Let $x,y,z$ be positive real numbers such that $x+y+z = 1$ and $xy+yz+zx = \frac{1}{3}$. Find the vlaue of $$\frac{4x}{y+1} + \frac{16y}{z+1} + \frac{64z}{x+1}.$$
So far using the fact that $(x+y+z)^2 = x^2+y^2+z^2 + 2(xy+yz+zx)$, I was able to get $x^2+y^2+z^2 = \frac{1}{3}$. Then I tried finding the common denominator of $\frac{4x}{y+1} + \frac{16y}{z+1} + \frac{64z}{x+1}$, which got very messy quickly and I think there's a nicer way to do this problem, but for now I am stuck.
The point is that $x=y=z=1/3$.
Too see this consider $$0=(x+y+z)^2-3(xy+xz+yz)=\frac{(x-y)^2+(x-z)^2+(y-z)^2}{2}.$$