Crystal Wheeler

2022-07-02

Proof that ${a}^{\mathrm{tan}x}+{a}^{\mathrm{cot}x}\le 2a$ where $\frac{1}{2}\le a\le 1$ and $0\le x\le \frac{\pi}{4}$

eurgylchnj

Beginner2022-07-03Added 14 answers

If we set $\mathrm{tan}(x)=z$ we have $z\in [0,1]$ and we may find the maximum of

$f(a)={a}^{z-1}+{a}^{\frac{1}{z}-1}$

over the interval $a\in [\frac{1}{2},1]$. 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

$f(a)=\frac{2}{{2}^{z}}+\frac{2}{{2}^{1/z}}\le 2$

hence $f(a)\le 2$ as wanted.

$f(a)={a}^{z-1}+{a}^{\frac{1}{z}-1}$

over the interval $a\in [\frac{1}{2},1]$. 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

$f(a)=\frac{2}{{2}^{z}}+\frac{2}{{2}^{1/z}}\le 2$

hence $f(a)\le 2$ as wanted.

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