$a,b,c > 0$ zodat $\frac{1}{a} + \frac{1}{b} + \frac{1}{c} = a+b+c$. Bewijs dat $$\frac{1}{(2a+b+c)^2}+\frac{1}{(a+2b+c)^2}+\frac{1}{(a+b+2c)^2}\leq \frac{3}{16}$$