\(x+y+z=3xyz\Leftrightarrow\frac{1}{xy}+\frac{1}{yz}+\frac{1}{zx}=3\).
Đổi biến: \(\left(\frac{1}{xy},\frac{1}{yz},\frac{1}{zx}\right)\rightarrow\left(a,b,c\right)\).
Khi đó ta có: \(\left\{{}\begin{matrix}y^2=\frac{c}{ab}\\z^2=\frac{a}{bc}\\x^2=\frac{b}{ca}\end{matrix}\right.\).
BĐT cần chứng minh tương đương với:
\(\frac{1}{\frac{b}{ca}+2\frac{1}{b^2}+1}+\frac{1}{\frac{c}{ab}+2\frac{1}{c^2}+1}+\frac{1}{\frac{a}{bc}+2\frac{1}{a^2}+1}\le\frac{3}{4}\)
\(\Leftrightarrow\frac{b^2ca}{b^3+2ca+b^2ca}+\frac{c^2ab}{c^3+2ab+c^2ab}+\frac{a^2bc}{a^3+2bc+a^2bc}\le\frac{3}{4}\). (*)
Áp dụng BĐT AM - GM:
\(b^3+2ca+b^2ca=b^3+ca+ca+b^2ca\ge4\sqrt[4]{b^5\left(ca\right)^3}\).
Do đó \(\frac{b^2ca}{b^3+2ca+b^2ca}\le\frac{b^2ca}{4\sqrt[4]{b^5\left(ca\right)^3}}=\frac{\sqrt[4]{b^3ca}}{4}\).
Tiếp tục áp dụng bất đẳng thức AM - GM ta có:
\(\sqrt[4]{b^3ca}=\sqrt[4]{bc.b.c.a}\le\frac{bc+b+c+a}{4}\).
Do đó: \(\frac{b^2ca}{b^3+2ca+b^2ca}\le\frac{\sqrt[4]{b^3ca}}{4}\le\frac{bc+b+c+a}{16}\).
Tương tự: \(\frac{c^2ab}{c^3+2ab+c^2ab}\le\frac{ca+c+a+b}{16};\frac{a^2bc}{a^3+2bc+a^2bc}\le\frac{ab+a+b+c}{16}\).
Cộng vế với vế ta có:
\(\Leftrightarrow\frac{b^2ca}{b^3+2ca+b^2ca}+\frac{c^2ab}{c^3+2ab+c^2ab}+\frac{a^2bc}{a^3+2bc+a^2bc}\le\frac{ab+bc+ca+3\left(a+b+c\right)}{16}\le\frac{\frac{\left(a+b+c\right)^2}{3}+3\left(a+b+c\right)}{16}=\frac{3+9}{16}=\frac{3}{4}\).
Do đó (*) luôn đúng.
BĐT được chứng minh.
