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 $\sqrt[11]{2015}$.
Problem
Source: Moldova TST Problem 5
Tags: trigonometry, Trigonometric inequality, algebra, inequalities, function
tchebytchev
01.04.2015 22:17
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 $b \geq 2 >\sqrt[11]{2015}$
Quantum_fluctuations
11.08.2020 07:52
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?
Quantum_fluctuations
19.12.2020 19:01
/bump..........
augustin_p
11.05.2021 13:35
$k =\sqrt[11]{2015} < 2$ 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.