<s>[INST] If $x,y,z \in \mathbb{R^+}$ such that $x+y+z=3$. Prove the inequality $\sqrt x+\sqrt y+\sqrt z\ge xy+yz+zx$ 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.
 [/INST] Hint: Now, one way to proceed: show $x^2-3x+2\sqrt x>0$.   Let $x=w^2$ and factor.
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