Inequality for sum of squares

Question

let $a_t$ and $b_t$ be any real number. Is the following inequality true? \begin{equation} \frac{1}{T} \sum_{t=1}^{T}2a_t^2 + \frac{1}{T} \sum_{t=1}^{T}2b_t^2 > \frac{1}{T} \sum_{t=1}^{T}(a_t + b_t)^2 \end{equation} Any comments is appreciated.

Answer 1

Just observe that

$2a_t^2+2b_t^2-(a_t+b_t)^2=(a_t-b_t)^2\geq0$

That proves your inequality (if you exchange the "$>$" with a "$\geq$").

Answer 2

As I showed in the comments, it doesn't have to be necessarily "bigger", so I'll prove the inequality $$\frac{1}{T} \sum_{t=1}^{T}2a_t^2 + \frac{1}{T} \sum_{t=1}^{T}2b_t^2 ≥ \frac{1}{T} \sum_{t=1}^{T}(a_t + b_t)^2\;\; a_t,b_t\in\mathbb R \;\text{and} \;T\in\mathbb N$$ Note that $$\frac{1}{T} \sum_{t=1}^{T}2a_t^2 + \frac{1}{T} \sum_{t=1}^{T}2b_t^2 ≥ \frac{1}{T} \sum_{t=1}^{T}(a_t + b_t)^2\iff\frac{1}{T}\sum^T_{t=1}2a_t^2+2b_t^2≥\frac{1}{T}\sum_{t=1}^T(a_t + b_t)^2$$ $$\iff \sum^T_{t=1}2a_t^2+2b_t^2≥\sum_{t=1}^T(a_t + b_t)^2$$ $$\iff2a_t^2+2b_t^2≥(a_t+b_t)^2\iff a_t^2+b_t^2≥2a_tb_t\iff a_t^2+b_t^2-2a_tb_t≥0$$ Which obviously holds for every $a_t,b_t\in\mathbb R$ since $$a_t^2+b_t^2-2a_tb_t=(a_t-b_t)^2≥0$$