Let a_1, a_2, a_3, dots, a_{100} be positive | vedclass

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(A) Let $n = 100$. The sum of products of $n$ numbers taken $k$ at a time is denoted by $S_k$. By Newton's inequality or the Maclaurin's inequality,we

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$\binom{100}{2}^2$

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(A) Let $n = 100$. The sum of products of $n$ numbers taken $k$ at a time is denoted by $S_k$. By Newton's inequality or the Maclaurin's inequality,we know that for positive real numbers,the elementary symmetric means $E_k = \frac{S_k}{\binom{n}{k}}$ satisfy $E_k^2 \ge E_{k-1} E_{k+1}$. Alternatively,using the property $S_k S_{n-k} \ge \binom{n}{k}^2 (a_1 a_2 \dots a_n)$,we set $n = 100$ and $k = 2$. Then $S_2 S_{100-2} = S_2 S_{98} \ge \binom{100}{2}^2 (a_1 a_2 \dots a_{100})$. Comparing this with $S_{98} S_2 \ge \lambda (a_1 a_2 \dots a_{100})$,we get $\lambda = \binom{100}{2}^2$. Calculating the value: $\binom{100}{2} = \frac{100 	imes 99}{2} = 4950$. Thus,$\lambda = (4950)^2$.

Let $a_1, a_2, a_3, \dots, a_{100}$ be positive real numbers and $S_k$ be the sum of products of $a_1, a_2, \dots, a_{100}$ taken $k$ at a time. If $S_{98} S_2 \ge \lambda (a_1 a_2 \dots a_{100})$,then $\lambda$ is

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