# f(x)+f(y)<f(x+y)

#### Sasbe

Let's assume that f:[0,∞[ -> R is differentiable twice and f(0)=0 and f''(x)>0, when x>0.

Show that f(x)+f(y)<f(x+y) when x,y>0.

It's really important that I get this done, so any help is highly appreciated!!

#### Opalg

MHF Hall of Honor
Let's assume that f:[0,∞[ -> R is differentiable twice and f(0)=0 and f''(x)>0, when x>0.

Show that f(x)+f(y)<f(x+y) when x,y>0.
For fixed y, let $$\displaystyle g(x) = f(x+y)-f(x)-f(y)$$. Then $$\displaystyle g(0) = 0$$. Show that $$\displaystyle g'(x)>0$$ for all x>0, and deduce that $$\displaystyle g(x)>0$$ for all x>0.

#### Sasbe

Thanks Opalg for the hint. Is f a (strictly) convex function because f''(x)>0? Or is it not because f''(x)=0 when x=0 (because f(0)=0)?

I'm still pretty confused with this.

#### Opalg

MHF Hall of Honor
Thanks Opalg for the hint. Is f a (strictly) convex function because f''(x)>0? Or is it not because f''(x)=0 when x=0 (because f(0)=0)?
The only reason that you need the condition f''(x)>0 (for x>0) is that it implies that f'(x) is a strictly increasing function, so that f'(x+y) > f'(x) when y>0.

The value of f''(x) when x=0 is not relevant. The fact that f''(x) > 0 when x>0 does in fact imply that f is strictly convex.

• Sasbe