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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How do I show that $\frac a{1 - a^2} + \frac b{1 - b^2} + \frac c{1 - c^2} \ge \frac {3 \sqrt 3}2$

For $0 \lt a, b, c \lt 1$, if $ab + bc + ca = 1$, show that $$\frac a{1 - a^2} + \frac b{1 - b^2} + \frac c{1 - c^2} \ge \frac {3 \sqrt 3}2.$$ I want to use trigonometric substitution: For the angles $A, B, C$ of any acute triangle, $$\tan A + \tan…
Yuxiao Xie
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When does Poincaré inequality hold?

Poincaré inequality is given by $$\int_\Omega u^2\le C\int_\Omega|\nabla u|^2dx ,$$ where $\Omega$ is bounded open region in $\mathbb R^n$. However this inequality is not satisfied by all the function. Take for example a constant function $u=10$…
Theorem
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Let $x,y,z>0$ and $x+y+z=1$, then find the least value of ${{x}\over {2-x}}+{{y}\over {2-y}}+{{z}\over {2-z}}$

Let $x,y,z>0$ and $x+y+z=1$, then find the least value of $${{x}\over {2-x}}+{{y}\over {2-y}}+{{z}\over {2-z}}$$ I tried various ways of rearranging and using AM > GM inequality. But I couldn't get it. I am not good at inequalities. Please help…
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Prove that $0 \leq ab^2-ba^2 \leq \frac{1}{4}$ with $0 \leq a \leq b \leq 1$.

Let $a$ and $b$ be real numbers such that $0 \leq a \leq b \leq 1$. Prove that $0 \leq ab^2-ba^2 \leq \dfrac{1}{4}$. Attempt We can see that $ab^2-ba^2 = ab(b-a)$, so it is obvious that it is greater than or equal to $0$. But how do I show it is…
user19405892
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Find the Range of $f(a,b)=2a+b-3ab$

Let $a,b>0$,and such $$a^2+b^2-ab=4$$ Find the range $$f(a,b)=2a+b-3ab$$ I try let $a=x+y,b=x-y$,then $$a^2+b^2-ab=4\Longrightarrow x^2+3y^2=16,x>y,x>-y$$ so we Let $$\begin{align} x &=4\cos{t}, \qquad y=\dfrac{4}{\sqrt{3}}\sin{t} \\ f(a,b) &=…
user246688
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How can one show that $|(a-b)(b-c)(c-d)(d-a)|\le\frac{abcd}{4}$

Given that a,b,c,d real numbers in the interval [1,2] How can one show that : $|(a-b)(b-c)(c-d)(d-a)|\le\frac{abcd}{4}$ And when equality occurs I tried using AM GM but no results
user233658
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Upper bound for $2(x+y+z)-3(xy+yz+zx)+4xyz$

Let $x,y,z\geq 0$ and $x+y+z\leq\frac12$. What is the maximum of $$S=2(x+y+z)-3(xy+yz+zx)+4xyz?$$ When $x=y=z\leq\frac{1}{6}$, we have $S=6x-9x^2+4x^3$, which is an increasing function in $[0,\frac16]$, so the maximum is attained when…
Alexi
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Absolute value inequality, where am I wrong?

Tried to solving $|x^2-5x+5|<1$ using the square method, but I don't know what I did…
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Multiplicating inequalities

I have two inequalities: $|x|\leq\sqrt{x^2+y^2}$ and $|y|\leq\sqrt{x^2+y^2}, \forall x,y \in \Bbb R$, can I multiply these inequalities to get $|xy|\leq x^2+y^2$? If yes, what is the justification? If not, why?
YoTengoUnLCD
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Application of Jensen's inequality to $x^x+y^y+z^z$

Claim: If $x, y, z >0$ and $x+y+z = 3\pi, $ then $x^x + y^y + z^z > 81.$ My attempt: Let $f(w) = w^w$, so $f$ is convex on $(0, \infty).$ By Jensen's inequality, $f(x\frac{x}{3\pi}+ y\frac{y}{3\pi} + z\frac{z}{3\pi}) \leq \frac{x}{3\pi}f(x) + …
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$\frac{1}{n}+\frac{1}{n+1}+\cdots+\frac{1}{2n}>\frac{2}{3}$

This is from the book Problems in Mathematical Analysis I by Kaczor and Nowak: Show that, for $n\in \mathbb{N}$, $$\frac{1}{n}+\frac{1}{n+1}+\cdots+\frac{1}{2n}>\frac{2}{3}$$ The solution in the back of the book says to apply the Arithmetic…
user140776
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$\frac{x}{\sqrt{yz}+\sqrt{3}}+\frac{y}{\sqrt{xz}+\sqrt{3}}+\frac{z}{\sqrt{yx}+\sqrt{3}}\leq \frac{1}{4\sqrt{3}xyz}$

Let $x;y;z>0$ such that: $xy+yz+zx=1$. Prove that: $\frac{x}{\sqrt{yz}+\sqrt{3}}+\frac{y}{\sqrt{xz}+\sqrt{3}}+\frac{z}{\sqrt{yx}+\sqrt{3}}\leq \frac{1}{4\sqrt{3}xyz}$ I think: $1=xy+yz+zx\geq 3\sqrt[3]{x^2y^2z^2}\Rightarrow xyz\leq…
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If $a+b+c\le 1$ then $3(a+b+c)-(a^2+b^2+c^2-ab-bc-ac)\ge (\sqrt{a}+\sqrt{b}+\sqrt{c})^2$

Let $a,b,c\ge 0$ such that $a+b+c\le 1$, prove that $$3(a+b+c)-(a^2+b^2+c^2-ab-bc-ac)\ge (\sqrt{a}+\sqrt{b}+\sqrt{c})^2\tag{1}$$ I conjecture: Let $a_{i}\ge 0$, $i=1,2,\cdots$, $a_{1}+a_{2}+\cdots+a_{n}\le 1$, $n\ge 3$,then…
user237685
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Prove the following inequality $\sqrt[n]{x_1x_2...x_n}\ge (n-1)$

If we have that $\frac{1}{1+x_1}+\frac{1}{1+x_2}+...+\frac{1}{1+x_n}=1$, then prove that $\sqrt[n]{x_1x_2...x_n}\ge (n-1)$. Where $x_1,x_2,...,x_n$ are all non negative real numbers. I got $x_1+x_2+...+x_n\ge n(n-1)$. Also we have…
CryoDrakon
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The inequality. Regional olympiad 2015

Let $a, b, c$ - the positive real numbers, and $ab+bc+ca=1$ Prove that $\sqrt{a+\frac{1}{a}}+\sqrt{b+\frac{1}{b}}+\sqrt{c+\frac{1}{c}} \geqslant 2(\sqrt{a}+\sqrt{b}+\sqrt{c})$ Probably, we should use these facts: $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}…