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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Exponential of Average Less than Average of Exponential

I am aiming to show the following inequality: $$\exp\left[\frac{1}{b-a}\int_{a}^bf(x)dx\right] \leq \frac{1}{b-a}\int_a^b \exp[f(x)]dx$$ where $f(x) \in C([a,b])$. Intuitively, this makes sense since for some real numbers $r$ and…
Charles
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Prove that $\left (1+\frac ba \right )\left (1+\frac ac \right )\left ( 1+ \frac cb\right )\ge 4+3\sqrt2$

Given $a,b,c>0$ and $a^2\ge b^2+c^2$. Prove that $$\left (1+\frac ba \right )\left (1+\frac ac \right )\left ( 1+ \frac cb\right )\ge 4+3\sqrt2$$ This is my try: I expanded the LHS, and I have to show that $\displaystyle\frac a b +\frac a c +\frac…
mja
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Knowing that $x+y+z=3$ and $x,y,z\ge0$, how to prove the following inequality?

Let $x,y,z\ge 0$, if $x+y+z=3$, show that $$\dfrac{1}{\sqrt{x^2+xy+y^2}}+\dfrac{1}{\sqrt{y^2+yz+z^2}}+\dfrac{1}{\sqrt{z^2+zx+x^2}}\ge \dfrac{12+2\sqrt{3}}{9}$$ Using this …
user246384
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Proving an Inequality using a Different Method

Is there another way to prove that: If $a,b\geq 0$ and $x,y>0$ $$\frac{a^2}{x} + \frac{b^2}{y} \ge \frac{(a+b)^2}{x+y}$$ using a different method than clearing denominators and reducing to $(ay-bx)^2 \ge 0$?
Guest
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Proving the inequality $\frac{a+b}{2} - \sqrt{ab} \geq \sqrt{\frac{a^2+b^2}{2}} - \frac{a+b}{2}$

Show that for any two positive real numbers $a$ and $b$, $\frac{a+b}{2} - \sqrt{ab} \geq \sqrt{\frac{a^2+b^2}{2}} - \frac{a+b}{2}$ My attempt: $(\sqrt a-\sqrt b)^2\geq0\\\frac{a+b}{2}\geq \sqrt{ab}$ $(a-b)^2\geq0\\a^2+b^2\geq2ab…
user73195
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Does $\frac{x+y}{2}>\frac{a+b}{2}$ hold?

$a$ and $b$ are two real positive numbers. Given that $x=\sqrt{ab}$ and $y=\sqrt{\frac{a^2+b^2}{2}}$, which one has a higher value, $\frac{x+y}{2}$ or $\frac{a+b}{2}$? We know that $y=\sqrt{\frac{a^2+b^2}{2}}>\frac{a+b}{2}>x=\sqrt{ab}$ by…
user233658
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Inequality for the difference of two products

Suppose $a_1,\ldots,a_k$ and $b_1,\ldots,b_k$ are complex numbers bounded in absolute value by $1$. Is it true that $$ \left| \prod_{i=1}^k a_i - \prod_{i=1}^k b_i\right|\leq \sum_{i=1}^k |a_i-b_i|? $$
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Prove that $xy+yz+zx \leq x^2+y^2+z^2$

Prove that $xy+yz+zx \leq x^2+y^2+z^2$ . Hint: Use $\frac{a+b}{2}\geq\sqrt{ab}$ First I tried using the hint by setting $a=x$ and $b=y+z$, however this results in the inequality: $$x^2+y^2+z^2 \geq 2xy-2yz+2zx $$ which isn't quite the same…
Jean
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If $\frac{a}{b}<\frac{c}{d}$ and $\frac{e}{f}<\frac{g}{h}$, then $\frac{a+e}{b+f} < \frac{c+g}{d+h}$.

If $a, b, c, d, e, f, g, h$ are positive numbers satisfying $\frac{a}{b}<\frac{c}{d}$ and $\frac{e}{f}<\frac{g}{h}$ and $b+f>d+h$, then $\frac{a+e}{b+f} < \frac{c+g}{d+h}$. I thought it is easy to prove. But I could not. How to prove this? Thank…
Sang
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Let $A \subset \mathbb Z^3$ / $|A| < \infty$. Prove that: $|A| \le \sqrt{|A_x| |A_y| |A_z|}$

Here is the problem statement word by word: $1)$ Prove that if $a_{ij}$, $b_{jk}$ and $c_{ki}$ are non-negative reals with $1 \le i,j,k \le n$, then: $$\sum_{i,j,k = 1}^n \sqrt{a_{ij} \times b_{jk} \times c_{ki}} \le \sqrt{ \sum_{i,j = 1}^n a_{ij}…
user230734
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Inequality: $\sum_{i} \frac1{\alpha_i} \ge n^2$

$\alpha_1, \ldots, \alpha_n$ are positive reals whose sum does not exceed one. It is required to prove that: $$\sum_{i} \frac1{\alpha_i} \ge n^2$$ I would show my work, but I am certain that it does not offer any insight because I feel that there is…
user230734
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Do more generalizations of Schur's inequality exist?

I meet this following problem If $$n\ge 3,\sum_{i=1}^{n}\left(\prod_{j\neq i}(a_{i}-a_{j})\right)\ge 0$$ where $a_{i}$ are real numbers. when $n=3$, it is Schur's inequality so which $n$ such this inequality? but for more generalization …
user237685
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Show inequality holds

Let $z \geq y \geq 1$. Show that $$ \sqrt{\frac{y}{1+z}} + \sqrt{\frac{z}{1+y}} + \sqrt{\frac{1}{y+z}} > 2 $$ These are actually the last steps of a bigger inequality, but I can't think of a nice way to prove it (no differentiation please).
nabla
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When does equality hold in this inequality?

The following inequality can be proven as follows: Let $n\geq3$ and $0=a_0