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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Prove that $\left(\frac{a}{b}+ \frac{b}{c}+\frac{c}{a}+x \right) \left(\frac{a^2}{b} + \frac{b^2}{c}+\frac{c^2}{a}+y \right) \ge z$ for $a+b+c=1$

Let $a,b,c>0$ such that $a+b+c=1$. Prove that for all $m>0$, the following inequality holds: \begin{align*} \left(\frac{a}{b} + \frac{b}{c} + \frac{c}{a} -\frac{m^4+2m^3-8m-4}{m^2} \right)…
Baby LE
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Show that $\frac{a^2+1}{b+c}+\frac{b^2+1}{a+c}+\frac{c^2+1}{a+b} \ge 3$

If $a,b,c$ are positive numbers then show that $$\dfrac{a^2+1}{b+c}+\dfrac{b^2+1}{a+c}+\dfrac{c^2+1}{a+b} \ge 3$$ I am stuck at the first stage. Please give me some hints so that I can solve the problem. Thanks in advance.
A.D
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How to find the minimum of the expression?

Let $a$, $b$, $c$ be three real positive numbersand $a^2 + b^2 + c^2 =3$. Find the minimum of the expression $$P = \dfrac{a^2}{b + 2c} +\dfrac{b^2}{c + 2a}+ \dfrac{c^2}{a + 2b}.$$ I tried $$\dfrac{a^2}{b + 2c} +\dfrac{b^2}{c + 2a}+ \dfrac{c^2}{a +…
minthao_2011
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Prove $\sum_{k=1}^{n-1}\frac{(n-k)^2}{2k} \geq \frac{n^2\log(n)}{8}$

As the title says, prove $$\sum_{k=1}^{n-1}\frac{(n-k)^2}{2k} \geq \frac{n^2\log(n)}{8},$$ for $n>1$. This inequality is from Erdős, "Problems and results on the theory of interpolation". I, Lemma 3. My attempt: since…
V.S.e.H.
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variance inequality

Show that. for any discrete random variable X that takes on values in the range [0,1]. Var[X] $\le$ 1/4. I translate it into a inequality like this: $x_1, x_2, x_3 \cdots ,x_n$ where $0 \le x_i \le 1$, and $p_1, p_2, p_3 \cdots ,p_n$ where $p_1+…
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How to show $\frac{a_1+\cdots+a_n}{n}\le S_h\sqrt[n]{a_1\cdots a_n}$

For positive numbers $a_i$ with $0
Fischer
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Prove inequality: $\sqrt {(x^2y+ y^2z+z^2x)(y^2x+z^2y+x^2z)} \ge \sqrt[3]{xyz(x^2+yz)(y^2+xz)(z^2+xy)} + xyz$

Let $x,y,z\in \mathbb R^+$ prove that: $$\sqrt {(x^2y+ y^2z+z^2x)(y^2x+z^2y+x^2z)} \ge xyz + \sqrt[3]{xyz(x^2+yz)(y^2+xz)(z^2+xy)}$$ The inequality $\sqrt {(x^2y+ y^2z+z^2x)(y^2x+z^2y+x^2z)} \overset{C-S}{\ge} 3xyz$ will not help because $3xyz…
Xeing
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The weighted average absolute value of pairwise sum of any finite sequence is larger than the weighted average of absolute value.

How do we prove for any given real number $a_1,\ldots,a_n$, we have $$\frac{1}{n^2}\sum_{i=1}^n\sum_{j=1}^n|a_i+a_j|\geq \frac{1}{n}\sum_{i=1}^n|a_i|$$ I have checked several easy cases for this: When $n=1$, LHS $=2|a_1|$ and RHS $=|a_1|$, so it…
William
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How prove this $\left(\sqrt{a^2+b^4}-a\right)\left(\sqrt{b^2+a^4}-b\right)\le a^2b^2$

let $a,b\in R$,and such that $$\left(\sqrt{a^2+b^4}-a\right)\left(\sqrt{b^2+a^4}-b\right)\le a^2b^2$$ prove that $$a+b\ge 0$$ I think this is very beatifull problem, have you nice methods? Thank you, I have see this…
math110
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Proving $\frac{x}{y} +\frac{y}{z} + \frac{z}{x} \ge 3$ for positive $x,y,z$

Suppose $x,y,z$ are real positive numbers, prove that: $$\dfrac{x}{y} +\dfrac{y}{z} + \dfrac{z}{x} \ge 3$$ with equality when $x=y=z$. Can someone help me find an easier solution? I started with assume only two are equal, without loss of…
cowchee
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Prove that $\frac{1}{2} \lt \sum_{r=1}^{n} \frac{1}{n+r} \lt \frac{3}{4} , n>1$

I have proven the left hand side inequality. Can someone give me a hint for the right hand side ? Proof for the left hand side: $\sum_{r=1}^{n} \frac{1}{n+r} >\sum_{r=1}^{n} \frac{1}{2n}$.
Aditya
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Proof of the inequality $1+1!+\frac{1}{2!}+\cdots+\frac{1}{n!} >\left( 1+\frac{1}{n}\right)^n$

I would appreciate if somebody could help me with the following problem: Q: Proof $$1+1!+\frac{1}{2!}+\cdots+\frac{1}{n!} >\left( 1+\frac{1}{n}\right)^n (n\geq2, n\in \mathbb{N})$$
Young
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A tricky Inequality problem

$ \dfrac{1}{1+a_1} + \dfrac{1}{1+a_2} + \cdots + \dfrac{1}{1+a_n} = 1;\ a_1 , a_2 , \ldots , a_n > 0 $ show that $ \sqrt{a_1} + \sqrt{a_2} + \cdots + \sqrt{a_n} \ge (n-1) \left(\dfrac{1}{\sqrt{a_1}}+ \cdots + \dfrac{1}{\sqrt{a_n}}\right)$ I have…
user683949
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Does $y(y+1) \leq (x+1)^2$ imply $y(y-1) \leq x^2$?

Can anyone see how to prove the following? If $x$ and $y$ are real numbers with $y\geq 0$ and $y(y+1) \leq (x+1)^2$ then $y(y-1) \leq x^2$. It seems it is true at least according to Mathematica.
user66151
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How can prove this inequality(8)

Let $a,b,c,x,y,z >0$ and $A=a^2+b^2+c^2,\ B=x^2+y^2+z^2,\ C=ax+by+cz$. By Cauchy-Schwarz inequality, we always have $C^2\le AB$. If $C^2
math110
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