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If $a, b, c$ are positive, and $a+b+c = 3$ show $$ 9 + 3 \sum_{\mbox{cyc}}a\cos\left( \frac{2b}{c}\right)\geq 2\left( \sum_{\mbox{cyc}}a\cos\left( \frac{b}{c}\right) \right)^2 $$

This is yet another one of those cyclic symmetric inequalities in three positive variables, with cyclic symmetric constraints, but it has a twist: The inequality can be saturated (the equality holds) at $a=b=c=1$ but it can also be saturated at other values of $a,b,c$.

I did not get this from a contest problem, but it might make a good one at the moderate-skill high-school level.

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We know that $\cos(2x)=2\cos^2(x)-1$

$$\Rightarrow\sum_{\mbox{cyc}}a\cos\left( \frac{2b}{c}\right)=2\left [\sum_{\mbox{cyc}}a\cos^2\left( \frac{b}{c}\right)\right ]-(a+b+c)=2\left [\sum_{\mbox{cyc}}a\cos^2\left( \frac{b}{c}\right)\right ]-3$$

$$\Rightarrow9 + 3 \sum_{\mbox{cyc}}a\cos\left( \frac{2b}{c}\right)=6\left [\sum_{\mbox{cyc}}a\cos^2\left( \frac{b}{c}\right)\right ]=2(a+b+c)\sum_{\mbox{cyc}}a\cos^2\left( \frac{b}{c}\right)$$

And by using Cauchy-Schwarz

$$2(a+b+c)\sum_{\mbox{cyc}}a\cos^2\left( \frac{b}{c}\right)\geq 2\left( \sum_{\mbox{cyc}}a\cos\left( \frac{b}{c}\right) \right)^2$$

So,

$$9 + 3 \sum_{\mbox{cyc}}a\cos\left( \frac{2b}{c}\right)=2(a+b+c)\sum_{\mbox{cyc}}a\cos^2\left( \frac{b}{c}\right)\geq 2\left( \sum_{\mbox{cyc}}a\cos\left( \frac{b}{c}\right) \right)^2$$

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