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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:$$x^t(x-y)(x-z)+y^t(y-z)(y-x)+z^t (z-x)(z-y)\ge 0$$

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 $$\sum_{\text{sym}}x_1^{a_1}x_2^{a_2}\dots x_n^{a_n}\geq \sum_{\text{sym}}x_1^{b_1}x_2^{b_2}\dots x_n^{b_n}$$