant \sqrt{ab}+\sqrt{bc}+\sqrt{ac}
$$
if $a, b, c$ are positive numbers. | Instruction. We have:
$$
a+b \geqslant 2 \sqrt{a b}, \quad b+c \geqslant 2 \sqrt{b c}, \quad a+c \geqslant 2 \sqrt{a c}
$$
Add these inequalities and reduce by 2.
6 i. v. Baranova and S. E. Lyapin
81 | proof | Inequalities | proof | Yes | Yes | olympiads | false | 42,476 |
298. Prove that for positive $a, b, c$, the following inequality holds:
$$
(a+b)(b+c)(c+a) \geqslant 8 a b c
$$ | Instruction. We have:
$$
a+b \geqslant 2 \sqrt{a b}, \quad b+c \geqslant 2 \sqrt{b c}, \quad a+c \geqslant 2 \sqrt{a c}.
$$
By multiplying these inequalities term by term, we obtain the inequality:
$$
(a+b)(b+c)(c+a) \geqslant 8 a b c
$$ | proof | Inequalities | proof | Yes | Yes | olympiads | false | 42,477 |
299. Prove the inequality:
$$
(a+1)(b+1)(a+c)(b+c)>16 a b c \text { for } a>1, b>1, c>1
$$ | Instruction. Consider the inequalities:
$$
a+1>2 \sqrt{a}, b+1>2 \sqrt{b}, a+c>2 \sqrt{a c}, b+c>2 \sqrt{b c}
$$
and multiply them term by term. | proof | Inequalities | proof | Yes | Yes | olympiads | false | 42,478 |
300. Prove that if $x+y+z=1$, where $x, y, z$ are positive numbers, then
$$
(1-x)(1-y)(1-z)>8 x y z
$$ | Consider the inequalities:
$$
x+y>2 \sqrt{x y}, x+z>2 \sqrt{x z}, y+z>2 \sqrt{y z}
$$
Taking into account the condition of the problem, we obtain the inequalities:
$$
1-z>2 \sqrt{x y}, 1-y>2 \sqrt{x z}, 1-x>2 \sqrt{y z}
$$
Multiplying these inequalities term by term, we obtain the required inequality. | proof | Inequalities | proof | Yes | Yes | olympiads | false | 42,479 |
301. Prove that for positive $a, b, c$, the inequality holds: $a^{4}+b^{4}+c^{4} \geqslant a b c(a+b+c)$. | Instruction. By adding the inequalities term by term
$$
a^{4}+b^{4} \geqslant 2 a^{2} b^{2}, b^{4}+c^{4} \geqslant 2 b^{2} c^{2}, a^{4}+c^{4} \geqslant 2 a^{2} c^{2}
$$
we transform the sum to the form:
$$
2\left(a^{4}+b^{4}+c^{4}\right) \geqslant a^{2}\left(b^{2}+c^{2}\right)+b^{2}\left(a^{2}+c^{2}\right)+c^{2}\lef... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 42,480 |
303. Prove the inequality $(\sqrt{a}+\sqrt{b})^{8} \geqslant 64 a b(a+b)^{2}$ for positive values of $a$ and $b$. | Given: $\frac{x+y}{2} \geqslant \sqrt{x y}$. Let
$$
x=\frac{a+b}{2}, y=\sqrt{a b}
$$
Then
$$
\frac{\frac{a+b}{2}+\sqrt{a b}}{2} \geqslant \sqrt{\frac{(a+b) \sqrt{a b}}{2}}
$$
or
$\frac{a+b+2 \sqrt{a b}}{4} \geqslant \sqrt{\frac{(a+b) \sqrt{a b}}{2}}$ or $\frac{(\sqrt{a}+\sqrt{b})^{2}}{4} \geqslant \sqrt{\frac{(a+b... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 42,482 |
304. Prove that $\frac{a+b+c+d}{4} \geqslant \sqrt[4]{a b c d}$, where $a, b, c, d-$ are positive numbers. | Instruction. Multiplying the inequalities term by term
$$
\frac{a+b}{2} \geqslant \sqrt{a b} ; \frac{c+d}{2} \geqslant \sqrt{c \bar{c}},
$$
we get:
$$
\sqrt{a b c d} \leqslant \frac{a+b}{2} \cdot \frac{c+d}{2}
$$
6
HO
$$
\frac{a+b}{2} \cdot \frac{c+d}{2} \leqslant\left(\frac{\frac{a+b}{2}+\frac{c+d}{2}}{2}\right)... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 42,483 |
305. Prove that the arithmetic mean of three non-negative numbers is not less than their geometric mean. | Instruction. The problem can be solved by transforming the expression: $(\sqrt[3]{a}+\sqrt[3]{b}+\sqrt[3]{c})^{3}$. However, the solution is significantly simplified if we use the result of problem 304. Consider four numbers: $a, b, c, \frac{a+b+c}{3}$.
