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 .
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:$g'(x)>0\\f(x)+f(x_2)−f(x+x_2)>0\\f(x)+f(x_2)>f(x+x_2)$