# Thread: Prove x^1/2 is concave

1. ## Prove x^1/2 is concave

I have to prove this inequality to show that the function$\displaystyle f(x)=x^{1/2}$ is concave.

I am just stucked with the algebra

$\displaystyle (\mu x+ ( 1- \mu ) y )^{1/2} \geq \mu x^{1/2}+(1-\mu ) y^{1/2}$

For x>0,y>0, 0<mu<1

2. Originally Posted by altave86
I have to prove this inequality to show that the function$\displaystyle f(x)=x^{1/2}$ is concave.

I am just stucked with the algebra

$\displaystyle (\mu x+ ( 1- \mu ) y )^{1/2} \geq \mu x^{1/2}+(1-\mu ) y^{1/2}$

For x>0,y>0, 0<mu<1

First of all, note that $\displaystyle x \geq 0$.

$\displaystyle y = x^{\frac{1}{2}} = \sqrt{x}$

$\displaystyle \frac{dy}{dx} = \frac{1}{2}x^{-\frac{1}{2}} = \frac{1}{2\sqrt{x}}$

$\displaystyle \frac{d^2y}{dx^2} = -\frac{1}{4}x^{-\frac{3}{2}}$

$\displaystyle = -\frac{1}{4\sqrt{x^3}}$.

The function does not have derivative at $\displaystyle x = 0$ (why), and for all $\displaystyle x > 0$, the second derivative is $\displaystyle < 0$. This means that the gradient is always decreasing. Hence, a concave function.

3. Thank you!

However, I was asked to prove concavity by using the inequality approach I specified in the OP.

It is the same one to prove that a set is convex, but with the inequality reversed.