Functional inequalities Let x,y,z be the lengths of the sides of a triangle, and let$$f(x,y,z)=\left|\frac {x-y}{x+y}+\frac {y-z}{y+z}+\frac {z-x}{z+x}\right|.$$ Find the upper limit of $f(x,y,z)$. I simply used the fact that $|x-y|\le z$ and the other 3 to prove that $f(x,y,z)\le \frac 18=0.125$. But the answer given is in terms of irrational numbers. Actually it is $f(x,y,z)\le \frac {8\sqrt2-5\sqrt5}{3}=0.04446.$ How do irrational numbers come into the picture...and well could you give the solution as well?

Solution:let $x\ge y\ge z$, First Note
$$I=\dfrac{x-y}{x+y}+\dfrac{y-z}{y+z}+\dfrac{z-x}{z+x}=\dfrac{(x-y)(x-z)(y-z)}{(x+y)(y+z)(x+z)}$$
then let $$x=c+b,y=c+a,z=a+b,c\ge b\ge a>0$$
so
$$I=\dfrac{(c-b)(c-a)(b-a)}{(2a+b+c)(2b+a+c)(2c+a+b)}<\dfrac{(c-b)cb}{(b+c)(2b+c)(2c+b)}=\dfrac{1}{F}$$
then we only find $F$ minimum
let $\dfrac{c}{b}=k\ge 1$,then
$$F=\dfrac{(1+k)(2+k)(1+2k)}{k(k-1)},k>1$$
since
$$F'_{k}=0\Longrightarrow \dfrac{2(k^4-2k^3-7k^2-2k+1)}{(k-1)^2k^2}=0$$
since $k>1$,this equation have only one roots,so seewolf
$$k=\dfrac{1}{2}(1+\sqrt{10}+\sqrt{7+2\sqrt{10}})=\dfrac{1}{2}(1+\sqrt{2}+\sqrt{5}+\sqrt{10})$$
and $$F(k)\ge F\left(\dfrac{1}{2}(1+\sqrt{2}+\sqrt{5}+\sqrt{10})\right)=\left(\dfrac{8\sqrt{2}-5\sqrt{3}}{3}\right)^{-1}$$
so
$$\left|\dfrac{x-y}{x+y}+\dfrac{y-z}{y+z}+\dfrac{z-x}{z+x}\right|\le \dfrac{8\sqrt{2}-5\sqrt{3}}{3}$$
