Question

Cho a,b,c > 0 và a+b+c=1.Tìm GTNN của \(X=\left(1+\frac{1}{a}\right)\left(1+\frac{1}{b}\right)\left(1+\frac{1}{c}\right)\)

Answer 1

Ta có:

\(1=a+b+c\ge3\sqrt[3]{abc}\)

\(\Rightarrow abc\le\frac{1}{27}\)

\(X=\left(1+\frac{1}{a}\right)\left(1+\frac{1}{b}\right)\left(1+\frac{1}{c}\right)\)

\(=\left(1+\frac{1}{3a}+\frac{1}{3a}+\frac{1}{3a}\right)\left(1+\frac{1}{3b}+\frac{1}{3b}+\frac{1}{3b}\right)\left(1+\frac{1}{3c}+\frac{1}{3c}+\frac{1}{3c}\right)\)

\(\ge\frac{4}{\sqrt[4]{27a^3}}.\frac{4}{\sqrt[4]{27b^3}}.\frac{4}{\sqrt[4]{27c^3}}\)

\(=\frac{4^3}{\sqrt[4]{27^3}.\sqrt[4]{a^3b^3c^3}}\ge\frac{4^3}{\sqrt[4]{27^3}.\sqrt[4]{\frac{1}{27^3}}}=64\)