# Prove $f(x_1)+f(x_2)\gt f(x_1+x_2)$

Suppose $f''(x)\lt0$ for $x\gt0$ , $f(0)=0$ . Prove $$f(x_1)+f(x_2)\gt f(x_1+x_2)$$ for all $x_1\gt0$ and $x_2\gt0$ .

Trying to perform Mean value Theorem, also considered the properties of convex up function. Still have no clue. Especially the condition $f(0)=0$, I don't how to use it .

• For a fixed $x_2>0$ define $g(x)=f(x)+f(x_2)-f(x+x_2)$. Then show $g(0)=0$ and $g'(x)>0$ to conclude that $g$ stays positive. – Marco Aug 30 '18 at 15:19
• More generally you could try to prove $f(x_1+x_2)-f(x_1) \lt f(x_2)-f(0)$ which almost says that the slope between $(x_1,f(x_1))$ and $(x_1+x_2,f(x_1+x_2))$ is less than between $(0,f(0))$ and $(x_2,f(x_2))$ something that is intuitively obvious if $f''(x)<0$. Setting $f(0)=0$ allows you to make $f(0)$ disappear so you can rearrange to $f(x_1)+f(x_2) > f(x_1+x_2)$ – Henry Aug 30 '18 at 15:28

Based on what Marco said: For a fixed $x_2>0$ define $g(x)=f(x)+f(x_2)−f(x+x_2)$.
Then $g(0)=f(0)+f(x_2)-f(0+x_2)=0+f(x_2)-f(x_2)=0$
and $\large{g′(x)\\=\mathop{lim}\limits_{h→0}\frac{g(x+h)-g(x)}{h}\\=\mathop{lim}\limits_{h→0}\frac{f(x+h)+f(x_2)−f(x+h+x_2)-f(x)-f(x_2)+f(x+x_2)}{h}\\=\mathop{lim}\limits_{h→0}\frac{[f(x+h)-f(x)]−[f(x+h+x_2)-f(x+x_2)]}{h}\\=\mathop{lim}\limits_{h→0}\frac{f(x+h)-f(x)}{h}-\mathop{lim}\limits_{h→0}\frac{f(x+h+x_2)-f(x+x_2)}{h}\\=f'(x)-f'(x+x_2)>0}$
This is because $f''<0$ gives us that $f'(a)>f'(b)$ for $0<a<b$.
Conclusion:$g'(x)>0\\f(x)+f(x_2)−f(x+x_2)>0\\f(x)+f(x_2)>f(x+x_2)$