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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Nice lower bounds for $a^n + b^n$ for $a,b,n \geq 1$?

I know this isn't a very typical question, but i was wondering if anyone knows any good lower bounds for $a^n + b^n$. I'm looking for something akin to $a^n + b^n \leq (a + b)^n$ for $n \geq 1$. The motivation is that I'm trying to find some nice…
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show this inequality $\sqrt{e^2r^2+es}+\sqrt{f^2r^2+fs}\ge\sqrt{2}(e+f)$

let $a,b,c,d\in R$ and $ad-bc=1$,$e,f\ge \frac{1}{2}$,and $r^2=a^2+b^2+c^2+d^2,s=ac+bd$.show that $$\sqrt{e^2r^2+es}+\sqrt{f^2r^2+fs}\ge\sqrt{2}(e+f)$$ This inequality is from Geometry inequality,I try square two side,we must to…
math110
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Easy Inequality Problem

Solve the given inequality by interpreting it as a statement about distances on the real line. $$\lvert x+1\rvert > \lvert x-3\rvert$$ I am confused on what this question is asking. Can anyone give me a quick explanation and answer? Please show…
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Solving basic inequality

I have the following steps to a solution $1.$ $\frac{10}{1 - x} ≥ 15 + \frac{-12x}{1 - x}$ $2.$ $10 ≥ 15 - 15x+ (-12)x + 27x ≥ 5$ $3.$ $x ≥ \frac 5{27}$ I am confused about the transition between step $1$ and $2$. Wouldn't multiplying $(1-x)$ to…
dev_el
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Explain why the solution to the inequality is wrong

Problem: $$x/2 - 4/x - 1 > 0$$ Simplified to: $$(x^2-8)/2x > 1$$ Right solution (put all at one side, bring to one fraction) is union of: $$x^2-8-2x>0, x > 0$$ and $$x^2-8-2x<0, x < 0$$ Wrong solution (multiply both sides by $2x$): $$x^2 - 8 >…
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Prove $|a + b|^p \leq (|a| + |b|)^p \leq (1 - \lambda)^{1 - p} |a|^p + \lambda^{1 - p} |b|^p$

Prove $$ |a + b|^p \leq (|a| + |b|)^p \leq (1 - \lambda)^{1 - p}|a|^p + \lambda^{1 - p} |b|^p $$ for any $0 < \lambda < 1$ and $p > 1$. Of course the first inequality is obvious. I have tried applying the definition of convexity for $|t|^p$ but have…
Cbc
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How come these two inequalities have different solution sets?

The following inequality has a solution set of $\{x \in \mathbb{R} | \frac{3}{2} < x < 5, x<\frac{-1}{4}\}$: $$\frac{2x + 1}{2x - 3}>\frac{x + 1}{x - 5}$$ However this inequality has a different solution set of $\{x \in \mathbb{R} | x > 5,…
Backslash
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How $x \ge 8$ represents every value of $x \ge 5$?

I'm new to quadratic inequalities. I was trying to solve this following problem - $$x^2 - 13x + 40 \ge 0 $$ $$(x-5)(x-8) \ge 0 $$ When we consider both of these expressions positive - $$(x-5) \ge 0$$ and $$(x-8) \ge 0 $$ we get $x \ge 5$ and $x…
Russell
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Inequality manipulation: $\frac{1}{\sqrt{4n + 1}} \cdot \frac{2n + 1}{2n + 2} > \frac{1}{\sqrt{4n + 5}}$

I'm reading Analytic Inequalities by Nicholas D. Kazarinoff. On page 5, we are trying to use induction to prove the inequality $$ \frac{1}{\sqrt{4n + 1}} < \frac{1}{2} \cdot \frac{3}{4} \cdots \frac{2n - 3}{2n - 2} \cdot \frac{2n - 1}{2n} <…
Jeremy Lindsay
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Why the result I obtain by squaring an inequality do not coincide?

I have $$ \left| \frac{f(x)}{\sqrt{g(x)}} \right| \leq 1 \quad\quad (1)$$ for $x>0$, and $f(x)$ for some values of $x$ is positive and for some is negative. Now, when I use the identity $\left|y\right|=\sqrt{y^2}$ in $(1)$, I obtain $$…
katy98
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An application of Minkowski iequality.

I am Reading this work on page 21. In the said page, the author reach the inequality: $$\int|\zeta\nabla u^{\beta/p}|^p\leq\Big(\frac{\beta}{\beta-p-1}\Big)^p\int|u^{\beta/p}\nabla\zeta|^p$$ By using the fact that $$|\nabla(\zeta u^{\beta/p})|\leq…
Tomás
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Can adding 1 to the numerator and denominator on both sides of an inequation change the outcome?

I'm computing a fraction from a database when both numerator and denominator can be zero. To solve this problem I thought of adding 1 to each. I know I can add 1 only to the denominator, but this is for optimization of resources and adding 1 to the…
Hachiloni
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Is $\frac{\lfloor{x}\rfloor+1}{2} \le \lfloor\frac{x}{2}\rfloor + 1$

The answer seems to be yes. Here's my reasoning: For $x < 1$, $\frac{1}{2} < 1$ For $1 < x < 2$, $1 = 1$ For $2 \mid x$, $\frac{x}{2} + \frac{1}{2} < \frac{x}{2} + 1$ For $2 \mid x-1$, $\frac{x+1}{2} = \frac{x-1}{2} + 1$ For all other values, the…
Larry Freeman
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How does $A$ relate to $B$ if $A - \lfloor A/B \rfloor - \lceil A/B \rceil \leq \lfloor A/B \rfloor \times (B+1)$?

For $A \geq B$, both are strictly positive integers, what is the relationship between $A$ and $B$ such that the following is true? $$A - \lfloor A/B \rfloor - \lceil A/B \rceil \leq \lfloor A/B \rfloor \times (B+1)$$ Previously I asked this question…
Nick
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Prove/Disprove: $A - \lfloor A/B \rfloor - \lceil A/B \rceil \leq \lfloor A/B \rfloor \times (B+1)$ for $A \geq B$

For $A \geq B$, both are strictly positive integers, is the following true? $$A - \lfloor A/B \rfloor - \lceil A/B \rceil \leq \lfloor A/B \rfloor \times (B+1)$$ I tried the technique used in proving a very similar question: Prove/Disprove: $A -…
Nick
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