Questions tagged [inequality]

Questions on proving, manipulating and applying inequalities. Do not use this tag just because an inequality appears somewhere in your question.

An inequality is a mathematical relation between two quantities that are not necessarily equal, but bigger or smaller.

To prove inequalities, a number of proven inequalities can be used, including:

  • The AM-GM inequality

    Let $x_i>0$, $\alpha_i>0$ such that $\alpha_1+\alpha_2+...+\alpha_n=1$. Prove that $$\alpha_1x_1+\alpha_2x_2+...+\alpha_nx_n\geq x_1^{\alpha_1}x_2^{\alpha_2}...x_n^{\alpha_n}$$

For $\alpha_1=\alpha_2=...=\alpha_n=\frac{1}{n}$ we obtain the well-known $$\frac{x_1+x_2+\cdots+x_n}{n} \ge \sqrt[n]{x_1x_2\cdots x_n}$$

  • The Power Mean inequality (P-M).

    Let $a_1, a_2,\cdots, a_n$ be positive numbers and $p>q$. Then $$\left(\frac{a_1^p+a_2^p+\cdots+a_n^p}{n}\right)^{\frac{1}{p}} \geq \left(\frac{a_1^q+a_2^q+\cdots+a_n^q}{n}\right)^{\frac{1}{q}}$$

  • The Rearrangement inequality (R).

    Let $a_1\le\dots\le a_n$ and $b_1\le\dots\le b_n$. For all permutations $\sigma\in S_n$, $$\sum_{i=1}^na_ib_{n-i+1}\le\sum_{i=1}^na_ib_{\sigma(i)}\leq\sum_{i=1}^na_ib_i.$$

The rearrangement generalizes similar for more than two sequences of numbers.

  • The Cauchy-Schwarz inequality (C-S).

    If $a_1, a_2, \cdots, a_n$ and $b_1, b_2,\cdots, b_n$ are two sequences of real numbers, then $$\sum^{n}_{i=1} a_i^2 \sum^{n}_{i=1} b_i^2\geq\left(\sum^{n}_{i=1} a_ib_i \right)^2$$

  • The H$\ddot o$lder inequality (H).

    Let $a_1$, $a_2$,..., $a_n$, $b_1$, $b_2$,..., $b_n$, $\alpha$ and $\beta$ be positive numbers. Then $$\left(\sum_{i =1}^n a_i\right )^\alpha \left(\sum_{i =1}^n b_i \right )^\beta\geq \left(\sum_{i =1}^n (a_ib_i)^\frac{1}{\alpha+\beta}\right )^{\alpha+\beta} $$

  • The Schur inequalities (S):

    Let $x$, $y$ and $z$ be positive numbers and $t$ is a real number. Prove that:$$x^t(x-y)(x-z)+y^t(y-z)(y-x)+z^t (z-x)(z-y)\ge 0$$

  • Muirhead inequalities

    A sequence $a_1 \geq a_2 \geq \dots \geq a_n$ majorizes a sequence $b_1 \geq b_2 \geq \dots \geq b_n$ if $$\sum_{i=1}^k a_i \geq\sum_{i=1}^k a_i $$ for all $1\leq k < n$ and $$\sum_{i=1}^n a_i =\sum_{i=1}^n a_i $$ If sequence $(a_i)$ majorizes $(b_i)$ (notated as $a_i \succ b_i$), then $$\sum_{\text{sym}}x_1^{a_1}x_2^{a_2}\dots x_n^{a_n}\geq \sum_{\text{sym}}x_1^{b_1}x_2^{b_2}\dots x_n^{b_n}$$

30160 questions
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show that $\sum\limits_{n|i+j+k}x_{i}y_{j}z_{k}\le n^2.$

This wrong question has been asked previously on math.SE ,Now I think now is right Being given an integer $n\ge 2$, and $x_{i},y_{i},z_{i}\in \mathbb{R}$ ($i=1,2,\cdots,n$) such that $$\sum_{i=1}^{n}(x^2_{i}+y^2_{i}+z^2_{i})=3n$$ show…
math110
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Solving $x - \frac{1}{x} < 0.$

I need to solve for x: $$x - \frac{1}{x} < 0.$$ I realize that $x\neq 0$. The following is always true: $$x-\frac{1}{x} < 0 \iff x < \frac{1}{x}.$$ If $x > 0$: $$x < \frac{1}{x} \iff x^2 < 1 \iff x < \pm 1$$ $$0
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Prove the next inequality

Let $ x,y,z \in \mathbb{R_{+}^{*}} $ , with $ xy+yz+zx=1 $. Prove that: $$4(x+y+z)\leq 9xyz+\frac{1}{xyz}$$ I tried to solve the inequality with the Cardano-Tartaglia formula but I havent't succeeded so far.
ztefelina
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Problem using Jensen's Inequality

For some convex function $f$, and elements $x_{1},x_{2},x_{3}$ of its domain, show that: $$f(x_{1})+f(x_{2})+f(x_{3})+f\left( \frac{1}{3}(x_{1}+x_{2}+x_{3}\right)\ge \frac{4}{3}\left( f\left(\frac{x_{1}+x_{2}}{2} \right)+f\left(\frac{x_{2}+x_{3}}{2}…
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Given nonnegative $x_1, \cdots, x_n \geq 0$, show that $x_1 + \cdots + x_n \geq \sqrt{1} + \sqrt{3} + \cdots + \sqrt{2n-1}$

Let $x_1, \cdots, x_n \geq 0$ be real nonnegative numbers satisfying $x_1 \leq x_2 \leq \cdots \leq x_n$. For all integer $1 \leq m \leq n$, let $$ x_1^2 + \cdots + x_m^2 \geq m^2.$$ Show that $x_1 + \cdots + x_n \geq \sqrt{1} + \sqrt{3} + \cdots…
aras
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If $1\leq a < b$ show that $\sqrt{b}-\sqrt{a}\leq \frac{1}{2}(b-a)$. Need help with proof.

