Question

Cho 3 số thực dương a, b, c thỏa mãn: \(\frac{1}{1+a}+\frac{21}{21+2b}\le\frac{4c}{4c+27}\). Tìm GTNN của biểu thức A=abc

Answer 1

\(\Leftrightarrow\frac{1}{1+a}+\frac{a}{1+a}+\frac{2b}{21+2b}+\frac{21}{21+2b}\le\frac{4c}{4c+27}+\frac{a}{1+a}+\frac{2b}{21+2b}\)

\(\Leftrightarrow2\le\frac{1}{1+\frac{1}{a}}+\frac{1}{1+\frac{21}{2b}}+\frac{1}{1+\frac{27}{4c}}\)

Đặt \(\left(\frac{1}{a};\frac{21}{2b};\frac{27}{4c}\right)=\left(x;y;z\right)\)

\(\Leftrightarrow\frac{1}{1+x}+\frac{1}{1+y}+\frac{1}{1+z}\ge2\)

\(\Leftrightarrow\frac{1}{1+x}\ge1-\frac{1}{1+y}+1-\frac{1}{1+z}=\frac{y}{1+y}+\frac{z}{1+z}\ge2\sqrt{\frac{yz}{\left(1+y\right)\left(1+z\right)}}\)

Tương tự: \(\frac{1}{1+y}\ge2\sqrt{\frac{zx}{\left(1+z\right)\left(1+x\right)}}\) ; \(\frac{1}{1+z}\ge2\sqrt{\frac{xy}{\left(1+x\right)\left(1+y\right)}}\)

Nhân vế với vế: \(1\ge8xyz\Rightarrow xyz\le\frac{1}{8}\)

\(\Leftrightarrow\frac{1}{a}.\frac{21}{2b}.\frac{27}{4c}\le\frac{1}{8}\Leftrightarrow abc\ge567\)

Dấu "=" xảy ra khi \(\frac{1}{a}=\frac{21}{2b}=\frac{27}{4c}=\frac{1}{2}\Rightarrow\left(a;b;c\right)=\left(2;21;\frac{27}{2}\right)\)