# Convex Function Inequality

1. Jun 11, 2012

### Contingency

1. The problem statement, all variables and given/known data
Givens: $$\forall x\ge 0:\quad f^{ \prime \prime }\left( x \right) \ge 0;\quad f\left( 0 \right) =0$$
Prove: $$\forall a,b\ge 0:\quad f\left( a+b \right) \ge f\left( a \right) +f\left( b \right)$$

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

3. 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.

Last edited: Jun 11, 2012
2. Jun 11, 2012

### micromass

Staff Emeritus
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.

3. Jun 11, 2012

### Contingency

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

4. Jun 11, 2012

### micromass

Staff Emeritus
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.

Last edited: Aug 2, 2013
5. Jun 11, 2012

### Contingency

just figured that bit out
thanks alot!

Last edited: Jun 11, 2012