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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Is this correct method to prove that $a^2 + b^2 + c^2 ≥ ab + bc + ac$, when $a,b,c \geq 0$?

Can I prove it like this: Let's say that $a=b=c$ so we get "If $a \geq 0$ then $3a^2 ≥ 3a^2$" Now I take the negation of that statement and get "If $a \geq 0$ then $3a^2 < 3a^2$" The anti-thesis is obviously wrong which makes the original thesis…
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Prove the Inequality with $t \in \left[\frac{\sqrt{3}}{3};1 \right)$.

Let $t \in \left[\frac{\sqrt{3}}{3};1 \right)$. Prove that $$\frac{{27{t^6}\left( {4{t^4} + 3{t^3} + 16{t^2} + 3t + 4} \right)}}{{(2t + 1)\left( {4{t^6} - 21{t^5} + 36{t^4} + 7{t^3} - 24{t^2} + 4} \right)}} \geq 3\left( {35\sqrt 3 - 48}…
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Prove $5(x_1^2+x_2^2+x_3^2+x_4^2+x_5^2)\geqslant(x_1+x_2+x_3+x_4+x_5)^2$

I can't figure out how to prove $$5(x_1^2+x_2^2+x_3^2+x_4^2+x_5^2)\geqslant(x_1+x_2+x_3+x_4+x_5)^2$$
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Prove that $\left(1+\frac{1}{k(k+2)}\right)^k \leq 1+\frac{1}{k+1}$

Is inequality $$\Big (1+\frac{1}{k(k+2)}\Big )^k \leq 1+\frac{1}{k+1}$$ is true for all $k \in \mathbb{N}$ ? In my previous question the $k+2$ in the LHS of the inequality was $k+1$. If it was the problem, then it can be settled using binomial…
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Prove that, if $a,b,c \geq 1$, then $\frac{3ab+2c+1}{a+b}+\frac{3bc+2a+1}{b+c}+\frac{3ca+2b+1}{c+a} \geq 9$.

the question Prove that, if $a,b,c \geq 1$, then $\frac{3ab+2c+1}{a+b}+\frac{3bc+2a+1}{b+c}+\frac{3ca+2b+1}{c+a} \geq 9$. the idea I know that these arent really helpful ideas, but as you know we have to use an inequality theorem to demonstrate…
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Proving Muirhead-like inequalities

Let $T_{m,n,p}(x,y,z)=\sum_{Sym} x^m y^n z^p$. For $x,y,z>0$, prove $2T_{6,3,0}(x,y,z)+T_{3,3,3}(x,y,z)+3T_{4,4,1}(x,y,z)\geq 6T_{5,2,2}(x,y,z)$. I tried to prove that by using AM-GM inequality, without success. Is there a general way to prove…
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How prove this $\prod_{i=1}^{n}a_{i}+\prod_{i=1}^{n}(1-a_{i})\ge\frac{1}{2^{n-1}}$

let $0<\le a_{i}\le \dfrac{1}{2},i=1,2,\cdots,n$.show that $$\prod_{i=1}^{n}a_{i}+\prod_{i=1}^{n}(1-a_{i})\ge\dfrac{1}{2^{n-1}}$$ my idea: I guess this problem will use Bernoulli inequality: $$(1+x_{1})(1+x_{2})\cdots (1+x_{n})\ge…
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proving this inequality $\left | \int_{\left | z \right |=2}^{}\frac{dz}{z^2+1} \right |\leq \frac{4\pi}{3}$

proving this inequality $$\left | \int_{\left | z \right |=2}^{}\frac{dz}{z^2+1} \right |\leq \frac{4\pi}{3}$$ I tried with $$\left | \int_{\left | z \right |=2}^{}\frac{dz}{z^2+1} \right |\leq \int_{\left | z \right |=2}^{} \left | \frac{dz}{z^2+1}…
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Find a $4$ digit number $N=abcd$ such that $(a+b+c+d)(a^2+b^2+c^2+d^2)^2 = N$

I want to find by hand a $4$ digit number $N=abcd$ such that $(a+b+c+d)(a^2+b^2+c^2+d^2)^2 = N$. Here is my work so far : Let $S = a+b+c+d$ and $X = a^2 + b^2+c^2+d^2$ (so $N = SX^2$), I searched for inequalities : $S^2\le 4X$ with cauchy…
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An elementary inequality involving exponentials

In a paper that I'm reading, the authors use the following "elementary inequality" to further derive other inequalities. For $ \sigma > 0, \gamma \in [0,1], x \geq y$, and $\langle \cdot \rangle$ is the Japanese bracket, i.e. $\ \langle x \rangle =…
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Proving $\frac{n(n-1)}{2}\le \sum_{k=1}^{n^2}\{\sqrt{k}\}\le \frac{n^2-1}{2}.$

For $n\in Z^{+}$ and $n>2$ prove that$$\frac{n(n-1)}{2}\le \sum_{k=1}^{n^2}\{\sqrt{k}\}\le \frac{n^2-1}{2},$$where $0\le k=[k]+\{k\}.$ I check that…
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Solving $(a^2+b^2+c^2)\left(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\right)\ge(3+2\sqrt{3})^2$

Given $a,b,c$ are real numbers such that: $abc\neq0$ and $$(a+b+c)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)=3.$$ Prove that: $$(a^2+b^2+c^2)\left(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\right)\ge(3+2\sqrt{3})^2$$ The problem is quite hard…
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About a cyclic inequality $a^2+b^2+c^2+3\ge 2\sqrt{2(a^3b+b^3c+c^3a)+3abc}.$

I am looking for a proof. Here is my question. Problem. Given non-negative real numbers $a,b,c$ satisfying $a+b+c=3.$ Prove that$$\color{black}{a^2+b^2+c^2+3\ge 2\sqrt{2(a^3b+b^3c+c^3a)+3abc}. }\tag{1}$$ Source: own Equality holds iff…
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Minimizing $P=\frac{a}{a+ab+2}+\frac{b}{b+bc+2}+\frac{c}{c+ca+2}$ when $ab+bc+ca=2.$

Problem. Given non-negative real numbers $a,b,c$ satisfying $ab+bc+ca=2.$ Find the minimum $$P=\frac{a}{a+ab+2}+\frac{b}{b+bc+2}+\frac{c}{c+ca+2}.$$ By $a=2,b=1,c=0$ I got a value $\dfrac{2}{3}.$ I've tried to prove…
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inequality $\sqrt[3]{(a^2+1)(b^2+1)(c^2+1)} \le \left(\frac{a+b+c}{3}\right)^2 + 1$

Let $a,b,c$ be positive real numbers such that $\min(ab,bc,ca) \ge 1$. Prove that $\sqrt[3]{(a^2+1)(b^2+1)(c^2+1)} \le \left(\frac{a+b+c}{3}\right)^2 + 1$. I tried to use the AM-GM inequality but that gives $\sqrt[3]{(a^2+1)(b^2+1)(c^2+1)} \le…