Following is the picture of the question regarding the application of Jensen Inequality.
Following is the picture my approach to proove the inequality.
Can anyone please check if my proof is acceptable...i feel that this is an elegant proof.
If there is any error, a detailed explanation would be helpful and appreciated.
Thank you
Also, appologies for the bad handwriting
The step $$\frac{n}{1-\frac{(a_1+...+a_n)^2}{n^2}}\geq\frac{n}{1-\frac{a_1^2+...+a_n^2}{n}}$$ is wrong: it should be $$\frac{n}{1-\frac{(a_1+...+a_n)^2}{n^2}}\leq\frac{n}{1-\frac{a_1^2+...+a_n^2}{n}},$$ which does not help.
My solution.
$$\sum_{i=1}^n\frac{a_i}{1-a_i^2}-\frac{na}{1-a^2}=\sum_{i=1}^n\left(\frac{a_i}{1-a_i^2}-\frac{a}{1-a^2}\right)=\frac{1}{1-a^2}\sum_{i=1}^n\frac{(a_i-a)(aa_i+1)}{1-a_i^2}=$$ $$=\frac{1}{1-a^2}\sum_{i=1}^n\left(\frac{(a_i-a)(aa_i+1)}{1-a_i^2}-\frac{(a_i^2-a^2)(a^2+1)}{2a(1-a^2)}\right)=$$ $$=\frac{1}{2a(1-a^2)^2}\sum_{i=1}^n\frac{(a_i-a)^2(a_i^2+(2a^3+a^2+2a)a_i+3a^2-1)}{1-a_i^2}\geq0.$$