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.
Open in App
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.