Cho a, b, c là ba số thực dương thỏa mãn điều kiện...

B

- Hướng dẫn giải

Phương pháp giải:

Giải chi tiết:

\(\begin{align}  & \frac{1}{\left( 5{{a}^{2}}+2ab+2{{b}^{2}} \right)}+\frac{1}{27}\ge \frac{2}{\sqrt{27\left( 5{{a}^{2}}+2ab+2{{b}^{2}} \right)}} \\  & \Leftrightarrow \frac{1}{\sqrt{5{{a}^{2}}+2ab+2{{b}^{2}}}}\le \frac{\sqrt{27}}{2}\left( \frac{1}{\left( 5{{a}^{2}}+2ab+2{{b}^{2}} \right)}+\frac{1}{27} \right) \\ \end{align}\)

Chứng minh tương tự ta có

\(\begin{align}  & \frac{1}{\sqrt{5{{b}^{2}}+2bc+2{{c}^{2}}}}\le \frac{\sqrt{27}}{2}\left( \frac{1}{\left( 5{{b}^{2}}+2bc+2{{c}^{2}} \right)}+\frac{1}{27} \right) \\  & \frac{1}{\sqrt{5{{c}^{2}}+2ca+2{{a}^{2}}}}\le \frac{\sqrt{27}}{2}\left( \frac{1}{\left( 5{{c}^{2}}+2ca+2{{a}^{2}} \right)}+\frac{1}{27} \right) \\ \end{align}\)

\(\Rightarrow P\le \frac{\sqrt{27}}{2}\left( \frac{1}{\left( 5{{a}^{2}}+2ab+2{{b}^{2}} \right)}+\frac{1}{\left( 5{{b}^{2}}+2bc+2{{c}^{2}} \right)}+\frac{1}{\left( 5{{c}^{2}}+2ca+2{{a}^{2}} \right)}+\frac{1}{9} \right)\)

Sử dụng BĐT \(\frac{1}{x+y+z}\le \frac{1}{9}\left( \frac{1}{x}+\frac{1}{y}+\frac{1}{z} \right)\)ta có:

\(\begin{align}  & \frac{1}{\left( 5{{a}^{2}}+2ab+2{{b}^{2}} \right)}=\frac{1}{3{{a}^{2}}+\left( 2ab+{{a}^{2}} \right)+\left( {{a}^{2}}+2{{b}^{2}} \right)}\le \frac{1}{9}\left( \frac{1}{3{{a}^{2}}}+\frac{1}{2ab+{{a}^{2}}}+\frac{1}{{{a}^{2}}+2{{b}^{2}}} \right) \\  & \le \frac{1}{9}\left( \frac{1}{3{{a}^{2}}}+\frac{1}{9}\left( \frac{1}{ab}+\frac{1}{ab}+\frac{1}{{{a}^{2}}} \right)+\frac{1}{9}\left( \frac{1}{{{a}^{2}}}+\frac{1}{{{b}^{2}}}+\frac{1}{{{b}^{2}}} \right) \right) \\  & =\frac{1}{9}\left( \frac{5}{9{{a}^{2}}}+\frac{2}{9ab}+\frac{2}{9{{b}^{2}}} \right) \\ \end{align}\)

Ta lại có \(\frac{2}{9ab}\overset{Cauchy}{\mathop{\le }}\,\frac{1}{9}\left( \frac{1}{{{a}^{2}}}+\frac{1}{{{b}^{2}}} \right)\)

\(\Rightarrow \frac{1}{5{{a}^{2}}+2ab+2{{b}^{2}}}\le \frac{1}{9}\left( \frac{5}{9{{a}^{2}}}+\frac{1}{9{{a}^{2}}}+\frac{1}{9{{a}^{2}}}+\frac{2}{9{{b}^{2}}} \right)=\frac{1}{9}\left( \frac{2}{3{{a}^{2}}}+\frac{1}{3{{b}^{2}}} \right)\)

CMTT ta có :

\(\begin{align}  & \frac{1}{5{{b}^{2}}+2bc+2{{c}^{2}}}\le \frac{1}{9}\left( \frac{2}{3{{b}^{2}}}+\frac{1}{3{{c}^{2}}} \right) \\  & \frac{1}{5{{c}^{2}}+2ca+2{{a}^{2}}}\le \frac{1}{9}\left( \frac{2}{3{{c}^{2}}}+\frac{1}{3{{a}^{2}}} \right) \\  & \Rightarrow \frac{1}{\left( 5{{a}^{2}}+2ab+2{{b}^{2}} \right)}+\frac{1}{\left( 5{{b}^{2}}+2bc+2{{c}^{2}} \right)}+\frac{1}{\left( 5{{c}^{2}}+2ca+2{{a}^{2}} \right)} \\  & \le \frac{1}{9}\left( \frac{2}{3{{a}^{2}}}+\frac{1}{3{{b}^{2}}} \right)+\frac{1}{9}\left( \frac{2}{3{{b}^{2}}}+\frac{1}{3{{c}^{2}}} \right)+\frac{1}{9}\left( \frac{2}{3{{c}^{2}}}+\frac{1}{3{{a}^{2}}} \right)=\frac{1}{9}\left( \frac{1}{{{a}^{2}}}+\frac{1}{{{b}^{2}}}+\frac{1}{{{c}^{2}}} \right)=\frac{1}{9} \\  & \Rightarrow P\le \frac{\sqrt{27}}{2}.\left( \frac{1}{9}+\frac{1}{9} \right)=\frac{\sqrt{3}}{3} \\ \end{align}\)

Dấu bằng xảy ra \(\Leftrightarrow \left\{ \begin{align}  & a=b=c \\  & \frac{1}{{{a}^{2}}}+\frac{1}{{{b}^{2}}}+\frac{1}{{{c}^{2}}}=1 \\ \end{align} \right.\Leftrightarrow a=b=c=\sqrt{3}\).

Vậy \({{P}_{\max }}=\frac{\sqrt{3}}{3}\)

Chọn B