<s>[INST] Let $a,b,c$ be non-negative real numbers , then $\frac{(a+b+c)^2}{3}\ge a\sqrt{bc}+b\sqrt{ca}+c\sqrt{ab}$? Let $a,b,c$ be non-negative  real numbers , then is it true that 
$\dfrac{(a+b+c)^2}{3}\ge a\sqrt{bc}+b\sqrt{ca}+c\sqrt{ab}$ ?
 [/INST] Yes, note that
$$ab+bc\ge 2 b\sqrt{ac},...$$
so
$$ab+bc+ca\ge a\sqrt{bc}+b\sqrt{ca}+c\sqrt{ab}.$$
The inequality then follows from 
$$\frac{(a+b+c)^2}3\ge ab+bc+ca.$$
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