Question

Using the reals $a_n(n=1,2,...,5)$

• A. None of these
• B. $\sum (a_n{)}^2\ge \sum (a_l{a}_m)$
• C. $\sum (a_n{)}^2\ge 2\sum (a_l{a}_m)$
• D. $2\sum (a_n{)}^2\ge \sum (a_l{a}_m)$

Answer 1

Answer: (D)

The correct option is B $2\sum (a_n{)}^2\ge \sum (a_l{a}_m)$
We know that, for five reals $a_n;(n=1,2,..,5)$
$\underset{\left(tenterms\right)}{\underset{}{(a_1-a_2{)}^2+(a_1-a_3{)}^2+...+(a_4-a_5{)}^2}}\ge 0$
$\Rightarrow 4(a_1^2+a_2^2+a_3^2+a_4^2+a_5^2)-2(a_1{a}_2+a_1{a}_3+a_1{a}_4+a_1{a}_5+a_2{a}_3+...+a_4{a}_5)\ge 0$
ie. $2(\sum a_n^2)\ge \sum \left(a_l{a}_m\right)$