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
5
votes
1 answer

Solving system of linear inequalities via elimination

Clearly, one can solve a system of linear equations by adding the constituent equations together or subtracting them from one another (via the so-called "elimination" method), but can one do the same with a system of linear inequalities. In other…
5
votes
2 answers

Given that $abc=1$, prove that $\frac{b}{c} + \frac{c}{a}+ \frac{a}{b} \ge \frac{1}{a} + \frac{1}{b} +\frac{1}{c}$.

I have tried to do a bunch of different versions of AM-GM on this and the equivalent versions that come from substituting in $abc=1$ and I always flip the inequality by doing it. Help is greatly appreciated.
TAJD
  • 141
5
votes
2 answers

Inequality under condition $a+b+c=0$

I don't know how to prove that the following inequality holds (under condition $a+b+c=0$): $$\frac{(2a+1)^2}{2a^2+1}+\frac{(2b+1)^2}{2b^2+1}+\frac{(2c+1)^2}{2c^2+1}\geqq 3$$
Tunger
  • 53
5
votes
0 answers

Inequality problems in different cases of $n$ [Difficult]

For $5\le n\le 19$, $a_1, a_2,\cdots, a_n\ge 0$ and $\displaystyle\sum\limits_{k=1}^na_k^2=1$, prove that: $$\sqrt{1-a_1a_2}+\sqrt{1-a_2a_3}+\sqrt{1-a_3a_4}+\cdots+\sqrt{1-a_na_1}\ge\sqrt{n(n-1)}$$ Update(counterexamples to $n$ beyond the scope) I…
5
votes
1 answer

Proof of ternary "Cauchy-Schwarz" inequality $\left(\sum a_i b_i c_i\right)^2\le\sum a_i ^2\sum b_i ^2\sum c_i ^2$?

Prove the following inequality. $$\left( \sum a_i b_i c_i \right)^2 \leq \left( \sum a_i ^2 \right ) \left( \sum b_i ^2 \right) \left( \sum c_i ^2 \right ). \ $$ As, it seems to be similar to Cauchy-Schwarz inequality, I thought of trying by…
Display name
  • 1,033
5
votes
1 answer

Power means inequality

Rummaging through old notebooks from high school I found the following inequality. Let $x, y, z$ and $t$ be positive numbers such that $x+y+z+t=1$. Then the following inequality holds: $$ 1 < \frac{\sqrt[3]{x^4 + y^4 + z^4 + t^4} - (x^2 + y^2 + z^2…
ivan
  • 1,812
5
votes
7 answers

Solution of Inequality $ \frac{1}{x-6}\le 3$

Solve the inequality: $\displaystyle \frac{1}{x-6}\le 3$ solution: \begin{align*}\frac{1}{x-6}& \le 3 \\ x-6& \le \frac{1}{3} \\x& \le 6+\frac{1}{3}\\ x&\le19/3\end{align*} but, for values of $x\le 6$ also, inequality holds true as left hand side…
5
votes
1 answer

Prove inequality $\sqrt[3]{\frac{a^3+b^3}{2}}+\sqrt[3]{\frac{b^3+c^3}{2}}+\sqrt[3]{\frac{c^3+d^3}{2}}+\sqrt[3]{\frac{d^3+a^3}{2}} \le 2(a+b+c+d)-4$

Let $a,b,c,d$ positive real numbers, such that $$\frac1a+\frac1b+\frac1c+\frac1d=4.$$ Prove inequality $$\sqrt[3]{\frac{a^3+b^3}{2}}+\sqrt[3]{\frac{b^3+c^3}{2}}+\sqrt[3]{\frac{c^3+d^3}{2}}+\sqrt[3]{\frac{d^3+a^3}{2}} \le 2(a+b+c+d)-4$$ My work so…
Roman83
  • 17,884
  • 3
  • 26
  • 70
5
votes
1 answer

Is it true that $\left(\frac{a^2+b^2+c^2}{a+b+c}\right)^{a+b+c}≥a^ab^bc^c$?

