I want determine whether the following function is concave:
$f,g: R^n \rightarrow R, x \rightarrow \sqrt{f(x)g(x)} $
$f,g $ are concave and positive
I due to the definition of concavity I know:
$ \sqrt{f(\lambda x + (1- \lambda)y) \cdot g(\lambda x + (1- \lambda)y} \geq \sqrt{(\lambda f(x)+(1-\lambda)f(y)) \cdot (\lambda g(x) +(1-\lambda)g(y))}$
To my knowledge the function is concave if i can prove that:
$\sqrt{(\lambda f(x)+(1-\lambda)f(y)) \cdot (\lambda g(x) +(1-\lambda)g(y))} \geq \lambda \sqrt{f(x)g(x)} + (1-\lambda) \sqrt{f(y)g(y)}$
I was not able to arrive at this result. Is there a easier way? Or is the function not convave? Thank you in advance for any tip.
Noting that $0\le\lambda\le 1\;\;$ and $f, g$ are positive. You may proceed as follows:
To prove:$\sqrt{f(\lambda x + (1- \lambda)y) \cdot g(\lambda x + (1- \lambda)y} \geq \sqrt{(\lambda f(x)+(1-\lambda)f(y)) \cdot (\lambda g(x) +(1-\lambda)g(y))}$
Note that $\sqrt{(\lambda f(x)+(1-\lambda)f(y)) \cdot (\lambda g(x) +(1-\lambda)g(y))}=\sqrt{\lambda ^2 f(x)g(x)+\lambda(1-\lambda)(f(x)g(y)+f(y)g(x)) +(1-\lambda)^2 f(y)g(y)}\ge \sqrt{\lambda ^2 f(x)g(x)+2\lambda(1-\lambda)\sqrt{f(x)g(y)f(y)g(x)} +(1-\lambda)^2 f(y)g(y)}\; \;\text {By A.M. $\ge$ G.M.} \\ = \sqrt{\lambda ^2 \sqrt{f(x)g(x)}^2+2\lambda(1-\lambda)\sqrt{f(x)g(y)f(y)g(x)} +(1-\lambda)^2 \sqrt{f(y)g(y)}^2}\\= \lambda \sqrt{f(x)g(x)} + (1-\lambda) \sqrt{f(y)g(y)}$