Dear,
ans:- We know that A.M>G.M>HM
or AM>HM
Using the above relation we get
1/n∑(1/1+a1)>n/∑(1+a1)
or ∑(1/1+a1)>n²/(n+∑a1)
Let ∑(1/1+a1)=S
then we Have S>n²/(n+∑a1)....................(1)
Now using AM>GM we get
(1/n)∑a1>(∏a1)^(1/n)...........................(2)
From 1 & 2 we get
S>n²/(n+n(∏a1)^1/n)
or S>n/(1+(n th root of a1a2a3......an) (Proved)
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