Based on problem 304
$$
a b c \cdot \frac{a+b+c}{3} \leqslant\le... | \sqrt[3]{}\leqslant\frac{+b+}{3} | Inequalities | proof | Yes | Yes | olympiads | false | 42,484 |
307*. Prove that for any positive numbers $a$ and $b$, $a \neq b$, the following inequality holds:
$$
\sqrt[n+1]{\overline{a b^{n}}}<\frac{a+n b}{n+1}
$$ | Indication.
$$
\sqrt[n+1]{a b^{n}}=\sqrt[n+1]{a \cdot \underbrace{b \cdot b \ldots b}_{n}}<\frac{a+\overbrace{b+b+\ldots+b}^{n}}{n+1}=\frac{a+n b}{n+1}
$$ | proof | Inequalities | proof | Yes | Yes | olympiads | false | 42,486 |
308*. Prove that $n a_{1} a_{2} \ldots a_{n} \leqslant a_{1}^{n}+a_{2}^{n}+\ldots+a_{n}^{n}$, where $n-$ is a natural number, $a_{i}$-positive numbers.[^8]
84 | Instruction. $\quad a_{1} \cdot a_{2} \ldots a_{n}=\sqrt[n]{a_{1}^{n} \cdot a_{2}^{n} \ldots a_{n}^{n}} \leqslant \frac{a_{1}^{n}+a_{2}^{n}+\ldots+a_{n}^{n}}{n}$.
The equality sign holds when $a_{1}=a_{2}=\ldots=a_{n}=a$.
Note. From this problem, it follows that twice the product of two positive numbers is not greate... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 42,487 |
309. Prove that for any positive numbers
$$
a_{1}, a_{2} \ldots a_{n}
$$
the following inequality holds:
$$
\frac{a_{1}}{a_{2}}+\frac{a_{2}}{a_{3}}+\ldots+\frac{a_{n}}{a_{1}}>n
$$ | Instruction. The geometric mean of the addends of the given sum is equal to one.
$310^{*}$. Prove that $1 \cdot 2 \ldots n \leqslant\left(\frac{n+1}{2}\right)^{n}$, where $n-$ is any natural number.
Instruction. $\frac{1+2+\ldots+n}{n} \geqslant \sqrt[n]{1 \cdot 2 \ldots n}$. But the numerator of the fraction on the ... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 42,488 |
312*. Prove that for positive numbers $a_{1}, a_{2}, \ldots, a_{n}$ and any positive integers $p_{1}, p_{2}, \ldots, p_{n}$, the following inequality holds:
$$
a_{1}^{p_{1}} \cdot a_{2}^{p_{2}} \ldots a_{n}^{p_{n}} \leqslant\left(\frac{p_{1} a_{1}+p_{2} a_{2}+\ldots+p_{n} a_{n}}{p_{1}+p_{2}+\ldots+p_{n}}\right)^{p_{1}... | Instruction. Using the inequality
$$
\begin{gathered}
\frac{a_{1}+a_{2}+\ldots+a_{n}}{n} \geqslant \sqrt[n]{a_{1} \cdot a_{2} \ldots a_{n}}, \\
a_{1}^{p_{1}} \cdot a_{2}^{p_{2}} \ldots a_{n}^{p_{n}}=\underbrace{a_{1} \cdot a_{1} \ldots a_{1}}_{p_{1}} \cdot \underbrace{a_{2} \cdot a_{2} \ldots a_{2}}_{p_{2}} \ldots \un... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 42,489 |
313*. Prove that for $p>0$ and for any rational $k>1$ the following inequality holds: $(1+p)^{k}>1+p k$ (Bernoulli's inequality). | Let $k=\frac{a}{b}$, where $a>b$. Taking into account the relationship between the arithmetic mean and the geometric mean, we can write:
$$
\sqrt[a]{(1+p k)^{b}}1$. For a natural $k$, Bernoulli's inequality can be proven using the binomial formula:
$$
(1+p)^{k}=1+p k+C_{k}^{2} p^{2}+\ldots+p^{k}
$$
Discarding the po... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 42,490 |
314. Prove that for real $a_{1}, b_{1}, a_{2}, b_{2}$, the inequality holds:
$$
a_{1} b_{1}+a_{2} b_{2} \leqslant \sqrt{a_{1}^{2}+a_{2}^{2}} \cdot \sqrt{b_{1}^{2}+b_{2}^{2}}
$$
or
$$
\left(a_{1} b_{1}+a_{2} b_{2}\right)^{2} \leqslant\left(a_{1}^{2}+a_{2}^{2}\right)\left(b_{1}^{2}+b_{2}^{2}\right)
$$ | Instruction. By expanding the brackets and moving all terms to one side of the inequalities, we obtain the obvious inequality:
$$
\left(a_{1} b_{2}-a_{2} b_{1}\right)^{2} \geqslant 0
$$
This inequality becomes an equality when
$$
a_{1}=k b_{1}, \quad a_{2}=k b_{2}
$$ | proof | Inequalities | proof | Yes | Yes | olympiads | false | 42,491 |
315*. Prove that for any $2 n$ real numbers $a, b, \ldots k, A, B, \ldots, K$ the following inequality holds:
$$
\begin{gathered}
(a A+b B+\ldots+k K)^{2} \leqslant\left(a^{2}+b^{2}+\ldots+k^{2}\right) \times \\
