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?

@GabrielRomon Yes, and there the OP was advised to post a new question. — saulspatz, Mar 31 '18 at 16:43
Just multiply the original inequality by $2$ and substitute the $2(ab+bc+ca)$ from observed equality. — Sil, Mar 31 '18 at 16:57
The cyc below the sum means that the sum will be of the permutations of the letter in that equation. For example: $\sum_{cyc} 2ab/c = 2ab/c + 2bc/a + 2ca/b$ — Gabriel Quintero, Mar 31 '18 at 17:00
I see what you are asking now... Didn't get it on first read. You have just confused word inequality and equality. What is really meant is that $\sum_{cyc} a^2 + 2\sum_{cyc} \sqrt{a} \geq 9$ is equivalent to the original inequality $\sqrt{a} + \sqrt{b} + \sqrt{c} \geq ab + bc + ca$, that's all. — Sil, Mar 31 '18 at 17:09
-Sil I got it now, your first comment was the one that helped me very much. My problem was when substituting the equality in the inequality and rearranging. Thanks a lot!!!! — Gabriel Quintero, Mar 31 '18 at 17:24

score 2 · Answer 1 · answered Mar 31 '18 at 17:58

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:

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

score 1 · Answer 2 · answered Mar 31 '18 at 17:09

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

answered Mar 31 '18 at 17:09

Michael Rozenberg

203,855

2

The OP actually did not ask for proving the inequality at all :) – Sil Mar 31 '18 at 17:13
I know. See please better my solution. In my answer there is the needed thing exactly. – Michael Rozenberg Mar 31 '18 at 17:15
It explains why it makes sense, however it did not tell me how Pham got to $\sqrt{etc}$. But still, I thank you very much, because your answer explained the last step of the proof. – Gabriel Quintero Mar 31 '18 at 17:26
You are welcome! – Michael Rozenberg Mar 31 '18 at 17:28

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

2 Answers2