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!

Hint:, As $t \mapsto t^3$ is increasing, we may cube both sides and replace $x, y$ with $a^3, b^3$ to equivalently prove:
$$ \left(\frac{\lambda}a +  \frac{1-\lambda}b \right)^3 (\lambda a^3 + (1-\lambda) b^3) \geqslant 1 \tag{$\star$}$$
which follows from Hölder's inequality ...
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P.S. Hölder's inequality for our case (positive numbers $a, b, x, y$ and $p, q > 0$ s.t. $1/p + 1/q = 1$.
$$(x_1^p + x_2^p)^{1/p} (y_1^q+y_2^q)^{1/q} \geqslant x_1y_1 + x_2y_2$$
Now if we let $x_1 = (\lambda/a)^{3/4}, x_2 = ((1-\lambda)/b)^{3/4}$, $y_1 = (\lambda a^3)^{1/4}, y_2 = ((1-\lambda)b^3)^{1/4}$ with $p = 4/3, q = 4$, we get:
\begin{align} 
\left(\frac{\lambda}a +  \frac{1-\lambda}b \right)^{3/4} (\lambda a^3 + (1-\lambda) b^3)^{1/4} & \geqslant \left(\frac{\lambda}{a}\right)^{3/4}\left( \lambda a^3\right)^{1/4} + \left(\frac{1-\lambda}{b}\right)^{3/4}\left((1- \lambda) b^3\right)^{1/4} \\&= \lambda+(1-\lambda) = 1
\end{align}
which is essentially $(\star)$.