\times\left(A^{2}+B^{2}+\ldots+K^{2}\right)
\end{gathered}
$$
The equality sign holds when the numbers of ... | 1) In the left part of the inequality to be proved, we will have a sum of the form
$$
a^{2} A^{2}+b^{2} B^{2}+\ldots+k^{2} K^{2}+2 a b A B+\ldots+2 b k B K
$$
in the right part - a sum of the form
$$
a^{2} A^{2}+b^{2} B^{2}+\ldots+k^{2} K^{2}+a^{2} B^{2}+\ldots+b^{2} K^{2}
$$
But
$$
2 a b A B \leqslant a^{2} B^{2}... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 42,492 |
317*. Prove that for any positive numbers $a$, $b$, $c$, the following inequality holds:
$$
(a+b+c)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right) \geqslant 9
$$ | By the Bunyakovsky-Cauchy inequality, we have:
$$
\begin{gathered}
(a+b+c)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right) \geqslant \\
\geqslant\left(\sqrt{ } \bar{a} \cdot \frac{1}{\sqrt{a}}+\sqrt{\bar{b}} \cdot \frac{1}{\sqrt{b}}+\sqrt{c} \cdot \frac{1}{\sqrt{c}}\right)^{2}=9
\end{gathered}
$$
Equality holds if an... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 42,493 |
319. Prove that if $x_{1}, x_{2}, \ldots, x_{n} ; y_{1}, y_{2}, \ldots, y_{n}$ are positive numbers, then
$$
\begin{gathered}
\sqrt{x_{1} y_{1}}+\sqrt{x_{2} y_{2}}+\ldots+\sqrt{x_{n} y_{n}} \leqslant \\
\leqslant \sqrt{x_{1}+x_{2}+\ldots+x_{n}} \cdot \sqrt{y_{1}+y_{2}+\ldots+y_{n}}
\end{gathered}
$$ | Instruction. Putting in the Cauchy-Bunyakovsky inequality
$$
a^{2}=x_{1}, \quad b^{2}=x_{2}, \ldots, k^{2}=x_{n}, A^{2}=y_{1}, \ldots, K^{2}=y_{n}
$$
and taking the square root of both sides of the inequality, we obtain the required inequality.
88 | proof | Inequalities | proof | Yes | Yes | olympiads | false | 42,495 |
320*. Prove that for any three positive numbers $a, b, c$ the following inequality holds:
$$
\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b} \geqslant \frac{3}{2}
$$ | Instruction. Transform the left side of the inequality to the form:
$$
\frac{1}{2}[(a+b)+(b+c)+(c+a)]\left(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}\right)-3
$$
then apply the Bunyakovsky-Cauchy inequality (see problem 317). | proof | Inequalities | proof | Yes | Yes | olympiads | false | 42,496 |
321**. Let $x$ and $y$ be positive rational numbers that sum to one: $x+y=1$. Prove that for any positive numbers $a_{1}, a_{2} \ldots a_{n} ; b_{1}, b_{2} \ldots b_{n}$, the following inequality holds:
$$
\begin{gathered}
a_{1}^{x} \cdot b_{1}^{y}+a_{2}^{x} \cdot b_{2}^{y}+\ldots+a_{n}^{x} \cdot b_{n}^{y} \leqslant \... | Given
$$
a_{1}+a_{2}+\ldots+a_{n}=a ; \quad b_{1}+b_{2}+\ldots+b_{n}=b
$$
and dividing the inequality to be proved by $a^{x} b^{y}$, we arrive at an inequality equivalent to the inequality of the problem:
$$
\left(\frac{a_{1}}{a}\right)^{x}\left(\frac{b_{1}}{b}\right)^{y}+\left(\frac{a_{2}}{a}\right)^{x}\left(\frac{... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 42,497 |
322*. Let $p>1$ and $\frac{1}{p}+\frac{1}{q}=1$, where $p$ and $q$ are rational numbers. Prove that for any positive numbers: $a, b, \ldots, k ; A, B, \ldots, K$ the following inequality holds:
$$
\begin{gathered}
(a A+b B+\ldots+k K) \leqslant \\
\leqslant\left(a^{p}+b^{p}+\ldots+k^{p}\right)^{\frac{1}{p}}\left(A^{q}... | Instruction. Substituting in the inequality of problem 321
$$
\begin{gathered}
x=\frac{1}{p}, \quad y=\frac{1}{q}, a_{1}=a^{p}, a_{2}=b^{p}, \ldots, a_{n}=k^{p} \\
b_{1}=A^{q}, b_{2}=B^{q}, \ldots, b_{n}=K^{q}
\end{gathered}
$$
we obtain the required inequality. | proof | Inequalities | proof | Yes | Yes | olympiads | false | 42,498 |
324*. If $a, b, c$ are the sides of a triangle, then prove that the following inequality holds:
$$
\frac{1}{b+c-a}+\frac{1}{c+a-b}+\frac{1}{a+b-c}>\frac{9}{a+b+c}
