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 $ab \leq \frac14$ and $ (1+1/a)(1+1/b)\ge 9$ when $a+b=1, a \ge 0, b \ge 0$

Our teacher gave us some identities firstly and said we could use one of them to prove it. The identities are: $$\frac{(a^2+b^2)}{2}≥\left(\frac{(a+b)}{2}\right)^2$$ $$(x+y)^2≥2xy$$ and $$\frac{(x+y)}{2} \ge \sqrt{xy}$$ Here's what I did to solve…
seda
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Proving $\sqrt{100,001}-\sqrt{100,000} < \frac{1}{2\sqrt{100,000}}$

Proving $\sqrt{100,001}-\sqrt{100,000} < \frac{1}{2\sqrt{100,000}}$ I squared both sides of the equation to get $100,001 + 100,000+-200\sqrt{10}\sqrt{100,001} < \frac{1}{400,000}$. I am just not sure how to justify this. I've tried multiplying both…
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Let a,b,c be positive real numbers numbers such that $ a^2 + b^2 + c^2 = 3 $

Let $a,b,c\in\mathbb{R^+}$ such that $ a^2 + b^2 + c^2 = 3 $. Prove that $$ (a+b+c)(a/b + b/c + c/a) \geq 9. $$ My Attempt I tried AM-GM on the symmetric expression so the $a+b+c \geq 3$, but I found $a+b+c \leq 3$.
novak
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Prove the inequality $a^3+2 \geq a^2+2 \sqrt{a}$

Prove the inequality $a^3+2 \geq a^2+2 \sqrt{a},a \geq 0.$ One way to do it is using the formula $$ a^3+2 - a^2-2 \sqrt{a}=(\sqrt{a}-1)^2(1+(a+1)(\sqrt{a}+1)^2) \geq 0. $$ But I hope there is a better way.
Leox
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Proving that if $xy + yz + zx \geq \frac{1}{\sqrt{x^2+y^2+z^2}}$, then $x+y+z\geq \sqrt{3}$

If $x, y, z$ are positive real numbers such that $$xy + yz + zx \geq \frac{1}{\sqrt{x^2+y^2+z^2}},$$ then prove that $x+y+z\geq \sqrt{3}$.
novak
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Inequality $\frac 1{\sqrt{1+xy}}+\frac 1{\sqrt{1+yz}}+\frac 1{\sqrt{1+zx}}\ge \frac 9{\sqrt {10}}$

Let $x,y,z$ be non-negative real numbers such that $x+y+z=1$ , then is the following true? $$\dfrac 1{\sqrt{1+xy}}+\dfrac 1{\sqrt{1+yz}}+\dfrac 1{\sqrt{1+zx}}\ge \dfrac 9{\sqrt {10}}$$
user102248
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Prove that $|x|\leq c\iff -c\leq x\leq c$

I want to show $|x|\le c$ is equivalent to $-c\le x\le c$. But I've taken this for granted so long I'm not actually sure where to start. Can someone give me some hints (not the full solution).
user2850514
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Inequality $ \vert \sqrt{a}-\sqrt{b} \vert \leq \sqrt{ \vert a -b \vert } $

I have the following inequality on my class notes that I haven't been able to prove, I was even wondering if it is actually true: $$ \forall a,b \in \mathbb{R}^{\ge0} \left( \left| \sqrt{a}-\sqrt{b} \right| \leq \sqrt{ \vert a -b \vert}\right)…
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Find the minimum value of the expression $P=\frac{x^2}{\sqrt{x^3+8}}+\frac{y^2}{\sqrt{y^3+8}}+\frac{z^2}{\sqrt{z^3+8}}$

Let $x,y,z$ be positive real number such that $xy+yz+zx=3$. Find the minimum value of the expression $$P=\frac{x^2}{\sqrt{x^3+8}}+\frac{y^2}{\sqrt{y^3+8}}+\frac{z^2}{\sqrt{z^3+8}}$$
Phi Linh
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Question on inequalities of symmetric means.

Let $\sum_\text{sym}a^xb^yc^z = [x,y,z]$. Prove that $\frac32[7/3,1/3,1/3]+2[4/3,4/3,1/3] \geq \frac32[1,1,1] + 2[5/3,2/3,2/3]$. I got this while trying to prove that for positive reals $a,b,c$ where $abc=1$ prove that $a+b+c \geq…
John Marty
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Let a,b $\in$ $\mathbb{R}.$ Show that $a^4+b^4+8\ge 8ab.$

Let $a,b \in \mathbb{R}.$ Show that $a^4+b^4+8\ge 8ab.$ The question is from the inequalities section of An Excursion in mathematics by Bhaskaraycharya Pratisthanan. My heuristics include using the AM-GM inequality. I am unable to design the…
gaufler
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Prove that $a+1 < a^2$ for all integers $a > 1$

I know it is true, but how could I prove it? $$a^2-(a+1)>0$$ $$a^2 - a -1 >0$$ via a graphical solution $a^2-1-1>0$ when $a>$ approx $1.68$...thus given $a$ is an integer $>1$ the statement is true. Can one do it without a graphical solution?
Greese
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An Inequality problem 123

Let $x_1,x_2,\ldots,x_n\in \left [ 0,1 \right ]$, prove that $(1+x_{1}+x_2+\cdots+x_n)^2\geq 4(x_1^2+x_2^2+\cdots+x_n^2)$
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Bernoulli's Inequality proof for $x\ge -1/6$

$$ \sqrt{1+6x}\le(1+x)^3 $$ I understood the AM is always greater or equal to the GM thing but it was for all values greater than $0$. This doesnt look like its going to be solved the same way. Do we use Cauchy-Schwarz or the AM-GM for this? Don't…
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Solve $x^3(x^2+1)^2(x-1)(x+1)>0$

I need to solve this inequality: $$x^3(x^2+1)^2(x-1)(x+1)>0$$ Note: I haven't learnt imaginary numbers. Does the $(x^2+1)^2$ affect the inequality? Thanks
Jasern. L
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