showing algebraic inequality with arithmetic and harmonic means Let x, y, z be positive real numbers. Prove that $$\frac{x}{y+z}+\frac{y}{z+x}+\frac{z}{x+y} \ge \frac{3}{2}$$
This problem appears to be simple, but upon further work and lots of failed attempts, I am stuck. I have tried using arithmetic and harmonic means (which I am sure are the key) to show that there exists some number which is fits between these two, thus proving the inequality. I have also tried multiplying it out and simplifying and obtained:
$$1 + \frac{x^3 + y^3 + z^3 + xyz}{(x+y)(y+z)(z+x)} \ge \frac{3}{2} $$ This didn't really seem to help. Any guidance is greatly appreciated!

$$A=\dfrac a{b+c}+\dfrac{b}{a+c}+\dfrac{c}{a+b}$$
$$B= \dfrac b{b+c}+ \dfrac{c}{c+a}+\dfrac{a}{a+b}$$
$$C=\dfrac c{b+c}+\dfrac{a}{a+c}+\dfrac{b}{b+a}$$
By A-G inequality
$$A+B=\dfrac{a+b}{b+c}+\dfrac{b+c}{a+c}+\dfrac{c+a}{b+a}\ge 3$$
$$A+C=\dfrac{a+c}{b+c}+\dfrac{a+b}{a+c}+\dfrac{b+c}{b+a}\ge3$$
so
+
$$(A+B)+(A+C)\ge6$$
$B+C=3$, we get $2A\ge3$, yeah
