Questions tagged [a.m.-g.m.-inequality]

For questions about proving and manipulating the AM-GM inequality. To be used necessarily with the [inequality] tag.

The AM-GM states that, given a finite number of non-negative real numbers, their arithmetic mean is always greater than or equal to their geometric mean (hence the name “AM-GM inequality”) and that the equality holds if and only if all the given numbers are equal. In other words, if $a_1,a_2,\ldots,a_n\in(0,+\infty)$, then$$\frac{a_1+a_2+\cdots+a_n}n\geqslant\sqrt[n]{a_1a_2\ldots a_n}$$and$$\frac{a_1+a_2+\cdots+a_n}n=\sqrt[n]{a_1a_2\ldots a_n}\iff a_1=a_2=\cdots=a_n.$$

A weighted version of AM-GM inequality is $$\frac{w_1a_1+w_2a_2+\cdots+w_na_n}w\geqslant\sqrt[w]{a_1^{w_1}a_2^{w_2}\ldots a_n^{w_n}}$$

where $w_i$ are nonnegative and $w=\sum_{i=1}^nw_i$.

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Prove the inequality $(a+b)(b+c)(a+c)\ge8$ if $a^2+b^2+c^2+abc=4$.

I noted $a=2\cos A,b=2\cos B,c=2\cos C$ because of the identity $\cos^2(x)+\cos^2(y)+\cos^2(z)+2\cos(x)\cos(y)\cos(z)=1$,but the calculations looked really bad, so I gave up. Any another suggestion?
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