Let $a,b,c\in\mathbb{R}_{>0}$. Is it true that: $$ \left(\frac{a^2+b^2+c^2}{a+b+c}\right)^{a+b+c}≥a^ab^bc^c $$ I remarked that the inequality is (a bit weirdly) homogeneous, but couldn't use it. Also directly taking the logarithm doesn't seem to…
Redundant Aunt
  • 12,030
  • 2
  • 22
  • 66
5
votes
3 answers

If $x,y,z\gt 0$ and $xyz=1$ Then minimum value of $\frac{x^2}{y+z}+\frac{y^2}{z+x}+\frac{z^2}{x+y}$

If $x,y,z\gt 0$ and $xyz=1$ Then find the minimum value of $\displaystyle \frac{x^2}{y+z}+\frac{y^2}{z+x}+\frac{z^2}{x+y}$ $\bf{My\; Try::}$Using Titu's Lemma $$\frac{x^2}{y+z}+\frac{y^2}{z+x}+\frac{z^2}{x+y}\ge \frac{(x+y+z)^2}{2(x+y+z)} =…
juantheron
  • 53,015
5
votes
3 answers

How to prove that $(x^2+y)(y^2+z)(z^2+x)+2xyz \leqslant 10$, where $x,y,z \in \mathbb{R}^+$ and $x^2+y^2+z^2+xyz=4$

$x,y,z \in \mathbb{R}^+$ and $x^2+y^2+z^2+xyz=4$, prove $$(x^2+y)(y^2+z)(z^2+x)+2xyz \leqslant 10$$ I try several trig substitutions but feel hopeless with the cyclic term here. The condition $x^2+y^2+z^2+xyz=4$ made it too difficult to homogenize…
HN_NH
  • 4,361
5
votes
1 answer

Prove that : $\frac{a+b+c+d}{a+b+c+d+f+g}+\frac{c+d+e+f}{c+d+e+f+b+g}>\frac{e+f+a+b}{e+f+a+b+d+g}$

Prove inequality for positive numbers: $$\frac{a+b+c+d}{a+b+c+d+f+g}+\frac{c+d+e+f}{c+d+e+f+b+g}>\frac{e+f+a+b}{e+f+a+b+d+g}$$ My work so far: Lemma: If $x>y>0, t>z>0$, then $$\frac{x}{x+z}>\frac y{y+t}.$$ Proof. Really, $\frac zx< \frac…
Roman83
  • 17,884
  • 3
  • 26
  • 70
5
votes
2 answers

Proving $\frac{x}{\sqrt{y^2+yz+z^2}}+\frac{y}{\sqrt{z^2+zx+x^2}}+\frac{z}{\sqrt{x^2+xy+y^2}} \ge \sqrt{3}$

Let $x,y,z >0$. Prove that:$$\dfrac{x}{\sqrt{y^2+yz+z^2}}+\dfrac{y}{\sqrt{z^2+zx+x^2}}+\dfrac{z}{\sqrt{x^2+xy+y^2}} \ge \sqrt{3}$$ My solution: By Hölder, $$\left(\sum\frac{x}{\sqrt{4y^2+yz+4z^2}}\right)^2(\sum x(4y^2+yz+4z^2)) \ge…
Frank
  • 2,738
5
votes
3 answers

$a,b,c,d,e$ are positive real numbers such that $a+b+c+d+e=8$ and $a^2+b^2+c^2+d^2+e^2=16$, find the range of $e$.

$a,b,c,d,e$ are positive real numbers such that $a+b+c+d+e=8$ and $a^2+b^2+c^2+d^2+e^2=16$, find the range of $e$. My book tells me to use tchebycheff's inequality $$\left(\frac{a+b+c+d}{4}\right)^2\le \frac{a^2+b^2+c^2+d^2}{4}$$ But this not the…
Aditya Dev
  • 4,774
5
votes
3 answers

prove this inequality $a^n>b^n+c^n$

We know a,b and c are positive and $a^2=b^2+c^2$ How we can conclude this inequality: $a^n>b^n+c^n$ , $n>2$ I tried Binomial Theorem but I can't prove this. Thanks
user224893