How can I prove the convexity of $f(x)=x^{-\frac{1}{3}}$ by using the definition I'd like to prove that the function $f(x)=x^{-\frac{1}{3}}$, with $x> 0$ is convex. Actually, I already know it's convex because I have studied its derivatives but I'd like to give a more "formal" prove by using convex definition, that is:
Let be $f:S \to \mathbb{R}$ a function. The function f is said to be convex in $S$ if $\forall x,y \in S$ and $\forall \lambda \in (0,1)$ the following holds:
$$f(\lambda x+(1-\lambda)y)\leq\lambda f(x)+(1-\lambda)f(y)$$
I know that I should prove this:
$$(\lambda x+(1-\lambda)y)^{-\frac{1}{3}} \leq \lambda x^{-\frac{1}{3}}+(1-\lambda)y^{-\frac{1}{3}} $$
But I don't know what can I do in order to do it. 
Any help would be aprecciated!

Instead of this, couldn't one just use the facts that 


*

*for $x_1,x_2 >0, t \in [0,1]: \\
 -  t x_1 + (1-t)  x_2 >0 \\
 -  t x_1 + (1-t)  x_2 \ge t x_1  \\
 -  t x_1 + (1-t) x_2 \ge (1-t) x_2  $

*$u>0 \rightarrow u^{1/3}>0$

*$u,v>0 , u \ge v \rightarrow u^{1/3} \ge v^{1/3}$
and finally 


*

*$u,v>0 , u \ge v \rightarrow u^{-1} \leq v^{-1}$

