<s>[INST] Russian MO 2004 $\sqrt{a} + \sqrt{b} + \sqrt{c} \geq ab + bc + ca$ I have a doubt on a proof included in "Secrets in Inequalities" by Pham Kim Hung. The exercise is to prove $$\sqrt{a} + \sqrt{b} + \sqrt{c} \geq ab + bc + ca$$ for a, b, c whose sum is 3.
His approach is the following
He observes that:
$$2(ab + bc + ca) = (a + b + c)^2 - (a^2 + b^2 + c^2)$$
And he says that the equation above is equivalent to the inequality below(which is what gives me doubt):
$$\sum_{cyc} a^2 + 2\sum_{cyc} \sqrt{a} \geq 9$$
How does he get to $2\sum_{cyc} \sqrt{a} $? Does he get this out of the blue? Or is there some logic behind this?
 [/INST] By AM-GM
$$\sum_{cyc}(a^2+2\sqrt{a})\geq3\sum_{cyc}\sqrt[3]{a^2\cdot(\sqrt{a})^2}=$$
$$=3\sum_{cyc}a=9=(a+b+c)^2=\sum_{cyc}(a^2+2ab),$$
which gives $$\sqrt{a}+\sqrt{b}+\sqrt{c}\geq ab+ac+bc.$$
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