# Mathematical Food for Thought

Serves a Daily Special and an All-You-Can-Eat Course in Problem Solving. Courtesy of me, Jeffrey Wang.

• ## Meta

Hey Now. Topic: Inequalities. Level: AIME. March 26th, 2007

Problem: Let be positive reals such that . Prove that .

Solution: We play around and guess that because that would be convenient. Indeed, replacing with , this is equivalent to

,

which is clearly true for . So we get the three inequalities

, , .

Adding them up, we have the desired . 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 be positive reals such that . Prove that .

### 2 Responses to “Hey Now. Topic: Inequalities. Level: AIME.”

1. ch1n353ch3s54a1l Says:

Hm…

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

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 $$a = 1/10, b = 10, c = 1$$.