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 $$\sum_{\text{sym}}x_1^{a_1}x_2^{a_2}\dots x_n^{a_n}\geq \sum_{\text{sym}}x_1^{b_1}x_2^{b_2}\dots x_n^{b_n}$$

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Prove $\frac{1+a^{2}}{1+b+c^{2}} +\frac{1+b^{2}}{1+c+a^{2}} +\frac{1+c^{2}}{1+a+b^{2}} \geq 2$

For positive integers Numbers $a, b, c $ prove that $$\frac{1+a^{2}}{1+b+c^{2}} +\frac{1+b^{2}}{1+c+a^{2}} +\frac{1+c^{2}}{1+a+b^{2}} \geq 2$$ This inequality above take long time to prove it and still couldn't complete it. How I can prove this…
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Given $a,b,c\ge1;abc\ge8$. Proving that $\sqrt{a^2-1}+\sqrt{b^2-1}+\sqrt{c^2-1}\ge 3\sqrt3$

Given $a,b,c\ge1;abc\ge8$. Proving that $$\sqrt{a^2-1}+\sqrt{b^2-1}+\sqrt{c^2-1}\ge 3\sqrt3$$ I have tried by using Jensen's inequality: We consider the inequality: $\displaystyle\sqrt{x^2-1}\ge\sqrt3+\frac4{\sqrt3}\ln\frac x2\tag{i}$ When…
mja
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Prove $\sum_{i=1}^{i=n}\frac{a_i}{S-a_i}\geqslant \frac{n}{n-1}$ and $\sum_{i=1}^{i=n}\frac{S-a_i}{a_i}\geqslant n(n-1)$.

Let $a_1,a_2,\ldots ,a_n$, be positive real numbers and S = $a_1+a_2+a_3+\cdots+a_n$. Use the Cauchy-Schwarz inequality to prove that $$\sum_{i=1}^{i=n}\frac{a_i}{S-a_i}\geqslant \frac{n}{n-1}$$ and $$\sum_{i=1}^{i=n}\frac{S-a_i}{a_i}\geqslant…
Faisal
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How to prove this inequality $(\sum_i x_i y_i)^2 - \sum_i x_i^2y_i^2 \leq 1-1/n$?

How to prove this inequality: $$(x_1y_1+x_2y_2+ \cdots + x_ny_n)^2 - (x_1^2y_1^2+x_2^2y_2^2+\cdots+x_n^2y_n^2)\leq 1-\frac{1}{n},$$ where $x_i,y_i \geq 0,i=1,2,\ldots,n$, and $x_1^2+x_2^2+\cdots+x_n^2=y_1^2+y_2^2+\cdots+y_n^2=1$? Clearly, the…
Doris
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Tough inequality in positive reals numbers.

Let $a, b, c$ be positive real. prove that $$\left(1+\frac{a}{b}\right)\left(1+\frac{b}{c}\right)\left(1+\frac{c}{a}\right)\geq\left(1+\frac{a+b+c}{\sqrt[3]{abc}}\right)$$ Thanks
Jarton
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My favorite proof of the generalized AM-GM inequality: where it came from?

I have already posted (most of) the present question as a (misplaced) answer to a question about understanding a particular proof of the AM-GM inequality. I sincerely hope I am not breaking the code of conduct on Stack Exchange. Let $n\geq 1$ be an…
chizhek
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Prove of Nesbitt's inequality in 6 variables

I was just reading "Pham kim hung secrets in inequalities,Volume 1" book and there was an interesting problem on it's Cauchy-Schwarz and Holder section that caught my eye. Prove that for all positive real numbers $a,b,c,d,e,f$,we always have…
user2838619
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How to divide two inequalities

I would like to know if somebody knows how to properly divide one inequality by another, as a resolution method similar to when we divide one equality by another. Take this as an example: $x^2 - y^2 \lt 8$ and $x + y \gt 3$. I want to know if it is…
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An inequality with exponentials

I am trying to show that for $t$ in $(0, T]$ for some $T > 0$ and $t |y| \leqslant C$ for some $C > 0$ that $$\tag{1} |x|^M \exp \left (-\frac{D}{1 - \rm e^{-2 t^2}} {|e^{-t^2} x - y|^2}{} \right )$$ is uniformly bounded in $x, y$ (in $\mathbf R^d$)…
JT_NL
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How find the maximum of the value $x^2_{1}+x^2_{2}+\cdots+x^2_{2014}$

Question: let $x_{i}\in[-11,5],i=1,2,\cdots,2014$,and such $$x_{1}+x_{2}+\cdots+x_{2014}=0$$ find the maximum of the value $$x^2_{1}+x^2_{2}+\cdots+x^2_{2014}$$ since $$(x_{i}-5)(x_{i}+11)\le 0$$ then $$x^2_{i}\le…
math110
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prove $\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a} \geq 3(a^2+b^2+c^2)$

If $a,b,c$ are positive real numbers and $a+b+c=1$,Prove: $$\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a} \geq 3(a^2+b^2+c^2)$$ Additional info:We can use AM-GM and Cauchy inequalities mostly.We are not allowed to use induction. Things I have done…
user2838619
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Show that $2 < e^{1/(n+1)} + e^{-1/n}$

I'm trying to show $2 < e^{1/(n+1)} + e^{-1/n}$. I can show that $ 2 < e^{1/n} + e^{-1/(n+1)}$ since $$2 \leq 2\cosh\left(\frac{1}{n}\right) = e^{1/n} + e^{-1/n} < e^{1/n} + e^{-1/(n+1)}$$ but I'm still having trouble with the other inequality. I…
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$ a+b+c+d=6 , a^2+b^2+c^2+d^2=12$ $\implies$ $ 36 \leq 4(a^3+b^3+c^3+d^3)-(a^4+b^4+c^4+d^4) \leq48 $

Let $a , b, c, d$ be real numbers such that $$ a+b+c+d=6 \\ a^2+b^2+c^2+d^2=12$$ How do we prove that $$ 36 \space \leq\space 4(a^3+b^3+c^3+d^3)-(a^4+b^4+c^4+d^4) \space\leq48\space$$ ?
user123733
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For all reals $x$, prove $2^x > x$

How can I prove that for all reals $x$, $2^x > x$? I can prove this for integers with induction, but I can't figure out how to prove it for reals. Perhaps you could say that since $2^x$ is strictly increasing, $2^x \ge 2^{floor(x)} > floor(x)$, but…
Axe
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How prove this inequality $\frac{1}{(a+1)^2+\sqrt{2(b^4+1)}}+\frac{1}{(b+1)^2+\sqrt{2(c^4+1)}}+\frac{1}{(c+1)^2+\sqrt{2(a^4+1)}}\le\frac{1}{2}$

let $a,b,c>0$,and such $abc=1$, show that $$\dfrac{1}{(a+1)^2+\sqrt{2(b^4+1)}}+\dfrac{1}{(b+1)^2+\sqrt{2(c^4+1)}}+\dfrac{1}{(c+1)^2+\sqrt{2(a^4+1)}}\le\dfrac{1}{2}$$ My idea : Use Cauchy-Schwarz inequality,we have $$2(b^4+1)=(1+1)(b^4+1)\ge…
user94270