# If $x,y,z$ are positive numbers, then prove that $\frac {x}{x+y} + \frac{y}{y+z} +\frac {z}{z+x} \le 2$ [duplicate]

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.

## marked as duplicate by Xander Henderson, user99914, Claude Leibovici, The Dead Legend, user91500Sep 25 '17 at 15:11

• @MichaelRozenberg I see, sorry for my foolishness. – A---B Sep 25 '17 at 14:37
• How is this question a duplicate of that question? – robjohn Mar 30 at 17:50
• $2$ is sharp, but unattainable. Let $(x,y,z)=(1,t,t^2)$ and let $t\to0$. – robjohn Mar 30 at 19:52

$$\sum_{cyc}\frac{x}{x+y}\leq\sum_{cyc}\frac{x+z}{x+y+z}=2$$

• How did you get $\frac {a}{b} \le \frac {a+c}{b+c}$? – ami_ba Sep 24 '17 at 18:27
• @Amitayas Banerjee I used $x(x+y+z)<(x+z)(x+y)$. – Michael Rozenberg Sep 24 '17 at 18:28
• Oh! I got it now. Thank you very much. – ami_ba Sep 24 '17 at 18:31
• @Amitayas Banerjee You are welcome! I have found also a solution by C-S. I can post it if you wish. – Michael Rozenberg Sep 24 '17 at 18:33
• Yes, please post that too – ami_ba Sep 24 '17 at 18:34

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: $$\sum_{cyc}\frac{y}{x+y}\geq\sum_{cyc}\frac{y}{x+y+z}=1.$$

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...

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.