If $1\leq a < b$ show that $\sqrt{b}-\sqrt{a}\leq \frac{1}{2}(b-a)$. Working: $\frac{1}{2}(1-a)\leq 0<\frac{1}{2}(b-a)$ and $(1-\sqrt{a})\leq 0<\sqrt{b}-\sqrt{a}$ Trying to show $\frac{1}{2}(b-a)-(\sqrt{b}-\sqrt{a})$ is positive but don't know what…
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If $x,y,z>0$ and $x^2+7y^2+16z^2=1\;,$ Then $\min(xy+yz+zx)$

If $x,y,z>0$ and $x^2+7y^2+16z^2=1\;,$ Then $\min(xy+yz+zx)$ $\bf{My\; Try::}$Using Multiplier Method: $$f(x,y,z,\lambda) = (xy+yz+zx)-\lambda(x^2+7y^2+16z^2-1)$$ So $$f'(x,y,z,\lambda)_{\bf{y,z,\lambda=const.}} = y+z-2\lambda x$$ and…
juantheron
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if $|f(x)|\leq M_0$ and $|f''(x)\leq M_2$ for all $x\in [0,\infty)$ then $|f'(x)|\leq M_1$ for all $x\in [0,\infty)$

I tried to prove that if $f\in C^2[0,\infty)$ and there exist two positive numbers $M_0$ and $M_2$ such that for all $x\in [0,\infty)$, $|f(x)|\leq M_0$ and $|f''(x)|\leq M_2$ then there exists $M_1\in [0,\infty)$ such that $|f'(x)|\leq M_1.$ If…
Jax
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Prove that $n! \geq n^3$, for all $n \geq 6$

I want to prove that $n! \geq n^3$, for all $n \geq 6$. I tried with induction, but it doesn't seem to help as it gets even more complicated.
Octys
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Prove the inequality $xyz \geq xy+yz+xz \implies \sqrt{x y z } \geq \sqrt{x}+\sqrt{y}+\sqrt{z}$

Let $x,y,x>0$ and $xyz \geq xy+yz+xz.$ Prove that $$\sqrt{x y z } \geq \sqrt{x}+\sqrt{y}+\sqrt{z}. $$ Solution. Using AM GM inequality we have \begin{gather*} x y+xz \geq 2 \sqrt{x y x z}=2 x \sqrt{yz},\\ xy+yz \geq 2 y \sqrt{x z},\\ x z +y z…
Leox
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Prove $n^{\ln(n)} < n!$ for $n$ big enough

I want to prove $n^{\ln(n)} < n!$ for $n$ big enough, but right now all my attempts failed... I tried mathematical induction, but that's not working, I get stuck at showing: $$(n + 1) \cdot n^{\ln(n)} \geq (n+1)^{\ln(n+1)}$$ I thought about trying…
Jonny
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How to prove: $e^x \ge x^e$ for $x>0$

I stumbled upon this problem and can't find a solution. I tried taking the derivative of this function $e^x - x^e$ and got this function: $e(e^{x-1} - x^{e-1})$ and I can't prove it is positive or equal to zero because it's almost my original…
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Inequality. $\sum{(a+b)(b+c)\sqrt{a-b+c}} \geq 4(a+b+c)\sqrt{(-a+b+c)(a-b+c)(a+b-c)}.$

Let $a,b,c$ be the side-lengths of a triangle. Prove that: I. $$\sum_{cyc}{(a+b)(b+c)\sqrt{a-b+c}} \geq 4(a+b+c)\sqrt{(-a+b+c)(a-b+c)(a+b-c)}.$$ What I have tried: \begin{eqnarray} a-b+c&=&x\\ b-c+a&=&y\\ c-a+b&=&z. \end{eqnarray} So $a+b+c=x+y+z$…
Iuli
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How to prove Cauchy-Schwarz Inequality in $R^3$?

I am having trouble proving this inequality in $R^3$. It makes sense in $R^2$ for the most part. Can anyone at least give me a starting point to try. I am lost on this thanks in advance.
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Inequality. $(n+1)\left(a^{n+1}+b^{n+1}\right) \geq (a+b)\left(a^n+a^{n-1}b+\ldots+b^{n}\right). $

Let $a$ and $b$ be positive numbers, and $n \in \mathbb{N}$. Prove that (using Rearrangement Inequality) $$(n+1)\left(a^{n+1}+b^{n+1}\right) \geq (a+b)\left(a^n+a^{n-1}b+\ldots+b^{n}\right). $$ Thanks :)
Iuli
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