$$ | Proof. According to the problem, $a+b>c, b+c>a$, $c+a>b$ and each fraction in the left part of the supposed inequality is a positive quantity. We will use the property of the arithmetic mean and the geometric mean of non-negative numbers:
$$
\frac{\frac{1}{b+c-a}+\frac{1}{c+a-b}+\frac{1}{a+b-c}}{3}>\sqrt[3]{\frac{1}{(... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 42,500 |
326*. Prove that $\frac{\sqrt{2}}{2} k \leqslant c$, where $k$ is the sum of the legs of any right triangle, and $c$ is its hypotenuse. | Let us denote the legs of a right triangle by $a$ and $b$. We will have:
$$
a^{2}+b^{2}=c^{2} ; \quad k=a+b
$$
It is known that
$$
a^{2}+b^{2} \geqslant 2 a b
$$
or
$$
\frac{a^{2}+b^{2}}{2} \geqslant\left(\frac{a+b}{2}\right)^{2}
$$
from which
$$
\frac{c^{2}}{2} \geqslant \frac{k^{2}}{4}
$$
or
$$
c \geqslant \... | \geqslant\frac{\sqrt{2}}{2}k | Inequalities | proof | Yes | Yes | olympiads | false | 42,501 |
327. Prove that among all rectangles with a given perimeter $p$, the one with the largest area is a square. | Let $a, b$ be the sides of a rectangle, then
$$
p=2(a+b) ; \quad S=a b
$$
where $S$ is the area of the rectangle. It is known that
$$
\frac{a+b}{2} \geqslant \sqrt{a b} \text { or } \quad a b \leqslant\left(\frac{a+b}{2}\right)^{2}
$$
from which
$$
S=a b \leqslant\left(\frac{a+b}{2}\right)^{2} \quad \text { or } \... | proof | Algebra | proof | Yes | Yes | olympiads | false | 42,502 |
328. Prove that among all rectangles with a given area, the one with the smallest perimeter is a square.
92 | Instruction. By analogy with the previous task, we obtain that $S \leqslant \frac{p^{2}}{16}$. The area is given, the perimeter is chosen to be the smallest, which is achieved when $a=b$. | proof | Geometry | proof | Yes | Yes | olympiads | false | 42,503 |
329. Prove that among all triangles with a given perimeter, the one with the greatest area is the equilateral triangle. | According to Heron's formula, the area of a triangle with sides $a, b, c$ is
$$
S=\sqrt{p(p-a)(p-b)(p-c)}
$$
where $p=\frac{a+b+c}{2}$, but
$$
(p-a)(p-b)(p-c) \leqslant\left(\frac{p-a+p-b+p-c}{3}\right)^{3},
$$
(see problem 305) or
$$
(p-a)(p-b)(p-c) \leqslant\left(\frac{p}{3}\right)^{3}
$$
equality will hold onl... | proof | Geometry | proof | Yes | Yes | olympiads | false | 42,504 |
331. Prove that for $p>n$ and $a>b>0$
$$
\frac{a^{p}-b^{p}}{a^{p}+b^{p}}>\frac{a^{n}-b^{n}}{a^{n}+b^{n}}
$$ | Instruction. Transform both sides of the assumed inequality by dividing the numerator and denominator of the fraction on the left side of the inequality by $a^{p}$, and the fractions on the right side of the inequality by $a^{n}$, and compare the resulting expressions. | proof | Inequalities | proof | Yes | Yes | olympiads | false | 42,505 |
335. Prove that if in an arithmetic progression, the same number of arithmetic means is inserted between every two consecutive terms, then the entire sequence will form an arithmetic progression. | Let $a, b, \ldots k$ be the terms of an arithmetic progression, $p$ be the number of arithmetic means inserted between each pair of consecutive terms of this progression. From the previous problem, the numbers $a$ and $b$ together with $p$ arithmetic means inserted between $a$ and $b$ form an arithmetic progression wit... | proof | Algebra | proof | Yes | Yes | olympiads | false | 42,508 |
336. Find the condition that links three numbers $a, b, c$, if they are members of the same arithmetic progression. | Instruction. Suppose that the given numbers are terms of the same arithmetic progression with the first term $a_{1}$ and common difference $d$. Then
$$
a=a_{1}+d n, \quad b=a_{1}+d \kappa, \quad c=a_{1}+d p
$$
where $n, p, k$ - are integers. From these equations, we will have
$$
