Find the value of $A$ for the inequality

Question

if $x,y,z$ are positive reals and

$$S=\frac{x}{(x+y)(x+z)}+\frac{y}{(y+x)(y+z)}+\frac{z}{(z+x)(z+y)} \le A$$ find $A$

i have taken L.C.M getting

$$S=\frac{2(xy+yz+zx)}{(x+y)(y+z)(z+x)}$$ dividing by $xyz$ we get

$$S=\frac{2\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)}{xyz\left((\frac{1}{x}+\frac{1}{y})(\frac{1}{y}+\frac{1}{z})(\frac{1}{z}+\frac{1}{x})\right)}$$

Letting $a=\frac{1}{x}$, $b=\frac{1}{y}$ and $c=\frac{1}{z}$ we get

$$S=\frac{2abc(a+b+c)}{(a+b)(b+c)(c+a)}$$ $\implies$

$$S=\frac{abc(a+b+b+c+c+a)}{(a+b)(b+c)(c+a)}$$ $\implies$

$$S=abc \times \left(\frac{1}{(a+b)(b+c)}+\frac{1}{(b+c)(c+a)}+\frac{1}{((a+b)(c+a)}\right)$$

Now by AM-GM inequality

$$(a+b) \ge 2 \sqrt{ab}$$ and $(b+c) \ge 2 \sqrt{bc}$ so

$$(a+b)(b+c) \ge 4c\sqrt{ab}$$ so

$$S \le \frac{abc}{4c\sqrt{ab}}+\frac{abc}{4b\sqrt{ac}}+\frac{abc}{4a\sqrt{bc}}$$

so

$$S \le \frac{\sqrt{ab}+\sqrt{bc}+\sqrt{ca}}{4} \le \frac{a+b+c}{4}$$ by cauchy inequality

but how can we get $A$ from here

Answer 1

For $x=y=z\rightarrow0^+$ we get $\sum\limits_{cyc}\frac{x}{(x+y)(x+z)}\rightarrow+\infty$,

which says that the needed value of $A$ does not exist.

Answer 2

Without loss of generality, we can consider $x\le y \le z$ or $x \le ax \le abx$ with $a,b\ge1$.

The given inequality becomes equivalent to:

$$S=\frac{x}{x(a+1)\cdot x(ab+1)}+\frac{ax}{x(b+1) \cdot ax(b+1)}+\frac{abx}{x(ab+1) \cdot ax(b+1)} \le A$$

If $x\to0^+$, the LHS $\to+\infty=A$.