**Problem**: Let $ a, b, c $ be positive reals such that $ a+b+c = 3 $. Prove that $ \sqrt{a}+\sqrt{b}+\sqrt{c} \ge ab+bc+ca $.

**Solution**: We play around and guess that $ \sqrt{a} \ge \frac{ab+ac}{2} $ because that would be convenient. Indeed, replacing $ b+c $ with $ 3-a $, this is equivalent to

$ \sqrt{a} \ge \frac{a(3-a)}{2} \Leftrightarrow 4a \ge a^2(3-a)^2 \Leftrightarrow (a-1)^2(4-a) \ge 0 $,

which is clearly true for $ 0 < a < 3 $. So we get the three inequalities

$ \sqrt{a} \ge \frac{ab+ac}{2} $, $ \sqrt{b} \ge \frac{bc+ba}{2} $, $ \sqrt{c} \ge \frac{ca+cb}{2} $.

Adding them up, we have the desired $ \sqrt{a}+\sqrt{b}+\sqrt{c} \ge ab+bc+ca $. QED.

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Comment: I found this solution to be pretty clever, as the initial "guess" is not trivially true. But the whole thing works out quite nicely with symmetry so it's all good. Inequality problems usually require several random ideas and inspiration before finding the crux step.

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Practice Problem: Let $ a, b, c $ be positive reals such that $ abc = 1 $. Prove that $ (a+b)(b+c)(c+a) \ge 2(1+a+b+c) $.

Hm...

ReplyDelete$ \displaystyle \sum_{sym} a^2b+2abc\ge 2+2(a+b+c)$

So maybe we can prove $a^2(b+c)\ge 2a$.

Or $\frac{1}{b}+\frac{1}{c}\ge 2$

This is correct for $a>1$ because of $\frac{1}{b}+\frac{1}{c}\ge 2\sqrt{a}\ge 2$ by AM-GM.

It's not always true though ;). Try $ a = 1/10, b = 10, c = 1$.

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