Russian MO 2004 $\sqrt{a} + \sqrt{b} + \sqrt{c} \geq ab + bc + ca$

Question

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?

Answer 1

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 $$\sqrt{a}+\sqrt{b}+\sqrt{c}\geq ab+ac+bc.$$

Answer 2

And he says that the equation above is equivalent to the inequality below (which is what gives me doubt): $$\sum_{cyc} a^2 + 2\sum_{cyc} \sqrt{a} \geq 9 \tag{1}$$

How does he get to $2\sum_{cyc} \sqrt{a}$?

The original inequality (not the equation) is equivalent to $(1)$ because, after multiplying by $\,2\,$, it is:

$$ 2\sum_{cyc} \sqrt{a} \ge 2 \sum_{cyc} ab = \sum_{cyc} a(b+c) = \sum_{cyc}a(3-a) = 3 \sum_{cyc}a - \sum_{cyc} a^2 = 9 - \sum_{cyc} a^2 $$