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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Subtraction of inequalities

Suppose I have two inequalities: $$k_1 \leq a$$ $$k_2 \leq b$$ where $k_1,k_2,a,b$ are all positive numbers I know that that summation of them can be written as: $$k_1+k_2 \leq a+b$$ But I want to find $$k_1-k_2 \leq ?$$ or $$? \leq k_1-k_2$$ Is…
kotoll
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If $p + q = 1$ prove that for any natural $n, m$ following is true: $(1 - p^n)^m + (1 - q^m)^n \ge 1$

Let $p, q \in \mathbb R$ be positive reals for which $p + q = 1$. How to prove that for any two natural numbers $n, m$ the following inequality is true? $(1 - p^n)^m + (1 - q^m)^n \ge 1$ I don't have a big knowledge about solving inequalities, so I…
J. Abraham
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Let $a,b,c>0$ and $a+b+c= 1$, how to prove the inequality $\frac{\sqrt{a}}{1-a}+\frac{\sqrt{b}}{1-b}+\frac{\sqrt{c}}{1-c}\geq \frac{3\sqrt{3}}{2}$?

Let $a,b,c>0$ and $a+b+c= 1$, how to prove the inequality $$\frac{\sqrt{a}}{1-a}+\frac{\sqrt{b}}{1-b}+\frac{\sqrt{c}}{1-c}\geq \frac{3\sqrt{3}}{2}$$?
yibotg
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Prove $\sqrt[2]{x+y}+\sqrt[3]{y+z}+\sqrt[4]{z+x} <4$

$x,y,z \geqslant 0$ and $x^2+y^2+z^2+xyz=4$, prove $$\sqrt[2]{x+y}+\sqrt[3]{y+z}+\sqrt[4]{z+x} <4$$ A natural though is that from the condition $x^2+y^2+z^2+xyz=4$, I tried a trig substitutions $x=2\cdot \cos A$, $y=2\cdot \cos B$ and $z=2\cdot…
HN_NH
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prove $\frac{1}{\sqrt[4]{a^3(a+b^2)}}+\frac{1}{\sqrt[4]{b^3(b+c^2)}}+\frac{1}{\sqrt[4]{c^3(c+a^2)}} \geqslant \frac{3}{\sqrt[4]{2}}$

$a,b,c >0$ and $abc=1$, prove $$\frac{1}{\sqrt[4]{a^3(a+b^2)}}+\frac{1}{\sqrt[4]{b^3(b+c^2)}}+\frac{1}{\sqrt[4]{c^3(c+a^2)}} \geqslant \frac{3}{\sqrt[4]{2}}$$ 1. I tried rearrangement and AM-GM but fail. 2. I think the power of $\frac14$ is tough. I…
HN_NH
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Proving $(ax+by-1)^2 \ge (x^2+y^2-1)(a^2+b^2-1)$

Let $x$, $y$, $a$, $b$ be real numbers such that $a^2+b^2 \leq 1$ and $x^2+y^2 \leq 1$. Show that $$(ax+by-1)^2 \ge (x^2+y^2-1)(a^2+b^2-1)$$ I am unable to find a solution to this problem. My initial thoughts were to have a trigonometric…
Das
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Another symmetric inequality

How would one show that for positive $a,b,c,d$ and $a+b+c+d = 4$ that $$ \frac{a}{b} + \frac{b}{c} + \frac{c}{d} + \frac{d}{a} \leq \frac{4}{abcd} $$
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prove that : $\frac{a^2+b^2}{a+b} + \frac{b^2+c^2}{b+c}+ \frac{c^2+a^2}{c+a} \geq 3$

For $a^2+b^2+c^2 =3$, with $a$, $b$ and $c$ are positive real numbers, prove that: $$\frac{a^2+b^2}{a+b} + \frac{b^2+c^2}{b+c}+ \frac{c^2+a^2}{c+a} \geq 3.$$ Can any one help me with this problem?
Aerrozard
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Very tough inequality I cannot prove

Suppose $x \geq 0 $ is real number, then $$ x^x \geq \sqrt{ 2x^x - 1 } $$ How can I show this? Also, what is the greometrical implication of such inequality ? thanks
user145801
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How prove this inequality $\sum\limits_{cyc}\frac{x^a\ln{x}}{(x^a+y+z)^2}\ge 0$

Question: let $x,y,z$ be postive numbers,and such $xyz\ge 1$,and such $a$ is real numbers.show that $$\dfrac{x^a\ln{x}}{(x^a+y+z)^2}+\dfrac{y^a\ln{y}}{(y^a+x+z)^2}+\dfrac{z^a\ln{z}}{( z^a+x+y)^2}\ge 0$$ I think this can use Integral to solve…
math110
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How prove this inequality $\frac{a+\sqrt{ab}+\sqrt[3]{abc}+\sqrt[4]{abcd}}{4}\le\sqrt[4]{\frac{a(a+b)(a+b+c)(a+b+c+d)}{24}}$

let $a,b,c,d>0$, show that $$\dfrac{a+\sqrt{ab}+\sqrt[3]{abc}+\sqrt[4]{abcd}}{4}\le\sqrt[4]{\dfrac{a(a+b)(a+b+c)(a+b+c+d)}{24}}$$ This post three -varible $a,b,c>0,a+b+c=21$ prove that $a+\sqrt{ab} +\sqrt[3]{abc} \leq 28$ How prove it Four-varible ?…
math110
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How to show that $a^ab^b+a^bb^a\leq 1$ for $a,b\in(0,1)$ with $a+b=1$.

A little while ago I came across this question on Yahoo! Answers and it had no replies. I have spent a while trying to figure this out as well, but I couldn't find an algebraic proof. The assumptions are $a+b=1$ and $a,b\in(0,1)$. We want to show…
Joffysloffy
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How prove this inequality $a+b+c+d=4$

let $a,b,c,d$ be postive numbers,and such $a+b+c+d=4$, show that $$\dfrac{a}{\sqrt{a+3b}}+\dfrac{b}{\sqrt{b+3c}}+\dfrac{c}{\sqrt{c+3d}}+\dfrac{d}{\sqrt{d+3a}}\le\dfrac{a^2}{a+b^2}+\dfrac{b^2}{b+c^2}+\dfrac{c^2}{c+d^2}+\dfrac{d^2}{d+a^2}$$
math110
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How prove this inequality $(\sum a_{1}^{1.5})^2\ge \sum a_{1}\sum a_{1}a_{2}$

Now my question let $a_{1},a_{2},\cdots,a_{n}$ are positive numbers,and $a_{n+i}=a_{i},i=1,2,\cdots$,show that $$(\sum a_{1}^{1.5})^2\ge \sum a_{1}\sum a_{1}a_{2}$$ my teacher (tian275461) have prove this $$(a^{1.5}+b^{1.5}+c^{1.5})^2\ge…
math110
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