How to prove that for a square matrix trace squared is less than $n$ times the sum of squared elements?

Question

How to prove that $${ \left( \sum_{i=1}^{n}{a_{ii}} \right)^2 \le n \sum_{i=1}^{n} \sum_{j=1}^{n}{a_{ij}}^2 }$$ for all ${A \in M_n( \Re ) }$?

I did manage to get to the point where I got the following expression

$${\mbox{tr}(A) \mbox{tr}(A^T) \le n \,\mbox{tr}(A^TA)}$$

but couldn't proceed further. Thanks in advance for your help!

Answer 1

Note that $\langle A, I \rangle = \operatorname{tr} A$ and so Cauchy Schwartz gives $| \operatorname{tr} A | \le \|A\|_F \|I\|_F$ or $(\operatorname{tr} A)^2 = n\sum_{i,j} A_{ij}^2 $.