Problem A. 566. (September 2012)

A. 566. (a) Prove that if n $\ge$ 2 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 $\ge$ 2 and positive real numbers $a_2,a_3,\ldots,a_n$ having product 1 that satisfy ${(1+a_2)}^2{(1+a_3)}^3\ldots{(1+a_n)}^n < 1.000001 \cdot \frac{n^n{(n-1)}^{n-1}}{4^{n-1}}$ .

(5 pont)

Deadline expired on October 10, 2012.

Solution. (a) For k $\ge$ 3, apply the AM-GM inequality to 1/(k-2) and a_k/2 with the weights k-2 and 2:

$\frac{1+a_k}{k} = \frac{(k-2)\cdot\frac1{k-2}+2\cdot\frac{a_k}{2}}{k} \ge \left(\frac{1}{k-2}\right)^{(k-2)/k}\cdot\left(\frac{a_k}{2}\right)^{2/k} = \left(\frac{a_k^2}{4(k-2)^{k-2}}\right)^{1/k}$

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

(1)

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

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

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

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

(2)

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

(b) We have equality in (2) if $a_k=\frac2{k-2}$ for k $\ge$ 3; then the constraint $a_2a_3\ldots a_n=1$ enforces $a_2=\frac1{a_3\ldots a_n}= \prod_{k=3}^n\frac{k-2}2=\frac{(n-2)!}{2^{n-2}}$ .

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

$n=16, \quad (a_2,a_3,\ldots,a_{16})= \left(\frac{14!}{2^{14}},\frac21,\frac22,\frac23,\ldots,\frac2{14}\right).$

Statistics:

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

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

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