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 $$\sum_{\text{sym}}x_1^{a_1}x_2^{a_2}\dots x_n^{a_n}\geq \sum_{\text{sym}}x_1^{b_1}x_2^{b_2}\dots x_n^{b_n}$$

30160 questions
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Prove hard inequality

Let $a,b,c>0$, prove that: $$\frac{1}{(2a+b)^2}+\frac{1}{(2b+c)^2}+\frac{1}{(2c+a)^2}\geq\frac{1}{ab+bc+ca}$$ I tried to use the inequality $\frac{a^2}{x}+\frac{b^2}{y}+\frac{c^2}{z}\geq\frac{(a+b+c)^2}{x+y+z} \forall x,y,z>0$ but that's all I can…
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how to prove this inequality?

Given $x>0$, $y>0$ and $x + y =1$, how to prove that $\frac{1}{x}\cdot\log_2\left(\frac{1}{y}\right)+\frac{1}{y}\cdot\log_2\left(\frac{1}{x}\right)\ge 4$ ?
user9587
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Set of positive integers

Let $A$ be a set of positive integers with the following properties: a) If $n \in A$ then $n \leq 2018$ b) If $S$ is a subset of $A$ with $|S|=3$ then there are two elements $m,n \in S$ such that $|n-m| \geq \sqrt{m}+\sqrt{n}$. What is the…
user680303
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Finding $n$ such that $\frac{3^n}{n!} \leq 10^{-6}$

This question actually came out of a question. In some other post, I saw a reference and going through, found this, $n>0$. Solve for n explicitly without calculator: $$\frac{3^n}{n!}\le10^{-6}$$ And I appreciate hint rather than explicit…
user45099
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Find $x,y,z$ if $3^{x^4+y^2}+3^{y^4+z^2}+3^{z^4+x^2}=3^7$.

How can I solve the following system with $x,y,z$ real numbers: $$x^2+y^2+z^2=6$$ $$3^{x^4+y^2}+3^{y^4+z^2}+3^{z^4+x^2}=3^7.$$ I observe that $x=y=z=\sqrt{2}$ and I feel it must apply the inequality $AM \geq GM$ but I don't know how to…
Iuli
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Is this a valid proof of the triangle inequality?

Inequality to prove: $|a+b|\leq |a| + |b|$ Proof: $-|a| \leq a \leq |a|$ $-|b| \leq b \leq |b|$ Add 1 and 2 together to get: $-(|a|+|b|)\leq a+b\leq|a|+|b|$ $|a+b|\leq|a|+|b|$ The reason I'm asking is because this looks like the simplest proof of…
user1242967
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Inequality $a^2+b^2+c^2 \leq a^2b+b^2c+c^2a+1. $

Prove that: $$a^2+b^2+c^2 \leq a^2b+b^2c+c^2a+1, (\forall) a,b,c \in [0,1].$$ I have no idea, I try $AM\geq GM$ but still nothing.
Iuli
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An arctangent inequality

As in the title, how can I prove $$ \frac{\arctan(x)}{x}\geq\frac{1}{2} $$ for $x\in(0,1]$? I think I can say: $$\frac{\arctan(x)}{x}$$ is monotonically decreasing in the interval, so its value is greater than the value in $1$, which is…
bateman
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How prove this inequality $ \sum_{i=1}^{m} \sum_{j=1}^{m}|A_i|\cdot |A_i \cap A_j|\geq \frac{1}{mn}\left(\sum_{i=1}^{m}|A_i|\right)^3$

Let $A_1,A_2,\cdots,A_m$ be $m$ subsets of the set of size $n$ . Prove that $$\sum_{i=1}^{m} \sum_{j=1}^{m}|A_i|\cdot |A_i \cap A_j|\geq \frac{1}{mn}\left(\sum_{i=1}^{m}|A_i|\right)^3.$$ Someone know what this inequality background? The article…
math110
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Simple inequality over positive reals: $2(x+y+z) \geq 3xyz + xy+yz+zx$ for $xyz=1$

Problem Let $x,y,z$ be real positive numbers with $xyz=1$. Prove: $$ 2(x+y+z) \geq 3xyz + xy+yz+zx$$ Note : I don't know whether the inequality is true or not. I couldn't find a prove in the place found it nor a solution to it. My try I firstly…
Erik T.
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Show $\sum_{k=1}^n (p_k + \frac{1}{p_k})^2\geq n^3 + 2n + \frac{1}{n}$ ; $p_k\geq 0 \forall k$ and $\sum_kp_k=1$

Attempt: $\sum_{k=1}^n (p_k + \frac{1}{p_k})^2 = 2n + \sum_k p_k^2 + \sum_k \frac{1}{p_k^2}$ I used the Cauchy inequality to decompose 1 as $\sqrt{p_k}(\frac{1}{\sqrt{p_k}})$ and got $n^2\leq \sum_k \frac{1}{p_k}$ I could, use the Cauchy…
Please Delete Account
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How prove this inequality $(a+b+c+d+e)^3\geq9(2abc+abd+abe+acd+ade+2bcd+bce+bde+2cde),$

Let $a,b,c,d$ and $,e$ are non-negatives .show that $$(a+b+c+d+e)^3\geq9(2abc+abd+abe+acd+ade+2bcd+bce+bde+2cde),$$ Michael Rozenberg says that this inequality proof is very ugly. I think this inequality seems nice and maybe it has simple methods…
math110
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How find this inequality with $x_{1}+x_{2}+\cdots+x_{n}=1$

show that: $$\sum_{1\le i
math110
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$\frac{a+b-2c}{b+c} + \frac{b+c-2a}{c+a} + \frac{c+a-2b}{a+b} \geq 0$

I recently met the inequality $\frac{a+b-2c}{b+c} + \frac{b+c-2a}{c+a} + \frac{c+a-2b}{a+b} \geq 0$ , where a , b , c are all positive real numbers. I wanted to prove it but had some…
Souvik Dey
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For $x+y=1$, show that $x^4+y^4\ge \frac{1}{8}$

As in the title. Let $x,y$ be two real numbers such that $x+y=1$. Prove that $x^4+y^4\ge \frac{1}{8}$. Any hints? Basically, the only method I am aware of is plugging $y=1-x$ into the inequality and investigating the extrema of the function, but I…
user263286