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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Proof of the Bergström inequality using Cauchy

$\sum_{i=1}^n \frac{a_i^2}{b_i} \geq \frac{(a_1+...+a_n)^2}{b_1+...+b_n}$ for real $a_i, b_i > 0$. I'm told it follows from Cauchy-Schwarz, but I just don't see it.
user2520938
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$2^x + 3^x - 4^x + 6^x - 9^x ≤ 1$ $\forall x \in R$

How can i prove that $2^x + 3^x - 4^x + 6^x - 9^x ≤ 1$ $\forall x \in R$. I tried $log(2^x + 3^x - 4^x + 6^x - 9^x) = log (1296^x) = x log(1296)$ i don't know if im correct here i stuck some help please
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Inequality $\left(a-1+\frac{1}b\right)\left(b-1+\frac{1}c\right)\left(c-1+\frac{1}a\right)\leq1$

Let $a,b,c\in\mathbb{R}^*_+$, $abc=1$. How can i show that $\left(a-1+\frac{1}b\right)\left(b-1+\frac{1}c\right)\left(c-1+\frac{1}a\right)\leq1$ ? I got $\left(ab-b+1\right)\left(bc-c+1\right)\left(ca-a+1\right)\leq abc=1$, but i can't go any…
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How prove $\frac{\sin{a}-\sin{c}}{a-c}>\frac{\sin{b}-\sin{d}}{b-d}$

Question: let $a,b,c,d$ such $0\dfrac{\sin{b}-\sin{d}}{b-d}$$ My idea: if we use Mean value theorem then there…
math110
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Prove that $x^3 + x^3 y^2 + x^2 y^3 + x^2 + y^2 + y^3 \geq 2xy(x+y+xy)$

Prove that $x^3 + x^3 y^2 + x^2 y^3 + x^2 + y^2 + y^3 \geq 2xy(x+y+xy)$ for $x,y \in \mathbb{R}^+$. I started by multiplying everything out on the RHS to get the equivalent statement \begin{align*} x^3 + x^3 y^2 + x^2 y^3 + x^2 + y^2 + y^3 \geq…
stochasm
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How prove $\frac{a^2}{a+2b^2}+\frac{b^2}{b+2c^2}+\frac{c^2}{c+2a^2}\ge 1$

let $a,b,c$ be postive real numbers ,and such $$ab+bc+ac=3$$ show that $$\dfrac{a^2}{a+2b^2}+\dfrac{b^2}{b+2c^2}+\dfrac{c^2}{c+2a^2}\ge 1$$ This problem is from Secrets In Inequalities volume 1 page 30,example 1.24. the comment.the author…
math110
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How to prove $(1+\frac{1}{k})^k < e < (1+\frac{1}{k})^{k+1}$?

I understand $(1+\frac{1}{k})^k$ converges as $e$ when $k$ goes to infinity. However, how to prove the inequality above? Please give me some hints.
athos
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A Very Hard Inequality

Find the smallest constant $c$ such that for any positive integers $a_1,a_2,\ldots,a_n$ for $n \geq 3$, the following inequality holds: \begin{align} \frac{a_1}{a_2+a_3}+\frac{a_2}{a_3+a_4}+\cdots+\frac{a_n}{a_1+a_2}\geq cn. \end{align}
James
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Show $a^2+b^2+c^2=1$ $\implies$ $ \dfrac {b^2c^2}{1+a^2} +\dfrac {c^2a^2}{1+b^2}+\dfrac {a^2b^2}{1+c^2} \le \dfrac 14 $

If $a$, $b$ and $c$ are real numbers such that $a^2+b^2+c^2=1$ , then how to prove that $ \dfrac {b^2c^2}{1+a^2} +\dfrac {c^2a^2}{1+b^2}+\dfrac {a^2b^2}{1+c^2} \le \dfrac 14 $ ( don't apply Schur's inequality )? Thanks.
user123733
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Tough Inequality

I was doing some problems for Olympiad training and encountered this: How would you prove that $(a+b+c+d)-(a+c)(b+d)\geq 1$? We are told that $0
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Show $\frac{b}{c} + \frac{c}{a} + \frac{a}{b} \ge 3$ for $a,b,c > 0$.

Disclaimer: The statement may be false, but for now I'm operating under the assumption it's true and trying to prove it. My workings: I got a common denominator and expressed it as: $$\frac{ab^2 + bc^2 + ca^2}{abc} \ge 3$$ $$ab^2 + bc^2 + ca^2 \ge…
Grid
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How prove this $|1+x|^a\ge 1+ax+\dfrac{1}{1000}|x|^a$

let $2\le a\le 13,a\in R$,and $x\in R$,show that: $$|1+x|^a\ge 1+ax+\dfrac{1}{1000}|x|^a\tag{1}$$ My try: let $$f(x)=|1+x|^a-1-ax-\dfrac{1}{1000}|x|^a$$ and since if $x>-1$,then $$|1+x|^a=(1+x)^a=1+ax+\dfrac{a(a-1)}{2}x^2+\cdots+x^a$$ and I…
math110
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