\frac{b-a}{c-b}=\frac{k-n}{p-k}=\frac... | \frac{b-}{-b}=\frac{k-n}{p-k} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 42,509 |
337. Find the condition under which the numbers $a, b, c$ would be the $p$-th, $n$-th, $k$-th terms of the same arithmetic progression. | Instruction. For the previous task, the required condition will be:
$$
\frac{b-a}{c-b}=\frac{n-p}{k-n}
$$
or
$$
a(n-k)+b(k-p)+c(p-n)=0
$$ | (n-k)+b(k-p)+(p-n)=0 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 42,510 |
338. Prove that $\sqrt{2}, \sqrt{3}, \sqrt{5}$ cannot be members of the same arithmetic progression. | If the given numbers are members of the same arithmetic progression, then
$$
\frac{\sqrt{3}-\sqrt{2}}{\sqrt{5}-\sqrt{3}}=\frac{p}{k}
$$
where $p$ and $k$ are integers (see problem 336), or
$$
\sqrt{3}-\sqrt{2}=\frac{p}{k}(\sqrt{5}-\sqrt{3})
$$
and
$$
(\sqrt{3}-\sqrt{2})(\sqrt{3}+\sqrt{2})=\frac{p}{k}(\sqrt{5}-\sqr... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 42,511 |
339. Prove that if the second term of an arithmetic progression is the mean proportional between the 1st and 4th terms, then the 6th term will be the mean proportional between the 4th and 9th terms. | According to the problem, the arithmetic progression has the form:
$$
a, 2a, 3a, 4a, \ldots, na
$$
from which
$$
a_{4} \cdot a_{9}=4a \cdot 9a=36a^{2}=(6a)^{2}=\left(a_{6}\right)^{2}
$$ | proof | Algebra | proof | Yes | Yes | olympiads | false | 42,512 |
340. If the positive numbers $x, y, z$ are the $k$-th, $p$-th, and $n$-th terms of an arithmetic progression, then prove that there exists an arithmetic progression in which the $x$-th, $y$-th, and $z$-th terms are respectively equal to $k, p$, and $n$ and that the product of the differences of these progressions is 1.... | Given the condition of the problem, we have:
$$
x=a+d(k-1), y=a+d(p-1), z=a+d(n-1)
$$
From which
$$
\begin{gathered}
k=\frac{1}{d}(x-a+d)=\frac{1}{d}(x-1+d-a+1)= \\
=\frac{d-a+1}{d}+\frac{1}{d}(x-1) \\
p=\frac{d-a+1}{d}+\frac{1}{d}(y-1) \\
n=\frac{d-a+1}{d}+\frac{1}{d}(z-1)
\end{gathered}
$$
The obtained equalities... | proof | Algebra | proof | Yes | Yes | olympiads | false | 42,513 |
341. Find the condition under which the sum of any two terms of an arithmetic progression represents a term of the same progression. | Instruction. According to the condition of the problem, we should have:
$$
(a+d n)+(a+d k)=a+d p
$$
where $a$ is the first term of the progression, $d$ is the common difference of the progression, and $n, p, k$ are positive integers. Then $a=d(p-k-n)$, i.e., the first term of the progression must be a multiple of the... | (p-k-n) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 42,514 |
342. Prove that the squares of the expressions
$$
x^{2}-2 x-1, \quad x^{2}+1, \quad x^{2}+2 x-1
$$
form three consecutive terms of an arithmetic progression. | Instruction. Prove that the square of the middle term will be the arithmetic mean of the squares of the other two terms. | proof | Algebra | proof | Yes | Yes | olympiads | false | 42,515 |
343. Prove that if $a, b, c$ are three consecutive terms of an arithmetic progression, then
$$
a^{2}+8 b c=(2 b+c)^{2}
$$ | Given the condition of the problem, $a+c=2 b$, from which
$$
a^{2}+8 b c=(2 b-c)^{2}+8 b c=4 b^{2}-4 b c+c^{2}+8 b c=(2 b+c)^{2}
$$ | proof | Algebra | proof | Yes | Yes | olympiads | false | 42,516 |
345. Prove that if the numbers $a, b, c$ form an arithmetic progression, then
$$
3\left(a^{2}+b^{2}+c^{2}\right)=6(a-b)^{2}+(a+b+c)^{2}
$$ | Given the condition of the problem, $a+c=2b$, hence
$$
ac=2ab-a^2 \quad \text{and} \quad a^2+c^2+2ac=4b^2
$$
from which
$$
\begin{gathered}
3\left(a^2+b^2+c^2\right)=3\left(4b^2-2ac+b^2\right)=6\left(2b^2-ac\right)+3b^2= \\
=6\left(2b^2-2ab+a^2\right)+3b^2=6(a-b)^2+9b^2= \\
