From this closed question: Let $x,y,z$ be positive real numbers that $x+y+z=3$. Prove: $\sqrt{x}+\sqrt{y}+\sqrt{z} \ge xy+yz+xz$
Show that for $x, y, z> 0$ such that $x+y+z = 3$:
$$\sqrt x + \sqrt y + \sqrt z \geq xy + xz+ yz$$
My attempt:
My first idea was to use cauchy schwarz, but that leads to a bound that is too coarse (for example take $x=3-2\epsilon, y=\epsilon, z=\epsilon$):
$$xy + yz + zx \leq x^2+y^2+z^2$$
I also tried looking at rearrangements of $xy + yz + zx$ but this led to similarly coarse bounds.
Finally I noticed that the constraint looks like the points along a circle if we use the substitution:
$$u = \sqrt 3 \sin\theta\cos\phi, v= \sqrt 3 \sin\theta\sin\phi, w = \sqrt 3 \cos\theta$$
So that: $u^2 + v^2 +w^2 = 3$. But after some work, I still havent made much progress:
$$\sqrt 3 \sin\theta\cos\phi + \sqrt 3 \sin\theta\sin\phi+ \sqrt 3 \cos\theta \stackrel{show}{\geq} 3^2 (\sin^2\theta\cos^2\phi\sin^2\theta\sin^2\phi + \sin^2\theta\cos^2\phi\cos^2\theta + \sin^2\theta\sin^2\phi\cos^2\theta)\\ = 3^2\sin^2\theta(\cos^2\phi\sin^2\theta\sin^2\phi + \cos^2\theta)$$
I tried to apply some double angle identities, but showing the inequality for all $\theta, \phi$ seems difficult still.
