If $x,y,z \in \mathbb{R^+}$ such that $x+y+z=3$. Prove the inequality $\sqrt x+\sqrt y+\sqrt z\ge xy+yz+zx$

Question

If $x,y,z \in \mathbb{R^+}$ such that $x+y+z=3$. Prove the inequality $\sqrt x+\sqrt y+\sqrt z\ge xy+yz+zx$.

My work: We have

$$3(x+y+z)=x^2+y^2+z^2+2(xy+yz+zx) \implies (xy+yz+zx)=\dfrac12(3x-x^2+3y-y^2+3z-z^2)$$

So, we have to prove, $\sqrt x+\sqrt y+\sqrt z-\dfrac12(3x-x^2+3y-y^2+3z-z^2)\ge 0$

Now, I cannot proceed further. Please help.

Answer 1

Hint: Now, one way to proceed: show $x^2-3x+2\sqrt x>0$. Let $x=w^2$ and factor.