Let  $f:R\to R$  be a continuous function defined by  $f\left(x\right)=\frac{1}{e^x+2e^{-x}}$
Statement-1:  $f\left(c\right)=1/3$, for some  $c\in R$
Statement-2 :  $0<f\left(x\right)\le \frac{1}{2\sqrt{2}}$, for all $x\in R$

Using A.M.-G.M. inequality
$\frac{e^x+2e^{-x}}{2}\ge \sqrt{e^x\cdot 2e^{-x}}\text{.}$
Thus,  $e^x+2e^{-x}\ge 2\sqrt{2}$ Then  $\frac{1}{e^x+2e^{-x}}\le \frac{1}{2\sqrt{2}}$
As  $\frac{1}{e^x+2e^{-x}}$ is always positive, we have  $0<\frac{1}{e^x+2e^{-x}}\le \frac{1}{2\sqrt{2}}$
Observe that  $f\left(0\right)=1/3$. Thus such that  $f\left(c\right)=1/3$
Using extreme-value theorem, we can say that as f is continuous, f will attain a value $1/3$ at some point. Here we are able to identify the point as well