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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Question about establishing an inequality

I am having trouble figuring out an inequality that is supposed to be simple. Let N be a natural number and $t\ge N$, $j$ is also a natural number. Assume that $\frac{2\log t}{\log N} \le j \le \frac{3\log t}{\log N}$ and $n\ge N$. Then from this…
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Demonstration with inequalities

I have these stated inequalities, true for all $i$. $$\begin{align} |P_{i}+q_{i}|\leq \epsilon \tag{1} \end{align}$$ with $\epsilon > 0$ and I know that $$\begin{align} |P_{i}-P_{j}|\leq \delta \tag{2} \end{align}$$ My problem is that by using (1) I…
mwoua
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Solving $1 - (1 - p)(1 - p^2)^{n - 2} = f(n)$ for $p$ (or else finding a tight upper bound on the value)

I have the equation: \begin{equation} 1 - (1 - p)(1 - p^2)^{n - 2} = f(n) \end{equation} where $p$ is a probability (thus $0\le p\le 1$ and, in fact, the l.h.s. of the equation is a probability function on $p$), and $f(n)$ is some function of $n$…
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How does the inequality $1 - (1 - x_i)^{\ln{\vert E \vert} + \ln{4}} \leq (\ln{\vert E \vert} +\ln {4})x_i$ stand?

The inequality is $$ 1 - (1 - x_i)^{\ln{\vert E \vert} + \ln{4}} \leq (\ln{\vert E \vert} +\ln {4})x_i $$ With the following constraints: $\vert E \vert \in \mathbb{N}^+$ $x_i \in [0, 1]$. This appeared during my approximation algorithm class and…
Uduru
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Prove $\frac{a}{\sqrt{b^2+c^2+7bc}}+\frac{b}{\sqrt{c^2+a^2+7ca}}+\frac{c}{\sqrt{a^2+b^2+7ab}} \le \frac{a^2+b^2+c^2}{ab+bc+ca} a,b,c \ge 0 ab+bc+ca>0$

Let $a,b,c \ge 0 : ab+bc+ca>0.$ Prove that $$\frac{a}{\sqrt{b^2+c^2+7bc}}+\frac{b}{\sqrt{c^2+a^2+7ca}}+\frac{c}{\sqrt{a^2+b^2+7ab}} \le \frac{a^2+b^2+c^2}{ab+bc+ca}.$$ I posted on AOPS here. I am the author of this problem. I hope someone prove it…
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Minimum of $\frac{x^2 + 7}{\sqrt{x^2 + 6}}$ without using calculus?

If $x \in \mathbb{R} $, find the smallest value of $y=\frac{x^2 + 7}{\sqrt{x^2 + 6}}$ Attempted solution: $y$ can be expressed as $ y=\frac{(x^2 + 6)+1}{\sqrt{x^2 + 6}} = \sqrt{x^2+6} + \frac{1}{\sqrt{x^2 + 6}}$. However, we can't use AM-GM here…
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$x,y,z\ge 0:x^2+y^2+z^2=1.$ Find maximum value $P=\sqrt{\frac{x}{2x^3+2yz+x}}+\sqrt{\frac{y}{2y^3+2zx+y}}+\sqrt{\frac{z}{2z^3+2xy+z}}.$

Problem. Let $x,y,z\ge 0:x^2+y^2+z^2=1.$ Find maximum value of expression: $$P=\sqrt{\frac{x}{2x^3+2yz+x}}+\sqrt{\frac{y}{2y^3+2zx+y}}+\sqrt{\frac{z}{2z^3+2xy+z}}.$$ I saw it in a book of Vo Quoc Ba Can, in a opened list problems. The problem is…
TATA box
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Simple Inequality, maximisation.

Maximise $(a+1)(b+1)(c+1)$ when $a+b+c=1$. I know the answer is just set $a=b=c=\frac{1}{3}$, and you could prove that by comparing two terms at a time: fix one term, and the two other terms are maximised when equal because of squares, so you need…
user1162071
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Proving $\sum_{cyc}\sqrt{\frac{a^3}{a^2+8b^2}}\ge\sqrt{\frac{a^3+b^3+c^3}{a^2+b^2+c^2}}$ for positive $a$, $b$, $c$

Hard problem with inequalities: Let $a,b,c>0.$ Prove that $$\sqrt{\dfrac{a^3}{a^2+8b^2}}+\sqrt{\dfrac{b^3}{b^2+8c^2}}+\sqrt{\dfrac{c^3}{c^2+8a^2}} \ge \sqrt{\dfrac{a^3+b^3+c^3}{a^2+b^2+c^2}}$$ I saw it's on AOPS here. I just can prove the weaker…
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If $\sum\limits_{i=1}^{n+1} \frac{1}{ny_i+1}=1$, then $\prod\limits_{i=1}^{n+1}y_i\geq 1$?

Let $y_1,\ldots,y_{n+1}$ be positive real numbers satisfying $\displaystyle{\sum_{i=1}^{n+1} \frac{1}{ny_i+1}=1}$. Is it true that $y_1y_2\cdots y_{n+1}\geq 1$? Added: can we determine this inequality in terms of high-school math? (e.g.…
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Inequality based on square

We need to find the range of $x$ for the following problem: $$\frac{(x+6)(x-4)^2}{(x-6)(x+4)^2} + \frac{(x-6)(x+9)^2}{(x+6)(x-9)^2} < \frac{2x^2 + 72}{x^2 - 36}$$ They invlove the use of squares, which we have to avoid while solving…
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Average Inequality

I came across the following inequality: In an ordered field, suppose $a_1 \leq a_2 \leq \dots$ Let $b_n = (a_1+ \dots + a_n)/n$. Show that $b_1 \leq b_2 \leq b_3 \leq \dots$ This is equivalent to showing that $$a_1 \leq \frac{a_1+a_2}{2} \leq…
Damien
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If $a>0$, $b>0$, $\frac{1}{a}+\frac{8}{b}=1$, what is the minimum of $a^2+b^2$?

I change $\frac{1}{a}+\frac{8}{b}=1$ into $\frac{1}{a^2}+\frac{16}{ab}+\frac{64}{b^2}=1$, then I get $a^2+b^2=(a^2+b^2)(\frac{1}{a^2}+\frac{16}{ab}+\frac{64}{b^2})=65+\frac{16a}{b}+\frac{16b}{a}+\frac{64a^2}{b^2}+\frac{b^2}{a^2}$ though I get…
S.Y.Li
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Range of values of $k_1$ which satisfies the inequality

The core idea is to find the value of $k_1$ that satisfies the following inequality: \begin{equation} k_1\vert x\vert + k_2 > \vert x + k_3\vert \tag{1} \end{equation} where $k_2 > \vert k3 \vert > 0$, $k_1 \geq 0$ and $x \in \mathbb{R}$. I have…
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How find this $3\sqrt{x^2+y^2}+5\sqrt{(x-1)^2+(y-1)^2}+\sqrt{5}(\sqrt{(x-1)^2+y^2}+\sqrt{x^2+(y-1)^2})$

find this follow minimum $$3\sqrt{x^2+y^2}+5\sqrt{(x-1)^2+(y-1)^2}+\sqrt{5}\left(\sqrt{(x-1)^2+y^2}+\sqrt{x^2+(y-1)^2}\right)$$ I guess This minimum is $6\sqrt{2}$ But I can't prove,Thank you
math110
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