Russian MO 2004 Question involving the AM-GM inequality

Question

I'm reading Secrets in Inequalities by Pham Kim Hung, and I'm having trouble understanding this proof from a problem from the 2004 Russian MO.

Let a,b,c be positive real numbers and $a + b +c = 3$. Prove that $\sqrt{a} + \sqrt{b} + \sqrt{c} \ge ab + bc + ca$.

The proof is as follows.

First he notes that $2(ab + bc + ca) = (a + b +c)^2 - a^2 -b^2 -c^2$ which implies that $\sum_{cyc} a + 2 \sum_{cyc}\sqrt{a} \ge 9$, which must be true because of the AM-GM inequality because $\sum_{cyc} a + 2 \sum_{cyc}\sqrt{a} = \sum_{cyc}(a + \sqrt{a} + \sqrt{a}) \ge 3 \sum_{cyc} a = 9$

I don't quite understand how the last line follows from the AM-GM inequality.

Thanks

Related : https://math.stackexchange.com/questions/336362/prove-sqrta-sqrtb-sqrtc-ge-ab-bc-ca — Arnaud D., Dec 09 '19 at 16:04

score 3 · Accepted Answer · answered Aug 26 '15 at 05:55

3

I think there is a typo as I read that book and own one and some typos were spotted by me as well. It should be: $a^2+\sqrt{a}+\sqrt{a} \geq 3\sqrt[3]{a^2\sqrt{a}\sqrt{a}} = 3a$

answered Aug 26 '15 at 05:55

DeepSea

78,689

That makes a lot of sense. Ya he had a type on the first line of the proof as well. Thanks a lot. – Gargob Aug 26 '15 at 05:57
Fine, and it is a nice problem. – DeepSea Aug 26 '15 at 06:00

Russian MO 2004 Question involving the AM-GM inequality

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