﻿ Let z,y,z >0 such that xy+yz+zx+2xyz=1, show that \dfrac{1}{4x+1}+\dfrac{1}{4y+1}+\dfrac{1}{4z+1} \geq 1

#### Question

Let z,y,z >0 such that xy+yz+zx+2xyz=1, show that

\dfrac{1}{4x+1}+\dfrac{1}{4y+1}+\dfrac{1}{4z+1} \geq 1

Collected in the board: Inequality

Steven Zheng posted 5 days ago

Transform the given expression

xy+yz+zx+2xyz=1

(xyz+xy+xz+x)+(xyz+xy+yz+y)=xy+x+y+1

x(y+1)(z+1)+y(x+1)(z+1)=(x+1(y+1)

x(y+1)(z+1)+y(x+1)(z+1)+ z(x+1)(y+1) =(x+1(y+1)(z+1)

Divide both sides by x+1(y+1)(z+1)

Now we get

\dfrac{x}{x+1}+\dfrac{y}{y+1}+\dfrac{z}{z+1}=1

Let

a=\dfrac{x}{x+1}

b=\dfrac{y}{y+1}

c=\dfrac{z}{z+1}

then a+b+c=1

x=\dfrac{a}{1-a} , 4x+1=\dfrac{1+3a}{1-a}

y=\dfrac{b}{1-b} , 4y+1=\dfrac{1+3b}{1-b}

z=\dfrac{c}{1-c} , 4z+1=\dfrac{1+3c}{1-c}

\dfrac{1}{4x+1}+\dfrac{1}{4y+1}+\dfrac{1}{4z+1}

=\dfrac{1-a}{1+3a}+\dfrac{1-b}{1+3b}+\dfrac{1-c}{1+3c}

=\dfrac{4-1-3a}{3(1+3a)}+\dfrac{4-1-3b}{3(1+3b)}+\dfrac{4-1-3c}{3(1+3c)}

=\dfrac{4}{3}\dfrac{1}{1+3a} -\dfrac{1}{3} +\dfrac{4}{3}\dfrac{1}{1+3b} -\dfrac{1}{3} +\dfrac{4}{3}\dfrac{1}{1+3c} -\dfrac{1}{3}

=\dfrac{4}{3}( \dfrac{1}{1+3a} +\dfrac{1}{1+3b}+\dfrac{1}{1+3c})-1

=\dfrac{4}{3\cdot 6}( \dfrac{1}{1+3a} +\dfrac{1}{1+3b}+\dfrac{1}{1+3c})(6) -1

=\dfrac{2}{9}( \dfrac{1}{1+3a} +\dfrac{1}{1+3b}+\dfrac{1}{1+3c})[3+3(a+b+c)]-1

=\dfrac{2}{9}( \dfrac{1}{1+3a} +\dfrac{1}{1+3b}+\dfrac{1}{1+3c})[(1+3a)+(1+3b)+(1+3c)]-1

Apply the Cauchy–Schwarz inequality

\geq \dfrac{2}{9}(1+1+1)^2-1 = 1

Steven Zheng posted 5 days ago

Scroll to Top