Questions tagged [inequality]

Questions on proving, manipulating and applying inequalities.

19,212 questions
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If $\prod x_k\neq0$ and $\sum\frac{x_k}{x_{k+1}}=0$, then $|\sum x_kx_{k+1}|\le (\max|x_k|- \min|x_k|)\sum x_k$

Question: Suppose that $x_{1},x_{2},\cdots,x_{n}$ are real numbers, such that $$x_{1}x_{2}\cdots x_{n}\neq 0$$ and $$\dfrac{x_{1}}{x_{2}}+\dfrac{x_{2}}{x_{3}}+\cdots+\dfrac{x_{n}}{x_{1}}=0$$ ...
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Trace inequality for real matrices

Is there any general result characterizing real matrices $A$ such that $$[\mathrm{tr}(A)]^2\leq n\mathrm{tr}(A^2)?$$ I can see that the inequality holds if: all eigenvalues of $A$ are real (by the ...
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Prove inequality$\sum\limits_{n|i+j+k}x_{i}y_{j}z_{k}\le n^2$

Being given an integer $n\ge 2$, and $x_{i},y_{i},z_{i}\in \mathbb{R}$ ($i=1,2,\cdots,n$) such that $$\sum_{i=1}^{n}(x^3_{i}+y^3_{i}+z^3_{i})=3n$$ show that $$\sum_{i+j+k=n}x_{i}y_{j}z_{k}\le n^2.$$ ...
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Show that $\frac 1 {\sqrt{x+y}}+\frac 1 {\sqrt{y+z}}+\frac 1 {\sqrt{z+x}}\geq2+\frac 1 {\sqrt2}$.

Given $x,y,z\geq0$ and $xy+yz+zx=1$. Show that $\displaystyle\frac 1 {\sqrt{x+y}}+\frac 1 {\sqrt{y+z}}+\frac 1 {\sqrt{z+x}}\geq2+\frac 1 {\sqrt2}$. I've tried many things but all failed. The only ...
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$x^3-3x-3=0$, prove that $10^x<127$

$x$ is the real root of the equation $$3x^3-5x+8=0,\tag 1$$ prove that $$e^x>\frac{40}{237}.$$ I find this inequality in a very accidental way,I think it's very difficult,because the actual value ...
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CSB inquality: is $\|x\|^2\|y\|^2 - \langle x,y \rangle^2$ a square in any obvious way?

Suppose $x=(x_1,x_2),y = (y_1,y_2) \in \mathbb{R}^2$. I noticed that \begin{align*} \|x\|^2 \|y\|^2 - \langle x,y \rangle^2 &= x_1^2y_1^2 + x_1^2 y_2^2 + x_2^2 y_1^2 + x_2^2 y_2 ^2 - (x_1^2 y_1^2 +...
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If $\{a,b,c,d,e\}\subset[0,1]$ so $\sum\limits_{cyc}\frac{1}{1+a+b}\leq\frac{5}{1+2\sqrt{abcde}}$

Let $\{a,b,c,d,e\}\subset[0,1]$. Prove that: $$\frac{1}{1+a+b}+\frac{1}{1+b+c}+\frac{1}{1+c+d}+\frac{1}{1+d+e}+\frac{1}{1+e+a}\leq\frac{5}{1+2\sqrt{abcde}}$$ I tried C-S, convexity and more, but ...
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How to prove this inequality in Euclidean space?

Prove that \begin{align*}&|a+b||a+c|+|a+b||b+c|+|a+c||b+c|\\ \leq &(|a|+|b|+|c|) \cdot |a+b+c|+|a||b|+|a||c|+|b||c|\end{align*} in Euclidean space $\mathbb{R}^n$. I have been ...
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prove $f(x,y) \le 3$

$x \ge 0, y>0$ $f(x,y)=\sqrt{\dfrac{y}{y+x^2}}+4\sqrt{\dfrac{y}{(y+(x+1)^2)(y+(x+3)^2)}}+4\sqrt{\dfrac{y}{(y+(x-1)^2)(y+(x-3)^2)}}$ prove $f(x,y) \le 3$ I can prove when $x\ge 2, f(x,y) < 3$, ...
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If $a > b$, is $a^2 > b^2$?

Given $a > b$, where $a,b ∈ ℝ$, is it always true that $a^2 > b^2$?
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Proving that the sum of any two sides of the triangle is greater than the third side

I know it's easy to prove with the help of linear inequalities, but this time I want to prove it with the help of trigonometry. Is it possible? If yes, then how?
Showing $\sqrt{2}\sqrt{3}$ is greater or less than $\sqrt{2} + \sqrt{3}$ algebraically
How can we establish algebraically if $\sqrt{2}\sqrt{3}$ is greater than or less than $\sqrt{2} + \sqrt{3}$? I know I can plug the values into any calculator and compare the digits, but that is not ...