Problem 2

Let

a_1,a_2,\ldots,a_n

be positive real numbers, and let

S_r

be the sum of all products of

r

of them. Prove that

S_kS_{n-k}\ge\binom{n}{k}^{\!2}S_n

for

k=1,2,\ldots,n-1

Step 3 of 4: Rewrite the two factors over the same index set

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S_kS_{n-k}=\left(\sum_{|I|=k}a_I\right)\left(\sum_{|I|=k}a_{I^c}\right)

Detailed analysis

Complementation is a bijection from the $k$ -subsets to the $(n-k)$ -subsets. Therefore the second symmetric sum can be indexed by the same $k$ -subsets, giving the displayed product.