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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Inequality. $\sqrt[k]{\frac{a_{1}^{k}+a_{2}^{k}+\ldots +a_{n}^{k}}{n}}\geq \frac{a_{1}+a_{2}+\ldots + a_{n}}{n}$

Let $a_{1}, a_{2}, \ldots, a_{n}$ and $k \geq 1$. Prove that (using Chebyshev's inequality): $$\large \sqrt[k]{\frac{a_{1}^{k}+a_{2}^{k}+\ldots +a_{n}^{k}}{n}}\geq \frac{a_{1}+a_{2}+\ldots + a_{n}}{n}.$$ I think I have a (partial) solution but…
Iuli
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Showing that $ \frac{a}{a^2+3}+\frac{b}{b^2+3}\leq\frac{1}{2}$ for $a,b > 0$ and $ab = 1$ using rearrangement inequalities

Please help to solve the following inequality using rearrangement inequalities. Let $a \gt 0$, $b \gt0$ and $ab=1$. Prove that \begin{equation}\frac{a}{a^2+3}+\frac{b}{b^2+3}\leq\frac{1}{2}.\end{equation} Thanks.
Iuli
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inequality with the $\frac{b}{a}+\frac{c}{b}+\frac{d}{c}+\frac{a}{d}\ge\frac{a+b+c+d}{\sqrt[4]{abcd}}$

Let $a,b,c,d$ be side of quadrilateral $ABCD$,prove or disprove $$\dfrac{b}{a}+\dfrac{c}{b}+\dfrac{d}{c}+\dfrac{a}{d}\ge\dfrac{a+b+c+d}{\sqrt[4]{abcd}}$$
math110
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A simple inequality...or not?

I came upon this question, $\frac1x<\frac12$ thinking that the answer was $x>2$. But the answer turned out to be (-∞, 0) ∪ (2, +∞). Why is the answer in this form, and how do you get it?
suomynonA
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Inequality $(a+\frac{1}{b})^2+(b+\frac{1}{c})^2+(c+\frac{1}{a})^2\ge 16$

For every real positive number $a,b,c$ such that $ab+bc+ca=1$, how to prove that: $$\left(a+\frac{1}{b}\right)^2+\left(b+\frac{1}{c}\right)^2+\left(c+\frac{1}{a}\right)^2\ge 16$$
Dib
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Prove this inequality $\frac{1}{xy+z}+\frac{1}{yz+x}+\frac{1}{zx+y}\le\frac{1}{2}$

Let $x,y,z>0$ and such $xy+yz+xz\ge 2(x+y+z)$,show that $$\dfrac{1}{xy+z}+\dfrac{1}{yz+x}+\dfrac{1}{zx+y}\le\dfrac{1}{2}$$
math110
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Prove inequality $\frac{x^3}y+\frac{y^3}z+\frac{z^3}x+\frac{x^3}z+\frac{z^3}y+\frac{y^3}x\ge\frac{x^2+y^2+z^2+1}2$

Let $x,y,z>0$ and $x+y+z=1$. Prove that $$\frac{x^3}y+\frac{y^3}z+\frac{z^3}x+\frac{x^3}z+\frac{z^3}y+\frac{y^3}x\ge\frac{x^2+y^2+z^2+1}2$$ My work so far: I use Titu's…
Roman83
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Prove inequality $2\le\frac{a+b}{c+1} +\frac{b+c}{a+1} +\frac{c+a}{b+1} \le3$

Assume that $a,b,c$ are real numbers from the interval $(\frac{1}{2},1)$. What is the proof that $$2\le\frac{a+b}{c+1} +\frac{b+c}{a+1} +\frac{c+a}{b+1} \le3$$ holds?
rack
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Solve the following using AM-GM inequality

The least value of $a \in R$ for which $4ax^2 + \frac{1}{x} \ge 1 $for all $x \gt 0 $, is Using AM-GM inequality $$\frac{4ax^2 + \frac{1}{2x} + \frac{1}{2x}}{3} \ge \sqrt[3]{a}$$ $$4ax^2 + \frac{1}{x} \ge 3\sqrt[3]{a}$$ Now my question start from…
Aakash Kumar
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$3(x-\sqrt{xy}+y)^2\geq x^2+xy+y^2$

Prove that $3(x-\sqrt{xy}+y)^2\geq x^2+xy+y^2$ for all $x,y\geq 0$. Expanding, the inequality becomes $$3x^2+3xy+3y^2-6x\sqrt{xy}-6y\sqrt{xy}+6xy\geq x^2+xy+y^2$$ which is $$x^2+4xy+y^2\geq3\sqrt{xy}(x+y)$$ We can try using AM-GM: $$x^2+xy+xy\geq…
pi66
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Show that $x^2 \sin{x} + x \cos{x} + x^2 + \frac{1}{2} > 0$

Show that for any real number $x$: $$x^2 \sin{x} + x \cos{x} + x^2 + \frac{1}{2} > 0.$$ $\bf{My\; Try::}$ Using $a\sin x+b\cos x\geq -\sqrt{a^2+b^2}$ So $$x^2\sin x+x\cos x\geq -\sqrt{x^4+x^2}=-x\sqrt{1+x^2}$$ and…
juantheron
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Prove that $\sin x + 2x \ge \frac{3x.(x+1)}{\pi} $

Prove that $$\sin x + 2x \ge \frac{3x.(x+1)}{\pi}\quad\forall x \in \left[0,\frac{\pi}{2}\right].$$ My work: $$ 3x^2 + (3-2\pi )x - \pi \sin x \le 0 $$ $$ f(x) = 3x^2 + (3-2\pi )x -\pi \sin x $$ $$f(0)=0 $$ $$f\left({\pi\over 2 }\right)…
Aakash Kumar
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An inequality of three positive variables

Suppose that $x,y,z>0$ and $xyz=1$. Why does the inequality $$\frac{1}{x+y+1}+\frac{1}{y+z+1}+\frac{1}{x+z+1}\leq 1$$ holds? I couldn't see anything useful as I tried Jensen's inequality and calculus methods.
student
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The validity of normalization in homogeneous inequalities?

I'm going through a book on inequalities right now, and the author describes normalization with the following example. Prove that $a^2 + b^2 + c^2 \ge ab + bc + ca$ Of course the fundamental proof of this inequality has already been discussed…
Airdish
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Prove that $ A=\sqrt{\frac{(b-c)^2}{a^2}+\frac{(c-a)^2}{b^2}+\frac{(a-b)^2}{c^2}} $ is also rational number.

Let $a,b,c\in\mathbb{Q}$ distinct and none of them equal to $0$ satisfying $\frac{a^2}{(b-c)^2}+\frac{b^2}{(c-a)^2}+\frac{c^2}{(a-b)^2}\leq 2. $ Prove that $ A=\sqrt{\frac{(b-c)^2}{a^2}+\frac{(c-a)^2}{b^2}+\frac{(a-b)^2}{c^2}} $ is also rational…
piteer
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