If a, b, c>0 show that $\frac{a^2}{b^2+bc}+\frac{b^2}{c^2+ac}+\frac{c^2}{a^2+ab} \ge \frac{3}{2}$ 
For positive real numbers $a$, $b$, and $c$ prove that: $$\frac{a^2}{b^2+bc}+\frac{b^2}{c^2+ac}+\frac{c^2}{a^2+ab} \ge \frac{3}{2}.$$

I let $x=\frac{a}{b}$, $y=\frac{b}{c}$, and $z=\frac{c}{a}$. Then inequality becomes
$$A=\frac{x^2}{1+xz}+\frac{y^2}{1+xy}+\frac{z^2}{1+zy} \ge \frac{3}{2}$$
By Cauchy-Schwarz 
$$(3+xy+yz+zx)A \ge (x+y+z)^2 \ge 3(xy+yz+zx)$$
What then?

If you let $B=xy+yz+zx$, by AM-GM, $B \ge 3$ and 
$$A \ge \frac{3B}{B+3}=\frac{3}{1+3/B} \ge \frac32$$
