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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Strange Mean Inequality

This problem was inspired by this question. $\sqrt [ 3 ]{ a(\frac { a+b }{ 2 } )(\frac { a+b+c }{ 3 } ) } \ge \frac { a+\sqrt { ab } +\sqrt [ 3 ]{ abc } }{ 3 } $ The above can be proved using Hölder's inequality. $\sqrt [ 3 ]{ a(\frac { a+b }{ 2 }…
Chad Shin
  • 2,132
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How i show this beautiful inequality :$\frac{x^n}{x^m+y^m}+\frac{y^n}{y^m+z^m}+\frac{z^n}{z^m+x^m}\geq \frac{3} {2}(\frac{1}{\sqrt{3}})^{n-m}$?

let $m,n$ be integers, show that if $ n>m\geq 0 $ : $$\frac{x^n}{x^m+y^m}+\frac{y^n}{y^m+z^m}+\frac{z^n}{z^m+x^m}\geq \frac{3} {2}\left(\frac{1}{\sqrt{3}}\right)^{n-m}$$ where real $x,y,z > 0 $ and $xy + yz + zx = 1$ Thank you for your help .
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prove $ \frac{1}{13}<\frac{1}{2}\cdot \frac{3}{4}\cdot \frac{5}{6}\cdot \cdots \cdot \frac{99}{100}<\frac{1}{12} $

Without the aid of a computer,how to prove $$ \frac{1}{13}<\frac{1}{2}\cdot \frac{3}{4}\cdot \frac{5}{6}\cdot \cdots \cdot \frac{99}{100}<\frac{1}{12} $$
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How prove this Lower bound with $\sum_{i=1}^{n}\prod_{j=1}^{i}a_{j}$

Let $a_{i}\in [0,1]$,and $a_{1}\ge a_{2}\ge\cdots \ge a_{n}$,show that $$a_{1}+a_{1}a_{2}+a_{1}a_{2}a_{3}+\cdots+a_{1}a_{2}\cdots a_{n}\ge \sum_{i=1}^{n}\left(\dfrac{a_{1}+a_{2}+\cdots+a_{n}}{n}\right)^i$$ For $n=2$,we have $$a_{1}+a_{1}a_{2}\ge…
math110
  • 93,304
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Struggling with an inequality: $ \frac{1}{\sqrt[n]{1+m}} + \frac{1}{\sqrt[m]{1+n}} \ge 1 $

Prove that for every natural numbers, $m$ and $n$, this inequality holds: $$ \frac{1}{\sqrt[n]{1+m}} + \frac{1}{\sqrt[m]{1+n}} \ge 1 $$ I tried to use Bernoulli's inequality, but I can't figure it out.
haninja
  • 185
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Prove that $x_1^2+x_2^2+x_3^2=1$ yields $ \sum_{i=1}^{3}\frac{x_i}{1+x_i^2} \le \frac{3\sqrt{3}}{4} $

Prove this inequality, if $x_1^2+x_2^2+x_3^2=1$: $$ \sum_{i=1}^{3}\frac{x_i}{1+x_i^2} \le \frac{3\sqrt{3}}{4} $$ So far I got to $x_1^4+x_2^4+x_3^4\ge\frac{1}3$ by using QM-AM for $(2x_1^2+x_2^2, 2x_2^2+x_3^2, 2x_3^2+x_1^2)$, but to be honest I'm…
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How prove this inequality $\left(1-\frac{1}{3}\right)\left(1-\frac{1}{3^2}\right)\cdots\left(1-\frac{1}{3^n}\right)\ge\frac{14}{25}$

show that $$\left(1-\dfrac{1}{3}\right)\left(1-\dfrac{1}{3^2}\right)\cdots\left(1-\dfrac{1}{3^n}\right)\ge\dfrac{14}{25}\tag{1}$$ My try: I only prove following not strong…
math110
  • 93,304
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$a+b+c =0$; $a^2+b^2+c^2=1$. Prove that: $a^2 b^2 c^2 \le \frac{1}{54}$

If a,b,c are real numbers satisfying $a+b+c =0; a^2+b^2+c^2=1$. Prove that $a^2 b^2 c^2 \le \frac{1}{54}$.
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An inequality $ k\frac{\sum_{i\neq j} x_i x_j ( (1-x_i-x_j)^{k-1}- (1-x_i)^k(1-x_j)^k)}{\sum_{i=1}^n x_i (1-(1-x_i)^k)} \leq 2$

Let $x_1,\dots, x_n \geq 0$ be a sequence of numbers such that $\sum_{i=1}^n x_i = 1$. For every $k \geq 1$, I conjecture (and need to prove) that $$ \frac{\sum_{1\leq i\neq j\leq n} x_i x_j \left( (1-x_i-x_j)^{k-1}-…
Clement C.
  • 67,323
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Prove that $\frac{a}{b^2+5}+ \frac{b}{c^2+5} + \frac{c}{a^2+5} \le \frac 12$

Let $a,b,c>0$ and $a^3+b^3+c^3=3$. Prove that $$\dfrac{a}{b^2+5}+ \dfrac{b}{c^2+5} + \dfrac{c}{a^2+5} \le \dfrac 12$$ I have an ugly solution for this solution.
boxmath
  • 121
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Inequality:$ (a^{2}+c^{2})(a^{2}+d^{2})(b^{2}+c^{2})(b^{2}+d^{2})\leq 25$

For $ a,b,c,d\geq 0 $ with $ a+b = c+d = 2 $, how to prove that $$ (a^{2}+c^{2})(a^{2}+d^{2})(b^{2}+c^{2})(b^{2}+d^{2})\leq 25$$
Rib
  • 121
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Prove inequality $\frac{a}{b}+\frac{b}{c}+\frac{c}{a}+\frac{9\sqrt[3]{abc}}{a+b+c}\geq 6$

Let $a,b,c>0$. What is the proof that: $$\frac{a}{b}+\frac{b}{c}+\frac{c}{a}+\frac{9\sqrt[3]{abc}}{a+b+c}\geq 6$$
Rose
  • 139
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Prove this inequality $f(x_{1})+f(x_{2})+\cdots+f(x_{n})\ge\frac{n^{q+1}}{n^{q-p}-1}$

let $x_{i}>0,i=1,2,\cdots,n$,and such $x_{1}+x_{2}+\cdots+x_{n}=1$,Define function $$f(x)=\dfrac{1}{x^p-x^{q}},p>0,q\ge 1,-1
user253631
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with this inequality $\ln{x}\ln{(1-x)}<\sqrt{x(1-x)}$

If $0
user246384
11
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Prove $(1-a)(1-b)(1-c)(1-d)\geq abcd$ if $a^2+b^2+c^2+d^2=1$

Let $a,b,c,d\geq0$, $a^2+b^2+c^2+d^2=1$ Prove $\displaystyle (1-a)(1-b)(1-c)(1-d)\geq abcd$ I mutiplied both with $\displaystyle (1+a)(1+b)(1+c)(1+d)$ to use $1-a^2=b^2+c^2+d^2$ and try using the cauchy-schwarz and holder but it is doesn't work.