 Willow Pratt

2022-07-05

another inequality $\frac{1}{xy+z}+\frac{1}{yz+x}+\frac{1}{zx+y}\le \frac{1}{2}$
Let $x,y,z>0$ and such $xyz\ge 2+x+y+z$, show that
$\begin{array}{}\text{(1)}& \frac{1}{xy+z}+\frac{1}{yz+x}+\frac{1}{zx+y}\le \frac{1}{2}\end{array}$
The theory basis of speculation Use the following classical results
$xyz=2+x+y+z⟹xy+yz+zx\ge 2\left(x+y+z\right)$ Kaya Kemp

Expert

I agree with you. Your inequality is true!
Indeed, the condition gives $\sum _{cyc}\frac{1}{x+1}\le 1$
Let $x=a$, $y=b$ and $z=kc$, where $k>0$ and $\sum _{cyc}\frac{1}{a+1}=1$. Hence,
$\frac{1}{a+1}+\frac{1}{b+1}+\frac{1}{c+1}\ge \frac{1}{a+1}+\frac{1}{b+1}+\frac{1}{kc+1}$
which gives $k\ge 1$
Thus, $\sum _{cyc}\frac{1}{xy+z}=\frac{1}{ab+kc}+\frac{1}{kac+b}+\frac{1}{kbc+a}\le \frac{1}{ab+c}+\frac{1}{ac+b}+\frac{1}{bc+a}$
Id est, it remains to prove that $\sum _{cyc}\frac{1}{ab+c}\le \frac{1}{2}$ for positives $a$, $b$ and $csuch that\sum _{cyc}\frac{1}{a+1}=1$
Let $a=\frac{y+z}{x}$ and $b=\frac{x+z}{y}$, where $x$, $y$ and $z$ are positive numbers.
Hence, $c=\frac{x+y}{z}$ and we need to prove that $\sum _{cyc}\frac{1}{\frac{x+y}{z}\cdot \frac{x+z}{y}+\frac{y+z}{x}}\le \frac{1}{2}$ or
$\sum _{cyc}\frac{xyz}{x\left(x+y\right)\left(x+z\right)+yz\left(y+z\right)}\le \frac{1}{2}$
or
$\sum _{cyc}\frac{xyz}{\left(x+y+z\right)\left({x}^{2}+yz\right)}\le \frac{1}{2}$
or
$\sum _{cyc}\left(\frac{x}{2}-\frac{xyz}{{x}^{2}+yz}\right)\ge 0$
or
$\sum _{cyc}\frac{x\left({x}^{2}-yz\right)}{{x}^{2}+yz}\ge 0$
or
$\sum _{cyc}\frac{x\left(\left(x-y\right)\left(x+z\right)-\left(z-x\right)\left(x+y\right)\right)}{{x}^{2}+yz}\ge 0$
or
$\sum _{cyc}\left(x-y\right)\left(\frac{x\left(x+z\right)}{{x}^{2}+yz}-\frac{y\left(y+z\right)}{{y}^{2}+xz}\right)\ge 0$
or
$\sum _{cyc}z\left(x-y{\right)}^{2}\left({x}^{2}+{y}^{2}+xz+yz\right)\left({z}^{2}+xy\right)\ge 0$
Done!

Do you have a similar question?