If $x$, $y$ and $z$ are positive numbers, then prove that $$\frac {x}{x+y} + \frac{y}{y+z} +\frac {z}{z+x} \le 2.$$
Though I have solved a lot of problems on AM-GM inequality, I am unable to solve this one. I am also not showing my working because I do not think that they will be of any help.
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Solution by C-S.
We need to prove that: $$\sum_{cyc}\left(\frac{x}{x+y}-1\right)\leq-1$$ or $$\sum_{cyc}\frac{y}{x+y}\geq1,$$ which is true because by C-S $$\sum_{cyc}\frac{y}{x+y}=\sum_{cyc}\frac{y^2}{xy+y^2}\geq\frac{(x+y+z)^2}{\sum\limits_{cyc}(xy+y^2)}\geq1.$$
But the previous idea is still better: $$\sum_{cyc}\frac{y}{x+y}\geq\sum_{cyc}\frac{y}{x+y+z}=1.$$
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Alternatively: let $a = x+y, b = y+z, c = z+x$, then $a,b,c$ are the lengths of the sides of a... triangle From this: $x = (x+y+z) - (y+z) = \dfrac{a+b+c}{2} - b = \dfrac{a+c-b}{2}, y = \dfrac{a+b+c}{2} - c = \dfrac{a+b-c}{2}, z = \dfrac{a+b+c}{2}-a = \dfrac{b+c-a}{2}\implies LHS = \dfrac{a+c-b}{2a}+\dfrac{a+b-c}{2b}+\dfrac{b+c-a}{2c}\le 2=RHS\iff \dfrac{c-b}{a}+\dfrac{a-c}{b}+\dfrac{b-a}{c}\le 1$. Assume $0 < a \le b\le c$, then observe: $\dfrac{b-a}{c} < \dfrac{c}{c} = 1$, and $\dfrac{c-b}{a}+\dfrac{a-c}{b} = \dfrac{bc-b^2+a^2-ac}{ab}= \dfrac{(b-a)(c-a-b)}{ab}\le 0$ since $a \le b, c < a+b$. Thus $\dfrac{c-b}{a}+\dfrac{a-c}{b}+\dfrac{b-a}{c} < 1$, and the inequality is proven...
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If you want to do this more by hand write $$2=\frac {x+y}{x+y}+\frac {y+z}{y+z}$$ and the inequality becomes $$\frac z{x+z}\le\frac y{x+y}+\frac z{y+z}$$
If you now clear denominators (everything is positive) and expand the expression is homogeneous and not too difficult to work with. On the left you get $z(x+y)(y+z)=xyz+xz^2+y^2z+yz^2$ and it is easy to spot that these will appear on the right with other positive terms.
$$\sum_{cyc}\frac{x}{x+y}\leq\sum_{cyc}\frac{x+z}{x+y+z}=2$$