$a^{\tan x}+a^{\cot x} \leq 2a$ 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 :)


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.

  • 1
    $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.