Cho \(a,b,c\) là các số dương thay đổi thỏa mãn: \...

D

- Hướng dẫn giải

Phương pháp giải:

Áp dụng bất đẳng thức phụ:

\(\left( x+y+z+t \right).\left( \frac{1}{x}+\frac{1}{y}+\frac{1}{z}+\frac{1}{t} \right)\ge 16\text{ }hay\text{ }\frac{1}{x+y+z+t}\le \frac{1}{16}\left( \frac{1}{x}+\frac{1}{y}+\frac{1}{z}+\frac{1}{t} \right)\) (với \(x,y,z,t>0\))

Chứng minh:

\(\left\{ \begin{align} & x+y+z+t\overset{Cauchy}{\mathop{\ge }}\,4\sqrt[4]{xyzt} \\& \frac{1}{x}+\frac{1}{y}+\frac{1}{z}+\frac{1}{t}\overset{Cauchy}{\mathop{\ge }}\,4\sqrt[4]{\frac{1}{xyzt}} \\\end{align} \right.\Rightarrow \left( x+y+z+t \right).\left( \frac{1}{x}+\frac{1}{y}+\frac{1}{z}+\frac{1}{t} \right)\ge 16\)

Giải chi tiết:

Ta có: \(P=\frac{1}{2a+3b+3c}+\frac{1}{3a+2b+3c}+\frac{1}{3a+3b+2c}\)

\(\begin{align} & =\frac{1}{b+c+b+c+b+a+c+a}+\frac{1}{a+c+a+c+a+b+b+c}+\frac{1}{a+b+a+b+a+c+b+c} \\& \le \frac{1}{16}\left( \frac{1}{b+c}+\frac{1}{b+c}+\frac{1}{b+a}+\frac{1}{c+a} \right)+\frac{1}{16}\left( \frac{1}{a+c}+\frac{1}{a+c}+\frac{1}{a+b}+\frac{1}{b+c} \right) \\& +\frac{1}{16}\left( \frac{1}{a+b}+\frac{1}{a+b}+\frac{1}{a+c}+\frac{1}{b+c} \right) \\& =\frac{1}{16}\left( \frac{4}{b+c}+\frac{4}{a+b}+\frac{4}{c+a} \right) \\& =\frac{1}{4}\left( \frac{1}{b+c}+\frac{1}{a+b}+\frac{1}{c+a} \right)=\frac{2017}{4}. \\\end{align}\)

Dấu bằng xảy ra \(\Leftrightarrow a=b=c=\frac{3}{4034}\)

Vậy \(M\text{ax}P=\frac{2017}{4}\Leftrightarrow a=b=c=\frac{3}{4034}\) .

Chọn D