Willow Pratt

Answered

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

$\begin{array}{}\text{(1)}& {\displaystyle \frac{1}{xy+z}}+{\displaystyle \frac{1}{yz+x}}+{\displaystyle \frac{1}{zx+y}}\le {\displaystyle \frac{1}{2}}\end{array}$

The theory basis of speculation Use the following classical results

$xyz=2+x+y+z\u27f9xy+yz+zx\ge 2(x+y+z)$

Let $x,y,z>0$ and such $xyz\ge 2+x+y+z$, show that

$\begin{array}{}\text{(1)}& {\displaystyle \frac{1}{xy+z}}+{\displaystyle \frac{1}{yz+x}}+{\displaystyle \frac{1}{zx+y}}\le {\displaystyle \frac{1}{2}}\end{array}$

The theory basis of speculation Use the following classical results

$xyz=2+x+y+z\u27f9xy+yz+zx\ge 2(x+y+z)$

Answer & Explanation

Kaya Kemp

Expert

2022-07-06Added 18 answers

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

$\sum _{cyc}\frac{xyz}{x(x+y)(x+z)+yz(y+z)}\le \frac{1}{2}$

or

$\sum _{cyc}\frac{xyz}{(x+y+z)({x}^{2}+yz)}\le \frac{1}{2}$

or

$\sum _{cyc}(\frac{x}{2}-\frac{xyz}{{x}^{2}+yz})\ge 0$

or

$\sum _{cyc}\frac{x({x}^{2}-yz)}{{x}^{2}+yz}\ge 0$

or

$\sum _{cyc}\frac{x((x-y)(x+z)-(z-x)(x+y))}{{x}^{2}+yz}\ge 0$

or

$\sum _{cyc}(x-y)(\frac{x(x+z)}{{x}^{2}+yz}-\frac{y(y+z)}{{y}^{2}+xz})\ge 0$

or

$\sum _{cyc}z(x-y{)}^{2}({x}^{2}+{y}^{2}+xz+yz)({z}^{2}+xy)\ge 0$

Done!

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

$\sum _{cyc}\frac{xyz}{x(x+y)(x+z)+yz(y+z)}\le \frac{1}{2}$

or

$\sum _{cyc}\frac{xyz}{(x+y+z)({x}^{2}+yz)}\le \frac{1}{2}$

or

$\sum _{cyc}(\frac{x}{2}-\frac{xyz}{{x}^{2}+yz})\ge 0$

or

$\sum _{cyc}\frac{x({x}^{2}-yz)}{{x}^{2}+yz}\ge 0$

or

$\sum _{cyc}\frac{x((x-y)(x+z)-(z-x)(x+y))}{{x}^{2}+yz}\ge 0$

or

$\sum _{cyc}(x-y)(\frac{x(x+z)}{{x}^{2}+yz}-\frac{y(y+z)}{{y}^{2}+xz})\ge 0$

or

$\sum _{cyc}z(x-y{)}^{2}({x}^{2}+{y}^{2}+xz+yz)({z}^{2}+xy)\ge 0$

Done!

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