
Problem 10.1. Find all values of the real positive parameter $a$ such that the inequality $a^{\cos 2 x}+a^{2 \sin ^{2} x} \leq 2$ holds for any real $x$.


Solution: We know that $a^{\cos 2 x}+a^{2 \sin ^{2} x}=a^{1-2 \sin ^{2} x}+a^{2 \sin ^{2} x}=$ $\frac{a}{a^{2 \sin ^{2} x}}+a^{2 \sin ^{2} x}$. Substitute $t=a^{2 \sin ^{2} x}$. Since $0 \leq \sin ^{2} x \leq 1$, we obtain that $t$ is between 1 and $a^{2}$. Our inequality now becomes $\frac{a}{t}+t \leq 2 \Longleftrightarrow t^{2}-2 t+a \leq 0$. Since it holds true for any $x$ (i. e., for any $t$ between 1 and $a^{2}$ ), it follows that the roots of $f(t)=$ $t^{2}-2 t+a=0$ lie outside the open interval determined by 1 and $a^{2}$. Therefore $f(1) \leq 0$ and $f\left(a^{2}\right) \leq 0$. The first inequality gives $a \leq 1$ and the second one implies $a^{4}-2 a^{2}+a \leq 0 \Longleftrightarrow a^{3}-2 a+1 \leq$ $0 \Longleftrightarrow(a-1)\left(a^{2}+a-1\right) \leq 0$. Since $a \leq 1$, we obtain $a^{2}+a-1 \geq 0$.

The solution of this inequality is $a \in\left[\frac{-1-\sqrt{5}}{2}, \frac{-1+\sqrt{5}}{2}\right]$. So we obtain $a \in\left[1, \frac{-1+\sqrt{5}}{2}\right]$.
