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
2
votes
3 answers

If $a > 0$, $b > 0$ and $b^{3} - b^{2} = a^{3} - a^{2}$, where $a\neq b$, then prove that $a + b < 4/3$.

If $a > 0$, $b > 0$ and $b^{3} - b^{2} = a^{3} - a^{2}$, where $a\neq b$, then prove that $a + b < 4/3$. Now what I thought is to manipulate given result somehow to get something in the form of $a + b$: \begin{align*} b^{3} - b^{2} = a^{3} - a^{2} &…
2
votes
1 answer

Inequality proof $\frac{a_1a_2}{(1-a_1)(1-a_2)}\le\frac{1}{2^3}\frac{a_1+a_2}{1-(a_1+a_2)}$

This inequality can be generalized to n terms $a_1,a_2,...,a_n$, and here I first work on the basic case, which contain only $a_1,a_2$ terms: $\frac{a_1a_2}{(1-a_1)(1-a_2)}\le\frac{1}{2^3}\frac{a_1+a_2}{1-(a_1+a_2)}$ where $a_1>0, a_2>0,…
MathFail
  • 21,128
2
votes
3 answers

Prove $\sqrt{x-1} + \sqrt{y-1} \le xy, x \ge 1, y \ge 1$

Let $x$ and $y$ be real numbers, such that $x \ge 1$ and $y \ge 1$. Prove that this inequality is true: $\sqrt{x-1} + \sqrt{y-1} \le xy$ Can someone show me steps to solve it. PS:I need to give steps on how to solve it.
2
votes
1 answer

How to write condition in math?

How do I have to write condition in math? For example, Solve, $|x + 1| + |x - 1| \leq 2$ if x < -1 -x - 1 - x + 1 <= 2 -2x <= 2 x >= -1 No solution if -1 <= x < 1 x + 1 - x + 1 <= 2 2 <= 2 Infinitely many solutions -1 <= x <…
2
votes
3 answers

How to study positivity of $x^3 -x -1$?

Studying this inequality: $$x^3 -x -1 \ge 0 $$ Since I can't apply Ruffini's rule, I cannot recognize a method to study the function positivity. I could decompose it to: $$ x \cdot (x+1) \cdot (x-1) -1 \ge 0$$ But the $-1$ is a problem, how do I…
2
votes
4 answers

Find $x$ such that $\frac{1}{x} > -1 $

How do I solve this type of inequalities analytically? I know the answer is $ x<-1 $ and $ x> 0 $ but: $$\frac{1}{x} > -1 $$ $$1>-x $$ $$ x>-1 $$ Wat I'm doing wrong?
Msegade
  • 927
2
votes
1 answer

Inequality $|\sqrt[n]{x}-\sqrt[n]{y}|\le \sqrt[n]{C_n|x-y|}$ for odd $n$

I'm interested in a simple way to estimate $|\sqrt[n]{x}-\sqrt[n]{y}|$ from $|x-y|$. For $n=2$ and $n=3$, we have $|\sqrt{x}-\sqrt{y}|\le \sqrt{x-y}$ for $x,y\ge 0$ and $|\sqrt[3]{x}-\sqrt[3]{y}|\le \sqrt[3]{4(x-y)}$ for $x,y\in \mathbb{R}$. When…
2
votes
2 answers

Showing the AM-GM inequality, $\textit{Problem From The Book}$, 19.17, but using integrals

The chapter of this problem is Solving Elementary Inequality Using Integrals. After I typed the problem: I spent several hours trying to solve it, but to no avail, so I am hoping someone here can enlighten me. For part $a$,I can prove AM-GM…
math110
  • 93,304
2
votes
3 answers

Prove this $4(a+b+c+d)+(a^3+b^3+c^3+d^3)\le 20$

let $a,b,c,d\in R$,and such $a^2+b^2+c^2+d^2=4$, prove or disprove $$4(a+b+c+d)+(a^3+b^3+c^3+d^3)\le 20$$ I try use Cauchy-Schwarz inequality have $$4(a+b+c+d)\le 4\sqrt{4(a^2+b^2+c^2+d^2)}=16$$ but $$(a^3+b^3+c^3+d^3)^2(1+1+1+1)\ge…
math110
  • 93,304
2
votes
1 answer

Prove that: $ab\sqrt{ab}+bc\sqrt{bc}+ca\sqrt{ca}\le abc+\frac{1}{2}\sqrt[3]{\frac{(a^{2}+bc)^{2}(b^{2}+ca)^{2}(c^{2}+ab)^{2}}{abc}}$

Let $a,b,c>0$. Prove that: $$a b \sqrt{a b}+b c \sqrt{b c}+c a \sqrt{c a} \leqslant a b c+\frac{1}{2} \sqrt[3]{\frac{\left(a^{2}+b c\right)^{2}\left(b^{2}+c a\right)^{2}\left(c^{2}+a b\right)^{2}}{a b c}}$$ I really don't have many ideas in this…
trungbk
  • 83
2
votes
1 answer

Why the solution for $(x-1)(x+4) \geq 0$ is $x\geq 1$ or $x\leq -4$?

I start with an inequality such as: $$(x-1)(x+4) \geq 0$$ My understanding is that from this point it is solved using the null factor law where: \begin{align*} \begin{cases} x - 1 \geq 0\\\\ x + 4 \geq 0 \end{cases} \end{align*} This gives me $x…
Ronin
  • 21
2
votes
1 answer

An interesting inequality $(mn)^\alpha\le/(?\ge) p(np)^\alpha+q(mq)^\alpha$

Let positive real numbers $p$, $q$, $m$ and $n$ satisfy $p+q=1$ and $m+n=1$, and for real $\alpha$ when $\alpha\in[-1,0]$ \begin{align} p(np)^\alpha+q(mq)^\alpha\le…
MathArt
  • 1,053
2
votes
1 answer

SImple proof based on Inequalities

In a right-angled $\triangle ABC$, which is right-angled at $C$, prove that $a^n + b^n < c^n$ for all $n > 2$. I am able to prove this for powers of 2 as $a^2 + b^2 = c^2$ $$(a^2 + b^2)^n > (a^2)^n + (b^2)^n$$ But how to prove that it's true for…
2
votes
0 answers

How can I prove that $x_n=\frac{\sqrt[n]{n!}}{n}$ is decreasing?

I want to prove that $x_n=\frac{\sqrt[n]{n!}}{n}$ is decreasing, but I tried everything I know and it didn't work. At first, I tried to prove $x_n\ge x_{n+1}$ directly which is equivalent with $(n+1)^{n^2}n!\ge n^{n(n+1)}$. Then, I tried to prove…
2
votes
2 answers

Given real numbers $a,b,c >0$ and $a+b+c=3$. Prove that $\frac{a^2}{b}+\frac{b^2}{c} +\frac{c^2}{a} \ge3 + \frac{4}{3}\max\{(a-b)^2;(b-c)^2;(c-a)^2\}$

My first way is to put the inequality under the same degree, so I multipled both sides to $(a+b+c)$, however it leaded to a hard to solve result. Can anyone help me with this problem?
Tmt
  • 97