Mathematical and Physical Journal
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Problem A. 566. (September 2012)

A. 566. (a) Prove that if n\ge2 and the product of the positive real numbers a_2,a_3,\ldots,a_n is 1 then {(1+a_2)}^2{(1+a_3)}^3
\cdots {(1+a_n)}^n >
\frac{n^n{(n-1)}^{n-1}}{4^{n-1}}. (b) Show an example for an integer n\ge2 and positive real numbers a_2,a_3,\ldots,a_n having product 1 that satisfy {(1+a_2)}^2{(1+a_3)}^3\ldots{(1+a_n)}^n < 1.000001 \cdot

(5 pont)

Deadline expired on October 10, 2012.

Solution. (a) For k\ge3, apply the AM-GM inequality to 1/(k-2) and ak/2 with the weights k-2 and 2:

\frac{1+a_k}{k} =
\frac{(k-2)\cdot\frac1{k-2}+2\cdot\frac{a_k}{2}}{k} \ge
\left(\frac{1}{k-2}\right)^{(k-2)/k}\cdot\left(\frac{a_k}{2}\right)^{2/k} =

(1+a_k)^k \ge \frac{k^k}{4(k-2)^{k-2}}a_k^2.  (1)

We have equality if 1/(k-2)=ak/2, so a_k=\frac2{k-2}.

Taking the product for k=3,\ldots,n,

(1+a_2)^2 (1+a_3)^3 \ldots (1+a_n)^n =
(1+a_2)^2 \prod_{k=3}^n (1+a_k)^k \ge
(1+a_2^2) \prod_{k=3}^n \left(\frac{k^k}{4(k-2)^{k-2}}a_k^2\right) =

= \frac{(1+a_2)^2}{a_2^2} \cdot
\Big(\underbrace{a_2a_3\ldots a_n}_{1}\Big)^2 \cdot 
\prod_{k=3}^n \frac{k^k}{4(k-2)^{k-2}} =
\frac{(1+a_2)^2}{a_2^2} \cdot \frac{n^n(n-1)^{n-1}}{4^{n-2}\cdot1^1\cdot2^2} =
\frac{(1+a_2)^2}{a_2^2} \cdot \frac{n^n(n-1)^{n-1}}{4^{n-1}}.

The trivial estimate 1+a22>a22 completes the proof.

(b) We have equality in (2) if a_k=\frac2{k-2} for k\ge3; then the constraint a_2a_3\ldots a_n=1 enforces a_2=\frac1{a_3\ldots a_n}=

It is easy to find that for n=14 we have a_2=\frac{14!}{2^{14}}=5320940.625 and \frac{(1+a_2)^2}{a_2^2} = \left(1+\frac1{a_2}\right)^2<1,000\;001, so

(1+a_2)^2 (1+a_3)^3 \ldots (1+a_n)^n =
\frac{(1+a_2)^2}{a_2^2} \cdot \frac{n^n(n-1)^{n-1}}{4^{n-1}} <
1{,}000\;001 \cdot \frac{n^n(n-1)^{n-1}}{4^{n-1}}.

Hence, a possible example is

n=16, \quad


7 students sent a solution.
5 points:Herczeg József, Ioan Laurentiu Ploscaru, Janzer Olivér, Omer Cerrahoglu, Szabó 789 Barnabás, Szabó 928 Attila, Williams Kada.

Problems in Mathematics of KöMaL, September 2012