For $a\geq b\geq c >0$. Prove that $$\dfrac{a}{a+b}+\dfrac{b}{b+c}+\dfrac{c}{c+a}\ge\dfrac{3}{2}$$
$a=100;b=1;c=1/100$ it's wrong ???
$\dfrac{a}{a+b}+\dfrac{b}{b+c}+\dfrac{c}{c+a}\ge\dfrac{3}{2}$
$<=>\sum\frac{a}{a+b}-\frac{1}{2}+\frac{b}{b+c}-\frac{1}{2}+\frac{c}{c+a}-\frac{1}{2}\geq 0$
$<=>\sum \frac{a-b}{2(a+b)}\geq 0$$<=>\sum\frac{z-y}{2x}\geq 0$
$<=>xy^2+yz^2+xz^2-x^2y-y^2z-z^2x\geq 0$
$<=>(y-x)(z-y)(z-x)\geq 0$
And we have $y-x\le 0; z-x\le 0; z-y\geq 0$
we are done !?
For $a\geq b\geq c>0$ we obtain: $$\sum_{cyc}\frac{a}{a+b}-\frac{3}{2}=\sum_{cyc}\left(\frac{a}{a+b}-\frac{1}{2}\right)=\sum_{cyc}\frac{a-b}{2(a+b)}=$$ $$=\sum_{cyc}\frac{(a-b)(c^2+ab+ac+bc)}{2\prod\limits_{cyc}(a+b)}=\sum_{cyc}\frac{c^2a-c^2b}{2\prod\limits_{cyc}(a+b)}=$$ $$=\sum_{cyc}\frac{a^2b-a^2c}{2\prod\limits_{cyc}(a+b)}=\frac{(a-b)(a-c)(b-c)}{2\prod\limits_{cyc}(a+b)}\geq0$$