Find the minimum value of $f(x,y,z)=\frac{x^2}{(x+y)(x+z)}+\frac{y^2}{(y+z)(y+x)}+\frac{z^2}{(z+x)(z+y)}$.

Question

Find the minimum value of $f(x,y,z)=\frac{x^2}{(x+y)(x+z)}+\frac{y^2}{(y+z)(y+x)}+\frac{z^2}{(z+x)(z+y)}$ for all notnegative value of $x,y,z$.

I think that minimum value is $\frac{3}{4}$ when $x=y=z$ but I have no prove.

Answer 1

By C-S $$\sum_{cyc}\frac{x^2}{(x+y)(x+z)}\geq\frac{(x+y+z)^2}{\sum\limits_{cyc}(x+y)(x+z)}=\frac{\sum\limits_{cyc}(x^2+2xy)}{\sum\limits_{cyc}(x^2+3xy)}\geq\frac{3}{4},$$ where the last inequality it's $$\sum_{cyc}(x-y)^2\geq0.$$ The equality occurs for $x=y=z$, which says that we got a minimal value.

Another way:

We need to prove that: $$4\sum_{cyc}x^2(y+z)\geq3\prod_{cyc}(x+y)$$ or $$\sum_{cyc}(x^2y+x^2z-2xyz)\geq0$$ or $$\sum_{cyc}z(x-y)^2\geq0.$$

Answer 2

$f(x;y;z)=\frac{(xy+yz+zx)(y-z)^2 + (xy+xz-2yz)^2}{4(x+y)(z+x)(y+z)^2} +\frac{3}{4} \geq \frac{3}{4}$

Equality holds when $x=y=z$