Problem 2

Let

n\ge3

be an integer, and let

a_2,a_3,\dots,a_n

be positive real numbers such that

a_2a_3\cdots a_n=1

. Prove that

(1+a_2)^2(1+a_3)^3\cdots(1+a_n)^n>n^n

Step 4 of 4: Telescope and exclude equality

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In plain words

The desired lower bound is exactly n^n, and equality cannot occur simultaneously.

\prod_{k=2}^n\dfrac{k^k}{(k-1)^{k-1}}=n^n

Detailed analysis

The product telescopes to $n^n$ . Equality would require $a_k=1/(k-1)$ for every $k$ , whose product is $1/(n-1)!\ne1$ for $n\ge3$ . Therefore $(1+a_2)^2\cdots(1+a_n)^n>n^n$ .