**Problem**: (2004 USAMO - #5) Let $a,b,c > 0$. Prove that

$(a^5-a^2+3)(b^5-b^2+3)(c^5-c^2+3) \ge (a+b+c)^3$.

**Solution**: Note that $a^2-1$ and $a^3-1$ always have the same sign. Therefore $(a^2-1)(a^3-1) \ge 0 \Rightarrow a^5-a^2+3 \ge a^3+2$.

Then $(a^5-a^2+3)(b^5-b^2+3)(c^5-c^2+3) \ge (a^3+1+1)(1+b^3+1)(1+1+c^3) \ge (a+b+c)^3$

by Holder. QED.

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Comment: This is probably my favorite USAMO problem just because the solution is so short and simple. And the fact that I love inequalities. But honestly, can you get another USAMO problem (especially a #5) to give such a quick solution?

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Practice Problem: (2003 USAMO - #5) Let $a,b,c$ be positive real numbers. Prove that

$ \frac{(2a+b+c)^2}{2a^2+(b+c)^2}+\frac{(2b+c+a)^2}{2b^2+(c+a)^2}+\frac{(2c+a+b)^2}{2c^2+(a+b)^2} \le 8$.

Hey, I remember that question... but I know I didn't use Holder in the last step, lol. I think I expanded. XD

ReplyDeleteOh, there's an inequality I started on in Inequalities Unsolved Problems that I can't figure out, help me x.x