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}   $
$\blacksquare$
 A: Your proof is correct, but could be made a little briefer, with a gain in clarity.

Here's a proof using the same ideas as yours, but I think a little easier to read . . .

\begin{align*}
&a+b=1\\[4pt]
\implies\;&(a+b)^2 = 1\\[4pt]
\implies\;&(a+b)^2 + (a-b)^2 \ge 1\\[4pt]
\implies\;&2a^2 + 2b^2 \ge 1\\[4pt]
\implies\;&a^2 + b^2 \ge \frac{1}{2}\\[4pt]
\end{align*}
A: Ok, here is a different approach: Using graphs. In the $a,b$ coordinate plane, $a+b=1$   represents a line and $a^2+b^2=\frac{1}{2}$   represents a circle. Now when you graph (You can do that) you notice that the line falls completely outside the circle, except for one point where the line is tangent. Conclusion? Note: While there is absolutely nothing wrong with an algebraic approach, sometimes it is also good to consider a geometric approach.
A: It is ok. You could have also put $b=1-a$ and minimize a quadratic function
A: hint AM-GM
$$(a+b)^2-a^2-b^2=2ab\le a^2+b^2$$
$$\implies 1-a^2-b^2\le  (a^2+b^2) $$
$$\implies 1\le 2 (a^2+b^2). $$
A: For an alternative proof, let $a=\frac{1}{2}+u\,$, then $a+b=1 \implies b = 1-a = \frac{1}{2} - u$. It follows that:
$$\require{cancel}
a^2+b^2 = \left(\frac{1}{2}+u\right)^2+\left(\frac{1}{2}-u\right)^2 = \frac{1}{4}+\bcancel{u}+u^2+\frac{1}{4}-\bcancel{u}+u^2 = \frac{1}{2}+2u^2 \ge \frac{1}{2}
$$
Note that the condition that $a,b$ be positive is not used, or required for the conclusion to hold true.
A: Follows directly from Hölder's inequality on a finite measure space (with dual exponents $(2,2)$): $1 = (a+b)^2 \le (a^2+b^2)(1^2+1^2)$.
A: 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: $\sin(2t) = 2 \sin t \cos t$.
