A minimalistic approach?

Solution 1:

Since $a+b+c=1$ , we have $\sum_{cyc} \frac{a^2}{b+c}=\sum_{cyc} \frac{a^2}{1-a}$ let $f(x)=\frac{x^2}{1-x}$ . Since $f(x)$ is a convex function, by Jensen's Inequality, we have $f(\frac{a+b+c}{3}) \leq \frac{f(a)+f(b)+f(c)}{3}$ this implies that $f(a)+f(b)+f(c) \geq 3 \times \frac{1}{6}$ and hence the minimum value of this expression is $3 \times \frac{1}{6}=\boxed{0.5}$

Solution 2:

By Titu's lemma, we have $\sum_{cyc} \frac{a^2}{b+c} \geq \frac{(a+b+c)^2}{2(a+b+c)}=\boxed{0.5}$

Substitute $x_1$ = $\frac{a}{\sqrt{b+c}}$ , $x_2$ = $\frac{b}{\sqrt{a+c}}$ , $x_1$ = $\frac{c}{\sqrt{a+b}}$ , $y_1$ = $\sqrt{b+c}$ , $y_2$ = $\sqrt{a+c}$ and $y_3$ = $\sqrt{a+b}$ , into the Cauchy-Schwarz inequality. Rearranging gives: $\frac{a+b+c}{2}$ = $\frac{1}{2}$ $\leq(\frac{a^2}{b+c}$ + $\frac{b^2}{a+c}$ + $\frac{c^2}{a+b}$ )

by using am gm we come to know equality holds when a=b=c=1/3 apply the value in the expression then problem gets over.