**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 $.

Man, I was about to post a problem like that.

ReplyDeleteAnyway, consider the largest and smallest of the five variables; WLOG the largest is a and the smallest is e. If a is greater than 1/2, then either b or c must be greater than a, contradiction; hence a ≤ 1/2. Similarly, if e is less than 1/2, then either a or b must be less than e, contradiction; hence e ≥ 1/2. Together, this implies a = b = c = d = e = 1/2.

Hmm, I'm not sure you aren't losing any generality with that WLOG. Since the set of equations is cyclic, picking one to be the largest and another to be the smallest (especially ones that share an equation) should change the problem...

ReplyDeleteThere 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.

Sorry, the WLOG should be made separately for the largest and the smallest. That is, do the argument for the largest first, then the argument for the smallest.

ReplyDeleteOh, I see. That makes sense.

ReplyDelete