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
7
votes
1 answer

Proving Inequality with Square Roots and Algebraic Manipulation

Question: Edit: Original question is: Estimate $\sqrt{1}+\sqrt{2}+\cdots+\sqrt{10000}$ to nearest hundred. I used $\sqrt{1}+\sqrt{2}+\cdots+\sqrt{n}\geq \int_{0}^{n}\sqrt{x}dx$ and $\sqrt{1}+\sqrt{2}+\cdots+\sqrt{n}\leq \frac{4n+3}{6}\sqrt{n}$,…
7
votes
2 answers

Is $\frac{a_1+\cdots+a_n}{\sqrt{n(b_1+\cdots+b_n)}} \le \frac{1}{n}\left(\frac{a_1}{\sqrt{b_1}} +\cdots+\frac{a_n}{\sqrt{b_n}}\right)$?

For $a_1>0$, $a_2>0,\dots,a_n>0$, and $b_1>0$, $b_2>0,\dots,b_n>0$ I want to prove: $$\frac{a_1+a_2+\dots+a_n}{\sqrt{n(b_1+b_2+...+b_n)}} \le \frac{1}{n}\left(\frac{a_1}{\sqrt{b_1}}+\frac{a_2}{\sqrt{b_2}} +\cdots+\frac{a_n}{\sqrt{b_n}}\right)$$
Fan Zhang
  • 1,967
7
votes
4 answers

Looking for a slick way to prove this.

Suppose that $a_k \ge 0$, $k=0,1,2,3,…$ satisfies $a_k^2\le a_{k-1}a_{k+1}$ for all $k\ge 1$. Show that for all $k\in \{0,1,\dots, N\}$ we have $$ a_k \le a_0^{1-k/N} a_N^{k/N} $$ I am looking for a slick way to prove this statement. I have the…
Suugaku
  • 2,469
7
votes
0 answers

Proving the Power Mean Inequality using Chebyshev's sum inequality

Question: Can Chebyshev's sum inequality be used to prove the generalized mean inequality, or at least a portion of it? If we let $$P(r) = \sqrt[r]{\frac{x_1^r + x_2^r + \cdots +x_n^r}{n}}$$, we have that $P(r)$ is increasing, which is also known as…
7
votes
4 answers

Prove that $n^k < 2^n$ for all large enough $n$

If $k\ge 2$ is an integer, prove by elementary means (no calculus or limits) that there is a $N(k)$ such that $n^k < 2^n$ for all integers $n \ge N(k)$. Give an explicit form for $N(k)$.
marty cohen
  • 107,799
7
votes
6 answers

Prove that: $a^2+b^2+(1-a-b)^2\ge \frac {1}{3}$

Where $a$ and $b$ are any given real number. I have tried solving it using partial derivative. $$ s=a^2+b^2+(1-a-b)^2$$ $$\frac{\partial s}{\partial a}=2a-2(1-a-b) \tag{1}$$ $$\frac{\partial s}{\partial b}=2b-2(1-a-b) \tag{2}$$ for maxima both (1)…
Mustahid
  • 363
7
votes
3 answers

prove: $15(a+b)\ge17+14\sqrt{2ab}$

Let $a,b\ge0: a^4+b^4=17$. Prove that: $$15(a+b)\ge17+14\sqrt{2ab}$$ I am looking for a nice approach. My approach is ugly by replace: $b=\sqrt[4]{17-a^4} $ and the rest is working with fuction. I am quite sure $(1,2)$ is the only case equality so…
Sickness
  • 1
  • 4
  • 14
7
votes
3 answers

Prove $e^{a_1}+...+e^{a_n}\leqslant e^{b_1}+...+e^{b_n}$, where $a_1+...+a_n=b_1+...+b_n=0$, $|a_j|\leqslant |b_j|$

Let $\{a_1,\ldots,a_n\}$ and $\{b_1,\ldots,b_n\}$ $-$ sets of real numbers ($n\in \mathbb{N}$), such that $$ a_1+\cdots+a_n = b_1+\cdots+b_n=0, \qquad |a_j|\leqslant |b_j|, \; j=1,\ldots,n, $$ $a_j$ and $b_j$ $-$ of same sign ($a_j\cdot b_j\geqslant…
Oleg567
  • 17,295
7
votes
1 answer

Proving $\sum_{i=1}^n\sum_{j=1}^n\sqrt{|x_i-x_j|}\le \sum_{i=1}^n\sum_{j=1}^n\sqrt{|x_i+x_j|}$.

IMO 2021, Problem 2. Let $ n $ be an integer $ \ge 2$ and $x_1,x_2,...,x_n $ be $ n$ reals. prove that $$\sum_{i=1}^n\sum_{j=1}^n\sqrt{|x_i-x_j|}\le \sum_{i=1}^n\sum_{j=1}^n\sqrt{|x_i+x_j|}$$ I wrote the left sum…
7
votes
2 answers

Inequality Of Four Variables

Suppose $a,b,c,d$ are real numbers greater than $1$. Prove that $$abc+abd+acd+bcd-3abcd<1.$$
chloe_shi
  • 2,855
7
votes
3 answers

Minimum value of $A = (x^2 + 1)(y^2 +1)(z^2 +1)$?

Given: $\begin{cases}x;y;z \in\Bbb R\\x+y+z=3\end{cases}$ Then what is the minimum value of $A = (x^2 + 1)(y^2 +1)(z^2 +1)$ ? I start with $AM-GM$ So $A \geq 8xyz$ Then I need to prove $xyz\geq 1$ right? How to do that and if it's a wrong way,…
Solitarie
  • 531
7
votes
3 answers

solution to a root inequality

I have the inequality $$\sqrt{a^2+b^2+c^2}+2\sqrt{ab+ac+bc} \geq \sqrt{a^2+2bc}+\sqrt{b^2+2ac}+\sqrt{c^2+2ab}.$$I tried to do $u=a^2+b^2+c^2$ and $v=ab+ac+bc$ and $x=a^2+2bc$, $y=b^2+2ac$, $z=c^2+2ab$ ...but I did not find any solution. Any help…
user910
  • 383
7
votes
1 answer

Prove $ |\vec{a_1}-\vec{b}|+ \cdots +|\vec{a_n}-\vec{b}| > n $

There are $ \vec{a_1},\vec{a_2},\vec{a_3}, \ldots ,\vec{a_n},\vec{b}\; $ such that $ |\vec{a_i}|>1 $, $ |\vec{b}|<1 $, $ \vec{a_1}+\cdots+\vec{a_n}=0 $ .Prove : $ |\vec{a_1}-\vec{b}|+\cdots+ |\vec{a_n}-\vec{b}| > n $
7
votes
1 answer

Prove inequality: $\sum \frac{a^4}{a^3+b^3} \ge \frac{a+b+c}{2}$

Prove inequality with $a,b,c >0$ $$\frac{a^4}{a^3+b^3} + \frac{b^4}{b^3+c^3}+\frac{c^4}{c^3+a^3} \ge \frac{a+b+c}{2}$$ I tried the inequality: $\sum \frac {a^4+b^4}{a^3+b^3} \ge \sum \frac{a+b}2=a+b+c$, but seem like it doesn't help
Xeing
  • 2,967