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$a^{\tan x}+a^{\cot x} \leq 2a$ where $\frac{1}{2} \leq a \leq 1$ and $0 \leq x \leq \frac{\pi}{4}$

I would like to prove the inequality which is given above. I spent all day thinking about it - I tried some inequalities like relation between exponential and linear or Jensen, but it doesn't work. I tried also by calculus - but when we calculate the derivative and assume it is equal to 0 we have an equality which isn't easy to solve. Maybe anyone has an idea, maybe it's easy and I don't know why I have a problem... I would be grateful if you gave me a hint, I don't want a full solution :)

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If we set $\tan(x)=z$ 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 $a\in\left[\frac{1}{2},1\right]$. 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 $$ f(a) = \frac{2}{2^{z}}+\frac{2}{2^{1/z}}\leq 2 $$ hence $f(a)\leq 2$ as wanted.

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    $\begingroup$ How do we prove that $f(a)$ is convex in $a$ for all values of $z$? $\endgroup$ – Andreas Mar 20 '17 at 13:22
  • $\begingroup$ "At a=1 we have f(a)=1 " --> no, we have f(1) = 2. $\endgroup$ – Andreas Mar 20 '17 at 13:36
  • $\begingroup$ @Andreas: typo fixed. $\endgroup$ – Jack D'Aurizio Mar 20 '17 at 14:02
  • $\begingroup$ @MartinR: I am claiming what I am claiming, i.e. that $f(a)$ is convex on $[1/2,1]$. That can be checked by computing $f''(a)$ for instance. $\endgroup$ – Jack D'Aurizio Mar 20 '17 at 14:03
  • $\begingroup$ I do see now why $f''(a) \ge 0$, but not yet why $f(1/2) \le 2$. $\endgroup$ – Martin R Mar 20 '17 at 15:16

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