=6(a-b)^2+(3b)^2=6(a-b)^2+(2b+b)^2= \\
=6(... | proof | Algebra | proof | Yes | Yes | olympiads | false | 42,518 |
346. Prove that if $\frac{1}{b+c}, \frac{1}{c+a}, \frac{1}{b+a}$ form an arithmetic progression, then the numbers $a^{2}, b^{2}, c^{2}$ also form an arithmetic progression. | Indication. The task is to prove that
$$
2 b^{2}-a^{2}-c^{2}=0
$$
By the condition of the problem,
$$
\frac{2}{c+a}-\frac{1}{b+c}-\frac{1}{b+a}=0
$$
from which
$$
\begin{aligned}
& 2(b+c)(b+a)-(c+a)(b+a)-(c+a)(b+c)=0 \\
& \text { or } \\
& \begin{array}{c}
{[(b+a)(b+c)-(c+a)(b+a)]+[(a+b)(b+c)-(c+a)(b+c)]=} \\
=(b+... | 2b^{2}-^{2}-^{2}=0 | Algebra | proof | Yes | Yes | olympiads | false | 42,519 |
347. Given two progressions:
\[
\begin{gathered}
a, a+b, a+2 b, \ldots, a+(n-1) b \\
a, a+c, a+2 c, \ldots, a+(n-1) c
\end{gathered}
\]
such that the $p$-th term of the first is equal to the $p^2$-th term of the second. Prove that the difference between the sums of $n$ terms of these progressions is equal to $\frac{1... | Given the condition
$$
a+b(p-1)=a+c\left(p^{2}-1\right)
$$
from which $b=c(p+1)$. Let
$$
\begin{gathered}
a+a+b+\ldots+a+b(n-1)=S_{1} \\
a+a+c+\ldots+a+c(n-1)=S_{2} \\
S_{1}-S_{2}=\frac{2 a+b(n-1)}{2} \cdot n-\frac{2 a+c(n-1)}{2} \cdot n= \\
=\frac{1}{2} c n(n-1) p
\end{gathered}
$$
since $b=c(p+1)$. | \frac{1}{2}(n-1)p | Algebra | proof | Yes | Yes | olympiads | false | 42,520 |
348. Prove that for any arithmetic progression, the following equality holds:
$$
S_{n+3}-3 S_{n+2}+3 S_{n+1}-S_{n}=0
$$ | Instruction. Establish that
$$
S_{p}-S_{n}=(p-n)\left[a_{1}+\frac{d(p+n-1)}{2}\right]
$$ | proof | Algebra | proof | Yes | Yes | olympiads | false | 42,521 |
349. If $S_{p}, S_{n}, S_{k}$ are the sums of $p, n, k$ terms of an arithmetic progression, then prove that the following equality holds:
$$
\frac{S_{p}}{p}(n-k)+\frac{S_{n}}{n}(k-p)+\frac{S_{k}}{k}(p-n)=0
$$ | Indication. Take into account that
$$
\frac{S_{n}}{n}=\frac{a_{n}+a_{1}}{2}, a_{p}-a_{k}=d(p-k)
$$ | proof | Algebra | proof | Yes | Yes | olympiads | false | 42,522 |
351. Find the general form of natural numbers that can be represented as the sum of consecutive natural numbers, starting from one. | Instruction. The task is to determine the expression for the sum of $n$ terms of an arithmetic progression, where the first term and the difference are equal to one, i.e.
$$
S_{n}=\frac{n(n+1)}{2}
$$
Thus, the required numbers will be: $1,3,6,10,15,21 \ldots$ | S_{n}=\frac{n(n+1)}{2} | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 42,523 |
352. Find a natural number that is equal to the sum of all its preceding natural numbers. Does more than one such number exist, or is there only one? | Let $n$ be the desired natural number; the sum of the numbers preceding this natural number is expressed by the formula
$$
S_{n-1}=\frac{n(n-1)}{2}
$$
According to the problem, $\frac{n(n-1)}{2}=n$, from which $n^{2}-3 n=0$ or $n_{1}=3, n_{2}=0$. The desired number is 3. It is the only one, as the second root of the ... | 3 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 42,524 |
353. Prove that the sum of the squares of $n$ natural numbers is equal to $\frac{n(n+1)(2 n+1)}{6}$. | Consider the identities:
\[
\begin{aligned}
& n^{3}-(n-1)^{3}=3 n^{2}-3 n+1 \\
& (n-1)^{3}-(n-2)^{3}=3(n-1)^{2}-3(n-1)+1 \\
& (n-2)^{3}-(n-3)^{3}=3(n-2)^{2}-3(n-2)+1 \\
& 3^{3}-2^{3}=3 \cdot 3^{2}-3 \cdot 3+1
\end{aligned}
\]
\[
\begin{aligned}
& 2^{3}-1^{3}=3 \cdot 2^{2}-3 \cdot 2+1 \\
& 1^{3}-0^{3}=3 \cdot 1^{2}-3 ... | 1+2^{2}+\ldots+n^{2}=\frac{n(n+1)(2n+1)}{6} | Algebra | proof | Yes | Yes | olympiads | false | 42,525 |
354. Find the expressions for the sums:
\[
\begin{aligned}
& 2^{2}+4^{2}+\ldots+(2 n)^{2} \\
& 1+3^{2}+\ldots+(2 n+1)^{2}
