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.
 A: $f$ 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.
A: Let $$g(x)=\ln\left(\left(\frac{1}{3}\right)^x-\left(\frac{1}{4}\right)^x\right).$$
Thus, $$g'(x)=\frac{3^x\ln4-4^x\ln3}{4^x-3^x}<0$$
because $$3^x\ln4-4^x\ln3<0\Leftrightarrow\left(\frac{4}{3}\right)^x\geq\frac{\ln4}{\ln3},$$
for which it's enough to prove that
$$\frac{4}{3}\geq\frac{\ln4}{\ln3},$$ 
which is $3^4>4^3.$
Now, $$g''(x)=-\frac{12^x\ln^2\frac{4}{3}}{(4^x-3^x)^2}<0,$$
which says that $g$ is decreasing concave function.
Thus, by Jensen and AM-GM we obtain:
$$\frac{\ln\left(\left(\frac{1}{3}\right)^{x_1}-\left(\frac{1}{4}\right)^{x_1}\right)+\ln\left(\left(\frac{1}{3}\right)^{x_2}-\left(\frac{1}{4}\right)^{x_2}\right)}{2}\leq\ln\left(\left(\frac{1}{3}\right)^{\frac{x_1+x_2}{2}}-\left(\frac{1}{4}\right)^{\frac{x_1+x_2}{2}}\right)\leq$$
$$\leq\ln\left(\left(\frac{1}{3}\right)^{\sqrt{x_1x_2}}-\left(\frac{1}{4}\right)^{\sqrt{x_1x_2}}\right)$$
and we are done!
