Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question
4  
    
@ Angela Richardson: thanks for that link. The proof is very nice (marvellous)! –  Chris's sis Jul 21 '12 at 16:36
    
In fact, all proofs there are nice and simple. –  Chris's sis Jul 21 '12 at 17:01
    
Both displayed equations in the proof on the imomath site are marred by a serious, different, misprint. –  Did Jul 21 '12 at 17:07
    
@ 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 ... –  Chris's sis Jul 21 '12 at 17:12
show 3 more comments

1 Answer

up vote 8 down vote accepted

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 $n$.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.