Let $x,y,z \in \Bbb R^+$ such that $x+y+z=3$. Prove the inequality
$\sqrt x+\sqrt y+\sqrt z \ge xy+yz+zx$
I tried to prove that $\sqrt x+\sqrt y+\sqrt z-(xy+yz+zx)\ge 0$
I squared the equality, put the value of $xy+yz+zx$ (in terms of $x^2+y^2+z^2$).
Now I tried to prove that the above expression but failed.
Thanks for hints or solutions.
Note that $\frac{xy+yz}{2}=y\frac{x+z}{2}=y\frac{3-y}{2}\leq\sqrt{y}$. Similarly we have that $\frac{xy+zx}{2}\leq\sqrt{x}$ and $\frac{yz+zx}{2}\leq\sqrt{z}$.
Your idea works because for $\sqrt{x}=a$, $\sqrt{y}=b$ and $\sqrt{z}=c$ we obtain: $$\sum_{cyc}(\sqrt{x}-xy)=\sum_{cyc}\left(\sqrt{x}-\frac{3-x^2}{2}\right)=\frac{1}{2}\sum_{cyc}(x^2+2\sqrt{x}-3)=$$ $$=\frac{1}{2}\sum_{cyc}(x^2+2\sqrt{x}-3-3(x-1))=\frac{1}{2}\sum_{cyc}(x^2+2\sqrt{x}-3x)=$$ $$=\frac{1}{2}\sum_{cyc}(a^4-3a^2+2a)=\frac{1}{2}\sum_{cyc}a(a+2)(a-1)^2\geq0.$$