<s>[INST] show this inequality $f(\sqrt{x_{1}x_{2}})\ge \sqrt{f(x_{1})\cdot f(x_{2})}$ let $$f(x)=\dfrac{1}{3^x}-\dfrac{1}{4^x},x\ge 1$$
For any $x_{1},x_{2}\ge 1$,show that
$$f(\sqrt{x_{1}x_{2}})\ge \sqrt{f(x_{1})\cdot f(x_{2})}$$
or
$$\left(\dfrac{1}{3^{\sqrt{x_{1}x_{2}}}}-\dfrac{1}{4^{\sqrt{x_{1}x_{2}}}}\right)^2 \ge \left(\dfrac{1}{3^{x_{1}}}-\dfrac{1}{4^{x_{1}}}\right)\left(\dfrac{1}{3^{x_{2}}}-\dfrac{1}{4^{x_{2}}}\right)$$Any ideas?
Thanks.
 [/INST] $f$ is nonincreasing since
\begin{align}
f^\prime(x)&=-3^{-x}\log 3+4^{-x}\log 4<0,\\
f^{\prime\prime}(x)&=3^{-x}(\log 3)^2-4^{-x}(\log 4)^2.
\end{align}
Similarly, denoting with $g(x)=\log(f(x))$, you have
\begin{align}
g^\prime(x)&=f^\prime(x)/f(x)<0,\\
g^{\prime\prime}(x)&=\frac{f^{\prime\prime}(x)f(x)-(f^\prime(x))^2}{f^2(x)}=-\frac{12^x\log^2(4/3)}{(4^x-3^x)^2}<0,
\end{align}
hence $g$ is concave and nonincreasing.
It follows by AM-GM that
$$
\log f(\sqrt{xy})\ge \log f\left(\frac{x+y}{2}\right)\ge \frac{\log f(x)+\log f(y)}{2}.
$$
The claim follows taking the exponentiation to both sides.
</s>