\end{aligned}
\] | Solution. Let's transform the expression:
$$
\begin{gathered}
2^{2}+4^{2}+\ldots+4 n^{2}=4\left(1+2^{2}+\ldots+n^{2}\right)= \\
=\frac{2 n(n+1)(2 n+1)}{3}
\end{gathered}
$$
To determine the sum $1+3^{2}+\ldots+(2 n+1)^{2}$, consider the sum:
$$
\begin{gathered}
1+2^{2}+3^{2}+\ldots+n^{2}+\ldots+(2 n+1)^{2}= \\
=\fra... | \frac{2n(n+1)(2n+1)}{3} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 42,526 |
355*. Calculate the sum: $S_{k}=1^{k}+2^{k}+3^{k}+\ldots+n^{k}$. | Solution. Consider the difference:
$$
(p+1)^{k+1}-p^{k+1}=C_{k+1}^{1} p^{k}+C_{k+1}^{2} p^{k-1}+\ldots+1
$$
102
Setting $p=0,1,2 \ldots n$ successively, we get the equalities:
$$
\begin{aligned}
& 1^{k+1}=1 \\
& 2^{k+1}-1^{k+1}=C_{k+1}^{1} \cdot 1^{k}+C_{k+1}^{2} \cdot 1^{k-1}+\ldots+1 \\
& 3^{k+1}-2^{k+1}=C_{k+1}^... | 1+2^{3}+\ldots+n^{3}=[\frac{n(n+1)}{2}]^{2}=S_{1}^{2} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 42,527 |
356. Find the sum of all products of the form $1 \cdot 2, 1 \cdot 3, \ldots$ $n(n-1)$, that can be formed from the numbers $1, 2, 3, \ldots n$. | Solution. Let's take the sum of all natural numbers up to $n$ inclusive:
$$
S_{1}=1+2+3+\ldots+(n-1)+n=\frac{n(n+1)}{2}
$$
Square this equality:
$$
\begin{gathered}
S_{1}^{2}=1^{2}+2^{2}+\ldots+(n-1)^{2}+n^{2}+2[1 \cdot 2+1 \cdot 3+\ldots+ \\
+n(n-1)]
\end{gathered}
$$
Denote the sum of all double products by $\mat... | \frac{n(n+1)(n-1)(3n+2)}{24} | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 42,528 |
359. Prove that the common difference of an arithmetic progression is equal to the difference of the squares of the first and last terms, divided by the difference between twice the sum of all terms and the sum of the first and last terms. | Let's first prove the statement for an increasing progression by transforming the expression:
$$
\begin{gathered}
\frac{a_{n}^{2}-a_{1}^{2}}{2 S_{n}-\left(a_{1}+a_{n}\right)}=\frac{a_{n}^{2}-a_{1}^{2}}{\left(a_{1}+a_{n}\right) n-\left(a_{1}+a_{n}\right)}=\frac{a_{n}^{2}-a_{1}^{2}}{\left(a_{1}+a_{n}\right)(n-1)}= \\
=\... | proof | Algebra | proof | Yes | Yes | olympiads | false | 42,529 |
360. The sums of the terms of each of the arithmetic progressions, having $n$ terms, are equal to $n^{2}+p n$ and $3 n^{2}-2 n$. Find the condition under which the $n$-th terms of these progressions will be equal. | Given the problem,
$$
S_{1}=\frac{a_{n}+a_{1}}{2} \cdot n=n^{2}+p n, \quad S_{2}=\frac{b_{n}+b_{1}}{2} \cdot n=3 n^{2}-2 n
$$
from which we get
$$
a_{n}+a_{1}=2(n+p), \quad b_{n}+b_{1}=2(3 n-2)
$$
where $a_{i}$ are the terms of the first progression, and $b_{i}$ are the terms of the second progression. For $n=1$, $... | 4(n-1) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 42,530 |
361*. Prove that if in an arithmetic progression the sum of the first $n$ terms is equal to $n^{2} p$ and the sum of the first $k$ terms is equal to $k^{2} p$, then the sum of the first $p$ terms of this progression will be equal to $p^{3}$. | Indication. By the condition of the problem,
$$
\frac{a_{n}+a_{1}}{2} \cdot n=n^{2} p, \quad a_{n}+a_{1}=2 n p
$$
analogously
$$
a_{k}+a_{1}=2 k p
$$
from which
$$
a_{n}-a_{k}=2 p(n-k)
$$
But
$$
a_{n}-a_{k}=d(n-k)
$$
thus,
$$
d=2 p, \quad a_{1}=p
$$
Then
$$
S_{p}=\frac{2 p+2 p(p-1)}{2} \cdot p=p^{3}
$$
362 ... | p^3 | Algebra | proof | Yes | Yes | olympiads | false | 42,531 |
366. Find the value of $p$ geometric means inserted between the given positive numbers $a$ and $b$, where $a<b$. | Instruction. According to the problem, all numbers form a geometric progression, and the task is to determine the common ratio of the progression from the first term $a$, the last term $b$, and the number of terms $p+2$ (two terms are given and $p$ needs to be inserted), i.e.
