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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$|a-b|^p \le 2^{p-1} (|a|^p + |b|^p)$ for $a,b \in \mathbb{R}, p\ge 1$

$|a-b|^p \le 2^{p-1} (|a|^p + |b|^p)$ for $a,b \in \mathbb{R}, p \ge 1$. How do we get this inequality? I can get $|a-b|^p \le 2^{p} (|a|^p + |b|^p)$ but I don't know how to get rid of a factor of $1/2$ on the RHS.
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A conjecture from Cauchy–Schwarz inequality

we know the Cauchy–Schwarz inequality in $R^3$: $$(x_1^2+x_2^2+x_3^2)(y_1^2+y_2^2+y_3^2)\geq(x_1y_1+x_2y_2+x_3y_3)^2$$ $$(x_1+x_2+x_3)(y_1+y_2+y_3)\geq(\sqrt{x_1y_1}+\sqrt{x_2y_2}+\sqrt{x_3y_3})^2$$ i guess the following inequlity exists…
Charles Bao
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Using two inequality for bounding?

When we know $a \leq b - c$ $c\geq d$, then $a \leq b - d$ holds? If so, why does it?
mallea
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Range of an equation

I am trying to find the range of: $\Large\frac{4 - {\frac {4 + a^2}{a}}}{\frac {1}{2} - \frac {1}{a}}$ So, we have: $a \ne 0$ $\frac {1}{2} - \frac {1}{a} \ne 0 $, $-\frac {1}{a} \ne -\frac {1}{2}$, $\frac {1}{a} \ne \frac {1}{2}$ I am not sure…
Valeri
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Prove the inequality $b\le \sqrt{a^2 + b^2}$

How do i prove $b\le \sqrt{a^2 + b^2}$ ? What i think is : $\begin{align} a^2& \gt 0,\, b^2 \gt 0 \\ b &= \sqrt{b^2}\\ &\le\sqrt{a^2+b^2} \\ \therefore b &\le \sqrt{a^2 + b^2} \end{align}$ Am i right? Please do a correction.
user516076
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How to correctly solve $\sqrt{1+x}+\sqrt{1-x}>1$?

$$\sqrt{1+x}+\sqrt{1-x}>1, x\geq-1 \wedge x\leq1$$ $$\sqrt{1+x}>1-\sqrt{1-x}$$ $$1+x>1-2\sqrt{1-x}+1-x$$ $$0>-2\sqrt{1-x}+1-2x$$ $$2\sqrt{1-x}>1-2x$$ Lets say both sides of inequality are positive, then we can easily $2\sqrt{1-x}>1-2x$ square and…
thunder
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Solve a system of linear inequalities without graphing, i.e. using pure algebra?

I have a system of linear inequalities such as: $y\geq 2x+1$ $y>(x/2)-1$ I know how to solve this system of inequalities (in order to find possible values that will satisfy it) by graphing the two inequalities separately and finding values that will…
Joebevo
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For positive $a$ and $b$, and for $u\in[0,1]$, can we solve for $v$ in $a>b\sqrt{1-v^2}+\sqrt{1-(u+v)^2}$?

Let $a$ and $b$ be positive numbers and $u \in [0, 1]$. Can we solve this inequality for $v$? $$ a > b \sqrt{1-v^2}+ \sqrt{1-(u+v)^2} $$ I need the points for $v$ in the interval $[-1, 1-u]$ for which the corresponding equality holds, together with…
Paul
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$\sum_{i=1}^n \frac{1}{a+S-x_i} \leq M$

Given integer $n \geq 2$ and positive real number $a$, find the smallest real number $M = M (n, a)$, such that for any positive real numbers $x_1, x_2, \ldots, x_n$ with $x_1x_2 \ldots x_n=1$, the following inequality hold:…
user403160
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Problem about inequation with absolute value

I have this statement: It can be assured that | p | ≤ 2.4, if it is known that: (1) -2.7 ≤ p <2.3 (2) -2.2 < p ≤ 2.6 My development was: First, $ -2.4 \leq p \leq 2.4$ With $1)$ by itself, that can't be insured, same argument for $2)$ Now, i will…
ESCM
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Find the minimum value of $\frac{a}{b + 1} + \frac{b}{a + 1} + \frac{1}{a + b}$ where $a, b > 0$ and $a + b \le 1$.

$a$ and $b$ are two positives such that $a + b \le 1$. Find the minimum value of $$\large \frac{a}{b + 1} + \frac{b}{a + 1} + \frac{1}{a + b}$$ We have that $\dfrac{a}{b + 1} + \dfrac{b}{a + 1} = \dfrac{a^2 + b^2 + a + b}{ab + a + b + 1} \ge…
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How to prove these inequalities without using Faulhaber´s formula and without using the method of proof by mathematical induction?

In a book about calculus, I have seen these inequalities: $$1^k+2^k+ ... + (n-1)^k< \dfrac {n^{k+1}}{k+1}<1^k+2^k+ ... + n^k$$ and it is stated that they are valid for every integer $n \geq1$ and every integer $k\geq1$. I think that they can be…
user677585
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Prove that hat if $a,b \ge 2$ then $ab \ge a+b$

Prove that if $a,b \ge 2$ then $a+b \le ab$ so if $a \ge 2$ and $b \ge 2$ then $a-1 \ge 1$ and $b-1 \ge 1$ $(a-1)(b-1) \ge b-1$ $(a-1)(b-1) \ge 1$ $(a-1)(b-1) - 1 \ge 0$ $ab -a -b\ge 0$ $ab \ge a + b$ Thanks in advance, is this valid?
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Proving $\sqrt{a^2+c^2}+\sqrt{b^2+d^2}\ge \:\sqrt{\left(a+b\right)^2+\left(c+d\right)^2}$

The inequality: $$\sqrt{a^2+c^2}+\sqrt{b^2+d^2}\ge \:\sqrt{\left(a+b\right)^2+\left(c+d\right)^2}$$ But can someone help me with a nice elegant solution. This is an olympiad question I was trying to solve, but couldn't manage an elegant solution.
maria
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