Proving that for $\{ a,b\} \subset \Bbb{R^{+}}$; $a+b=1 \implies a^2+b^2 \ge \frac{1}{2}$ 
For $\{ a,b\} \subset \Bbb{R^{+}}$; $a+b=1 \implies a^2+b^2 \ge
 \frac{1}{2}$

I'm trying to prove this in the following way, but I'm not sure if it's correct. Could anyone please check it and see if it's okay?
$a+b=1 \implies (a+b)^2 = 1^2 = 1 \implies (a+b)-(a+b)^2 = 1-1 =0$   (1) 
$(a-b)^2 \ge 0$
So by (1) we have:
$(a-b)^2 \ge (a+b)-(a+b)^2$
$(a^2-2ab+b^2) \ge (a+b) - (a^2+2ab+b^2)$
$(a^2-2ab+b^2) + (a^2+2ab+b^2) \ge (a+b) $
$ a^2+a^2+b^2+b^2+2ab-2ab \ge (a+b)$
$2(a^2+b^2) \ge (a+b)$
$2(a^2+b^2) \ge 1$
$(a^2+b^2) \ge \frac{1}{2}   $
$\blacksquare$

For fun:
$a + b = 1$; 
$a := \cos^2t;$  $b := \sin^2t$ , $ 0 \le t \lt 2π$.
$(a+b)^2 = a^2 +2ab +b^2 = $
$\cos^4t +2\cos^2t \sin^2t + \sin^4t =1$.
$\cos^4t + \sin^4t = 1 - (1/2)(\sin2t)^2$.
$\Rightarrow \cos^4t + \sin^4t  \ge 1 - 1/2$ , 
since $ (\sin2t)^2 \le 1$.
Finally substituting back:  
$a^2 + b^2 \ge 1/2$.
Recall: $\sin(2t) = 2 \sin t \cos t$.
