5
$\begingroup$

Let $n \ge3$ be an integer, and let $a_{2},a_{3}, ... ,a_{n}$ be positive real numbers such that $a_{2} a_{3}\cdots a_{n}=1.$ Prove that: $$(1+a_{2})^{2}(1+a_{3})^{3}\cdots(1+a_{n})^{n}\ge n^n$$

This is the 2nd problem of the 53rd IMO and seems pretty interesting. How would we solve that?

  • 4
    http://www.imomath.com/index.php?options=9&lmm=02012-07-21
  • 0
    @ Angela Richardson: thanks for that link. The proof is very nice (marvellous)!2012-07-21
  • 0
    In fact, all proofs there are nice and simple.2012-07-21
  • 0
    Both displayed equations in the proof on the `imomath` site are marred by a serious, different, misprint.2012-07-21
  • 0
    @ did: that's true, but i've got the main idea proof. I do mistakes, as well. I wonder if there could also be other nice approaches ...2012-07-21
  • 0
    [AoPS](http://www.artofproblemsolving.com/Forum/viewtopic.php?f=834&t=488342) is also OK. Incidentally, I have discovered the asymptotic value for the minimum.2012-07-22
  • 0
    @Frank Science: the approach with derivatives is pretty nice, as well.2012-07-22
  • 0
    @Chris'sister Frankly speaking, I expect somebody to check my asymptotic value, though I computed it twice and got the same answer.2012-07-22

1 Answers 1

9

Another approach that I thought of :

Set $a_2 =\frac{x_2}{x_3}, a_3=\frac{x_3}{x_4},\ldots, a_n=\frac{x_n}{x_2}$. This is a very useful substitution that we use in cases when we have a product equal to one like in this one $ a_2 a_3 \cdots a_n=1 $.

Now we need to prove that $$ (x_2+x_3)^2 (x_3+x_4)^3 \cdots (x_n+x_2)^n > n^n x_3^2 x_4^3 \cdots x_{n}^{n-1}x_2^n$$

which become obvious since for each $k$ by applying the Arithmetic-Geometric Mean we have that: $$ (x_k+x_{k+1})^{k}=\left(x_k+(k-1)\frac{x_{k+1}}{k-1}\right)^k\geqslant k^k x_k\frac{x_{k+1}^{k-1}}{(k-1)^{k-1}} $$

Just multiply for $k$ from $2$ to $n$.