Let $c\in \Big(0,\dfrac{\pi}{2}\Big) , a = \Big(\dfrac{1}{sin(c)}\Big)^{\dfrac{1}{cos^2 (c)}}, b = \Big(\dfrac{1}{cos(c)}\Big)^{\dfrac{1}{sin^2 (c)}}$.
Prove that at least one of $a,b$ is bigger than $\sqrt[11]{2015}$.
the function $t \mapsto \frac{\log(1-t)}{t}$ is decreasing on $(0,1)$ so that if $t \geq \frac{1}{2}$ then $\frac{\log(1-t)}{t} \leq -2 \log(2)$ thus $1-t \leq 2^{-2t}$.
Now either $\cos^2(c)$ or $\sin^2(c)$ is greater than $ \frac{1}{2}$. If it is the case for $\sin^2(c)$ the result follows for $t=\sin^2(c)$ because in this case $b \geq 2 >\sqrt[11]{2015}$
is there a solution without using calculus?
Using bernoulli's inequality or weighted AM-GM only?
Is there some other inequality which can be used when variable is in power?
$k =\sqrt[11]{2015} < 2$
Suppose that $k \geq a$ and $k \geq b$. Then we have $k^{cos^2(c)} \geq \frac{1}{sin(c)}$ and $k^{sin^2(c)} \geq \frac{1}{cos(c)}$. By multiplying we have $k^{cos^2(c)+sin^2(c)} = k \geq \frac{1}{sin(c)cos(c)} = \frac{2}{2sin(c)cos(c)} = \frac{2}{sin(2c)} \geq 2$, which is a contradiction.