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 4 of 4: Apply Cauchy–Schwarz and finish

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S_kS_{n-k}\ge\left(\sum_{|I|=k}\sqrt{a_Ia_{I^c}}\right)^2=\binom nk^2S_n

Detailed analysis

Cauchy–Schwarz in the form $(\sum x_I)(\sum y_I)\ge(\sum\sqrt{x_Iy_I})^2$ gives the first inequality. By the complement identity, every square root equals $\sqrt{S_n}$ , and there are $\binom nk$ terms. Hence $S_kS_{n-k}\ge\binom nk^2S_n$ .