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Question

Prove the following inequality
[(a1+a2+....an)2(b1+b2+....+bn)2]1/2<a12+b12+a22+b22+....+an2+bn2
Where ar,br(r=1,2,.,n) are real.

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Solution

let zk=ak+ibk, k=1, 2.....n.
Then z1+z2+...+zn=(a1+a2+.....+an)+i(b1+b2+...+bn)
Now |z1+z2+...zn||z1|+|z2|+...|zn|....(1)
But |zk|=a2k+b2k and |z1+z2+....+zn|
=(a1+a2+...+an)2+(b1+b2+...+bn)2
Hence substitution in (1), we get the required inequality.

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