Proving Inequality for Convex Functions with Given Conditions

[Contingency]

Homework Statement

Givens: \forall x\ge 0:\quad f^{ \prime \prime }\left( x \right) \ge 0;\quad f\left( 0 \right) =0
Prove: \forall a,b\ge 0:\quad f\left( a+b \right) \ge f\left( a \right) +f\left( b \right)

Homework Equations

By definition, f is convex iff \forall x,y\in \Re \quad \wedge \quad \forall \lambda :\quad 0\le \lambda \le 1\quad \Rightarrow \quad f\left( \lambda x+(1-\lambda )y \right) \le \lambda f\left( x \right) +(1-\lambda )f\left( y \right)

The Attempt at a Solution

Intuition-wise I see that a convex function's values increase at an increasing rate, but that's equivalent to f^{ \prime \prime }\left( x \right) \ge 0
I also see that f\left( 0 \right) =0 is necessary for the inequality to hold, but I can't find any tools with which I can work on proving the inequality.
Also I figure \forall x\ge 0:\quad f^{ \prime }\left( x \right) \ge 0 and also monotonously increasing.

 

[micromass]

OK, so we wish to prove that a convex function is "superadditive".

First, can you prove that if t\in [0,1], that then

f(tx)\leq tf(x)

Just apply convexity and use that f(0)=0.

 

[Contingency]

alright, that's immediate from taking y=0.
So I now know that f(\lambda x)\leq \lambda f(x)

 

[micromass]

Now write

f(a)+f(b)=f\left(\frac{a}{a+b}(a+b)\right)+f\left(\frac{b}{a+b}(a+b) \right)

and apply that inequality you just obtained.

 

[Contingency]

just figured that bit out
thanks alot!