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Proof that ${a}^{\mathrm{tan}x}+{a}^{\mathrm{cot}x}\le 2a$ where $\frac{1}{2}\le a\le 1$ and $0\le x\le \frac{\pi }{4}$
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eurgylchnj
If we set $\mathrm{tan}\left(x\right)=z$ we have $z\in \left[0,1\right]$ and we may find the maximum of
$f\left(a\right)={a}^{z-1}+{a}^{\frac{1}{z}-1}$
over the interval $a\in \left[\frac{1}{2},1\right]$. Since the given function is convex (it is the sum of two convex functions) the maximum is attained at the boundary. At a=1 we have f(a)=2 and at $a=\frac{1}{2}$ we have
$f\left(a\right)=\frac{2}{{2}^{z}}+\frac{2}{{2}^{1/z}}\le 2$
hence $f\left(a\right)\le 2$ as wanted.