## Thursday, April 12, 2007

### Fishy Triangular Number. Topic: Inequalities. Level: AIME.

Problem: (2001 Poland Finals - #1) Prove the following inequality:

$x_1+2x_2+3x_3+\cdots+nx_n \le \frac{n(n-1)}{2}+x_1+x_2^2+x_3^3+\cdots+x_n^n$

where $x_i$ are positive reals.

Solution: Like the title says, that triangular number looks really fishy... let's write it as $1+2+\cdots+(n-1)$ and pair up the terms on the RHS.

$\frac{n(n-1)}{2}+x_1+x_2^2+x_3^3+\cdots+x_n^n = x_1+(x_2^2+1)+(x_3^3+2)+\cdots+(x_n^n+n-1)$.

We claim that $x_i^i+i-1 \ge ix_i$ for $i = 1, 2, \ldots, n$. We'll use our good friend AM-GM to show this; in fact, it is quite simple.

$\frac{x_i+1+1+\cdots+1 \text{(i-1 times)}}{i} \ge \sqrt[i]{x_i^i} = x_i$ so $x_i^i+i-1 \ge ix_i$

as desired. Sum them up to get our inequality. QED.

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Comment: Just a clever little application of AM-GM; apparantly not a strong inequality at all, and the only equality case is $x_i = 1$ for all $i$. Nevertheless, slightly on the tricky side which makes it sufficiently satisfying to solve.

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Practice Problem: (2006 Poland Finals - #1) Solve in reals:

$a^2 = b^3+c^3$
$b^2 = c^3+d^3$
$c^2 = d^3+e^3$
$d^2 = e^3+a^3$
$e^2 = a^3+b^3$.

There is a way to avoid that, in any case, because we can say WLOG $a$ is the maximum, but then $e^2 = a^3+b^3 \ge c^3+b^3 = a^2$ so $e \ge a$ and thus $e = a$. We can make similar arguments for the remaining variables.