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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$\frac{1}{15}<(\frac{1}{2}\cdot\frac{3}{4}\cdot\frac{5}{6}\cdot \cdot \cdot \cdot\frac{99}{100})<\frac{1}{10}$.

Show that $$\frac{1}{15}<(\frac{1}{2}\cdot\frac{3}{4}\cdot\frac{5}{6}\cdot \cdot \cdot\cdot\frac{99}{100})<\frac{1}{10}$$ My attempt: This problem is from a text book where is introduced as: https://en.wikipedia.org/wiki/Generalized_mean This…
josf
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Inequality. $abc(a+b+c) > 3abc+ab+bc+ca.$

I want to ask you a solution for the following problem. Let $a,b,c$ be real numbers, $a,b,c > \frac{1+\sqrt{5}}{2}$. Prove that: $$abc(a+b+c) > 3abc+ab+bc+ca.$$ I don't know how "to touch" this problem, I tried to use $AG \geq GM$, but also is a…
Iuli
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How to prove $\frac{a}{a+bc}+\frac{b}{b+ac}+\frac{c}{c+ab}\geq \frac{3}{2}$

With $a + b + c = 3$ and $a, b, c>0$ prove these inequality: 1)$$\frac{a}{a+bc}+\frac{b}{b+ac}+\frac{c}{c+ab}\geq \frac{3}{2}$$ 2)$$\frac{ab}{c}+\frac{bc}{a}+\frac{ca}{b}+abc\geq 4$$ 3)$$\frac{ab}{c}+\frac{bc}{a}+\frac{ca}{b}+\frac{9}{4}abc\geq…
Xeing
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How find the maximum of the value $\sum_{i=1}^{n}x_{i}|x_{i}-x_{i+1}|$

Let $n\ge 4$ be give postive integers,and $x_{i}\in [0,1](i=1,2,\cdots,n)$,find the maximum of the value $$\sum_{i=1}^{n}x_{i}|x_{i}-x_{i+1}|$$ where $x_{n+1}=x_{1}$ I think the maximum of the value for some $x_{i}=0$ and for some $x_{j}=1$.But How…
math110
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Inequality: $7a+5b+12ab\le9$

If we assume that $a,b$ are real numbers such that $9a^2+8ab+7b^2\le 6$, how to prove that : $$7a+5b+12ab\le9$$
rab
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Prove that $\frac{n^n}{3^n} < n! < \frac{n^n}{2^n} \forall n \geq 6 $

Prove that $$\frac{n^n}{3^n} < n! < \frac{n^n}{2^n} \forall n \geq 6 $$ I'm trying induction, this is what I have so far: Basecase $(n=6): 64<720<729$ is true. Inductive Case: Assume the inequality $\frac{n^n}{3^n} < n! < \frac{n^n}{2^n} $ holds $…
user39794
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Proving: $x\left ( 1-y \right )+y\left ( 1-z \right )+z\left ( 1-x \right )< 1$

What is the proof that: $x\left ( 1-y \right )+y\left ( 1-z \right )+z\left ( 1-x \right )< 1$ if: $0< x;y;z< 1$
seri
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Prove inequality $\sqrt{\frac{1}n}-\sqrt{\frac{2}n}+\sqrt{\frac{3}n}-\cdots+\sqrt{\frac{4n-3}n}-\sqrt{\frac{4n-2}n}+\sqrt{\frac{4n-1}n}>1$

For any $n\ge2, n \in \mathbb N$ prove that $$\sqrt{\frac{1}n}-\sqrt{\frac{2}n}+\sqrt{\frac{3}n}-\cdots+\sqrt{\frac{4n-3}n}-\sqrt{\frac{4n-2}n}+\sqrt{\frac{4n-1}n}>1$$ My work so far: 1) $$\sqrt{n+1}-\sqrt{n}>\frac1{2\sqrt{n+0.5}}$$ 2)…
Roman83
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Inequality problem, for positive $a,b,c$, if $abc=1$, then $\frac{1}{1+a+b^2}+\frac{1}{1+b+c^2}+\frac{1}{1+c+a^2}\leq1$

I need help or guidance in solving this inequality that I am battling for 3 days now. I have tried everything that comes to mind, but I am stuck. The inequality is as $$\sum_\textrm{cyc}\frac{1}{1+a+b^2}\leq1,\ abc=1, \ a,b,c>0$$ This can be…
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show that $\frac { 1 }{ 1-a } +\frac { 1 }{ 1-b } +\frac { 1 }{ 1-c } \ge \frac { 2 }{ 1+a } +\frac { 2 }{ 1+b } +\frac { 2 }{ 1+c } $

Let $a,b,c$ are positive numbers,if $$a+b+c=1$$ show that $$\frac { 1 }{ 1-a } +\frac { 1 }{ 1-b } +\frac { 1 }{ 1-c } \ge \frac { 2 }{ 1+a } +\frac { 2 }{ 1+b } +\frac { 2 }{ 1+c } $$ I am tried proving it but failed.Any hints will be…
unicornki
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Strengthen inequality

It is known that for positive integers $ a, b, c, d $ and $n $, the inequalities $ a + c
Roman83
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Prove that $0 \leq ab + ac + bc - abc \leq 2.$

Let $a,b,$ and $c$ be nonnegative real numbers such that $a^2+b^2+c^2+abc = 4$. Prove that $$0 \leq ab + ac + bc - abc \leq 2.$$ I tried using rearrangement to get $a^2+b^2+c^2+abc = 4 \geq ab+bc+ac+abc$. Then I just need to show that $0\leq ab +…
Puzzled417
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prove that $(x-y)^3+(y-z)^3+(z-x)^3 > 0.$

If $x 0.$$ Attempt We have that $(x-y)^3+(y-z)^3+(z-x)^3=-3 x^2 y+3 x^2 z+3 x y^2-3 x z^2-3 y^2 z+3 y z^2$. Grouping yields $3(x^2z+xy^2+yz^2)-3(x^2y+xz^2+y^2z)$. Then I was thinking of…
user19405892
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Prove that $ \sqrt{\frac{a}{b+3} } +\sqrt{\frac{b}{c+3} } +\sqrt{\frac{c}{a+3} } \leq \frac{3}{2} $

For all $a, b, c>0$ and $a+b+c=3$ Prove that $$ \sqrt{\frac{a}{b+3} } +\sqrt{\frac{b}{c+3} } +\sqrt{\frac{c}{a+3} } \leq \frac{3}{2} $$ I tried cauchy-schwarz inequality for the L. H. S like and I get $ [\left( \sqrt{a} \right) ^{2}+\left(…