Concavity of $x \rightarrow \sqrt{f(x)g(x)}$ with $f,g$ concave and positive

Question

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.

$f(x) = \sqrt{x}$ would be an example of a concave and positive function if I am not mistaken. — testuser, Apr 04 '20 at 20:41
But $\sqrt{x}$ is defined only for $x\geq 0$, not $\mathbb{R}$. — Mikal, Apr 04 '20 at 21:00

score 0 · Accepted Answer · answered Apr 04 '20 at 21:14

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)}$

Concavity of $x \rightarrow \sqrt{f(x)g(x)}$ with $f,g$ concave and positive

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