$$
b=a q^{p+1}
$$
or
$$
q=\sqrt[p+1]{\fr... | \sqrt[p+1]{\frac{b}{}} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 42,532 |
367. Prove that if between each two consecutive terms of a geometric progression with positive terms, the same number $n$ of geometric means is inserted, then the entire sequence will form a geometric progression. | Let the members of the progression be $a, b, c \ldots k$. From the previous problem, the numbers $a$ and $b$ together with $n$ geometric means inserted between them form a geometric progression with the common ratio $\sqrt[n+1]{\frac{b}{a}}$; the numbers $b$ and $c$ together with $n$ geometric means inserted between th... | proof | Algebra | proof | Yes | Yes | olympiads | false | 42,533 |
368. Prove that in a geometric progression, the product of terms equally distant from the ends is equal to the product of the extreme terms. | Given: $a_{1} \cdot a_{n}=a_{1} \cdot a_{1} \cdot q^{n-1}=a_{1}^{2} q^{n-1}, a_{2} \cdot a_{n-1}=$ $=a_{1} \cdot q \cdot a_{1} \cdot q^{n-2}=a_{1}^{2} q^{n-1}, \ldots$ and so on, i.e., each of the products is equal to the same value $a_{1}^{2} q^{n-1}$. Therefore, all these products are equal to each other. | proof | Algebra | proof | Yes | Yes | olympiads | false | 42,534 |
369. Prove that the product of $n$ terms of a geometric progression is equal to $a_{1}^{n} q^{\frac{n(n-1)}{2}}$, where $a_{1}$ is the first term of the progression, and $q$ is its common ratio. | Instruction. According to the previous task, the products of terms equally distant from the ends of a geometric progression are all equal to each other and equal to $a_{1}^{2} q^{n-1}$. In the product of $n$ terms of a geometric progression, there will be $\frac{n}{2}$ such pairs of terms, so the entire product will be... | a_{1}^{n}q^{\frac{n(n-1)}{2}} | Algebra | proof | Yes | Yes | olympiads | false | 42,535 |
371. Prove that in any geometric progression, the sum of the 4th, 5th, and 6th terms is the geometric mean between the sum of the 1st, 2nd, and 3rd terms and the sum of the 7th, 8th, and 9th terms. | Instruction. The sum of the 4th, 5th, and 6th terms of a geometric progression:
$$
a_{1} q^{3}+a_{1} q^{4}+a_{1} q^{5}=a_{1} q^{3}\left(1+q+q^{2}\right)
$$
or
$$
a_{1}^{2} q^{6}\left(1+q+q^{2}\right)^{2}=a_{1}\left(1+q+q^{2}\right) \cdot a_{1} \cdot q^{6}\left(1+q+q^{2}\right)
$$
but the first two factors of this p... | proof | Algebra | proof | Yes | Yes | olympiads | false | 42,536 |
372. Find the condition under which the numbers $a, b, c, d$, taken in any order, form consecutive terms of a geometric progression. | If the numbers $a, b, c, d$, taken in any order, form consecutive terms of a geometric progression, then $\frac{a}{b}=\frac{c}{d}$ and $\frac{a}{b}=\frac{d}{c}$, from which $\frac{a^{2}}{b^{2}}=1$ and $a= \pm b$. Similarly,
$b= \pm c, c= \pm d$, i.e., all four numbers must be equal in absolute value. | \b,\,\ | Algebra | math-word-problem | Yes | Yes | olympiads | false | 42,537 |
373. Find the condition under which the numbers $a, b, c$ represent the $k$-th, $n$-th, and $p$-th terms of the same geometric progression. | Given the problem,
$$
\frac{a}{b}=q^{k-n} ; \frac{a}{c}=q^{k-p}
$$
from which
$$
\begin{aligned}
& \left(\frac{a}{b}\right)^{k-p}=q^{(k-n)(k-p)} \\
& \left(\frac{a}{c}\right)^{k-n}=q^{(k-p)(k-n)}
\end{aligned}
$$
or
$$
\left(\frac{a}{b}\right)^{k-p}=\left(\frac{a}{c}\right)^{k-n}-\text { the desired condition. }
$... | (\frac{}{b})^{k-p}=(\frac{}{})^{k-n} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 42,538 |
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