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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$x,y,z \geqslant 0$ and $x^2+y^2+z^2+xyz=4$, prove $x^{\frac85}+y^{\frac85}+z^{\frac85} \geqslant 3$

$x,y,z \geqslant 0$ and $x^2+y^2+z^2+xyz=4$, prove $$x^{\frac85}+y^{\frac85}+z^{\frac85} \geqslant 3$$ 1) The equality occurs only at $x,y,z=1$. Let's assume $F=x^n+y^n+z^n$, I noticed that $n=1$ then $F \leqslant 3$ and $n=2$ then $F \geqslant 3$.…
HN_NH
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$a,b,c >0$, and $ab+bc+ca=3$, prove $(a^ab^bc^c)^{\frac{3}{a+b+c}} \geqslant \sqrt[3]{\frac{a^3+b^3+c^3}{3}}$

$a,b,c >0$, and $ab+bc+ca=3$, prove $$(a^ab^bc^c)^{\frac{3}{a+b+c}} \geqslant \sqrt[3]{\frac{a^3+b^3+c^3}{3}}$$ I think the equality is only achieve when $a=b=c=1$. The condition $ab+bc+ca=3$ is necessary. I used the estimation $x^x \geqslant…
HN_NH
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If $(x_1-a)(x_2-a)\cdots(x_n-a)=k^n$ prove by using the laws of inequality that $x_1x_2 \cdots x_n\geq (a+k)^n$

If $x_i>a>0$ for $i=1,2\cdots n$ and $(x_1-a)(x_2-a)\cdots(x_n-a)=k^n$, $k>0$, prove by using the laws of inequality that $$x_1x_2 \cdots x_n\geq (a+k)^n$$. Attempt: If we expand $(x_1-a)(x_2-a)\cdots(x_n-a)=k^n$ in the LHS, we get $x_1x_2 \cdots…
user1942348
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Prove that $2^{n(n+1)}>(n+1)^{n+1}\left(\frac{n}{1}\right)^n\left(\frac{n-1}{2}\right)^{n-1}\cdots \left(\frac{2}{n-1}\right)^{2}\frac{1}{n}$

If $n$ be a positive integer $>1$, prove that $$2^{n(n+1)}>(n+1)^{n+1}\left(\frac{n}{1}\right)^n\left(\frac{n-1}{2}\right)^{n-1}\left(\frac{n-2}{3}\right)^{n-2}\cdots \left(\frac{2}{n-1}\right)^{2}\frac{1}{n}$$ Please help me to prove the above. I…
user1942348
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Very strong inequality

Let $a$, $b$ and $c$ be non-negative numbers. Prove that: $$a^3+b^3+c^3+3abc\geq(a+b-c)\sqrt{ab(a+c)(b+c)}+(a+c-b)\sqrt{ac(a+b)(b+c)}+(b+c-a)\sqrt{bc(a+b)(a+c)}$$ I have a proof, but my proof is long. Maybe there is something simple? Thanks!
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Prove that $(a+b)(b+c)(c+a) \ge8$

Given that $a,b,c \in \mathbb{R}^{+}$ and $abc(a+b+c)=3$, Prove that $(a+b)(b+c)(c+a)\ge8$. My attempt: By AM-GM inequality, we have $$\frac{a+b}{2}\ge\sqrt {ab} \tag{1}$$ and similarly $$\frac{b+c}{2}\ge\sqrt {bc}…
Snehil Sinha
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Prove that $\left(\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\right)^2\ge{(a+b+c)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)}$

Let $a,b,c\in{\mathbb{R^+}}$. Prove that $$\left(\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\right)^2\ge{(a+b+c)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)}$$ I tried to expand both, but I did not get anything useful.
Satvik Mashkaria
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Find maximum without calculus

Let $f:(0,1]\rightarrow\mathbb{R}$ with $f(x)=2x(1+\sqrt{1-x^2})$. Is it possible to find the maximum of this function without calculus? Possibility through some series of inequalities?
1-___-
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Prove inequality using AM-GM inequality.

$$\sqrt{a}+\sqrt{b}\geq \sqrt{a+b}$$ I know it is trivial to prove straight forward but I need to prove it using AM-GM inequality.
Bakyr
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Prove two of $\frac{2}{a}+\frac{3}{b}+\frac{6}{c}\geq 6,\frac{2}{b}+\frac{3}{c}+\frac{6}{a}\geq 6,\frac{2}{c}+\frac{3}{a}+\frac{6}{b}\geq 6$ are True

if $a,b,c$ are positive real numbers that $a+b+c\geq abc$, Prove that at least $2$ of following inequalities are true. $\frac{2}{a}+\frac{3}{b}+\frac{6}{c}\geq 6, \space\space\space\space\frac{2}{b}+\frac{3}{c}+\frac{6}{a}\geq 6,…
user2838619
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Prove $\frac{a^2+b^2+c^2}{ab+bc+ca} + 8\frac{abc}{(a+b)(b+c)(c+a)} \ge 2$

Let $a,b,c>0$, prove that $$\frac{a^2+b^2+c^2}{ab+bc+ca}+\frac{8abc}{(a+b)(b+c)(c+a)}\ge 2.$$ I tried using the equality $(a+b)(b+c)(c+a)=(a+b+c)(ab+bc+ca)-abc$ and the Schur inequality but it's not very helpful. Thanks.
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Prove the inequality $\frac 1a + \frac 1b +\frac 1c \ge \frac{a^3+b^3+c^3}{3} +\frac 74$

Inequality Let $a$, $b$ and $c$ be positive numbers such that $a+b+c=3$. Prove the following inequality $$\frac 1a + \frac 1b +\frac 1c \ge \frac{a^3+b^3+c^3}{3} +\frac 74.$$ I stumbled upon this question some days ago and been trying AM-GM to…
Kalpan
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$a,b,c \geq 0$ and $a+b+c=3$ prove that $\frac{a^2+bc}{b+ac} + \frac{b^2+ac}{c+ab} + \frac{c^2+ab}{a+bc} \geq 3$

$a,b,c \geq 0$ and $a+b+c=3$ prove that $\frac{a^2+bc}{b+ac} + \frac{b^2+ac}{c+ab} + \frac{c^2+ab}{a+bc} \geq 3$ can anyone help me solve this problem,i've tried to use C-S and also AM-GM for couple in cycle.Sorry for my bad English that caused me…
Aerrozard
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prove $\frac {ab}{a+b} \geq \sum ^n_{i=1} \frac{a_ib_i}{a_i+b_i}$

If $a_i$ and $b_i$ are positive, and $b= \sum ^n_{i=1} b_i$,$a= \sum ^n_{i=1} a_i$ prove $$\frac {ab}{a+b} \geq \sum ^n_{i=1} \frac{a_ib_i}{a_i+b_i}$$ Additional: we should just use Cauchy inequality. However , if it's not possible, solve it…
user2838619
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How to prove $(1-\sum_{i=1}^{n}a^3_{i})^{1/3}\cdot (1-\sum_{i=1}^{n}b^3_{i})^{1/3}\ge 1-\sum_{i=1}^{n}a_{i}b_{i}-\sum_{i=1}^{n}|a_{i}-b_{i}|$?

let $a_{1},a_{2},\cdots,a_{n},b_{1},b_{2},\cdots,b_{n}$ be postive numbers, and such $$\sum_{i=1}^{n}a^2_{i}\le 1,\sum_{i=1}^{n}b^2_{i}\le 1$$ show that …
math110
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