Cho \(a,b,c\) là ba số thực dương thỏa mãn điều kiện

Với \(a,b,c > 0\), chứng minh được:

a+b+c1a+1b+1c≥9⇒1a+b+c≤191a+1b+1cx+y+z2≤3(x2+y2+z2)⇒1a+1b+1c≤31a2+1b2+1c2

Với \(a,b > 0\), ta có :

\(\begin{array}{l}5{a^2} + 2ab + 2{b^2} = (4{a^2} + 4ab + {b^2}) + ({a^2} - 2ab + {b^2})\\ = {(2a + b)^2} + {(a - b)^2} \ge {(2a + b)^2}\\ \Rightarrow \sqrt {5{a^2} + 2ab + 2{b^2}} \ge \sqrt {{{(2a + b)}^2}} = 2a + b\end{array}\): 

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

Tương tự: \(\frac{1}{{\sqrt {5{b^2} + 2bc + 2{c^2}} }} \le \frac{1}{9}\left( {\frac{2}{b} + \frac{1}{c}} \right){\rm{ ; }}\frac{1}{{\sqrt {5{c^2} + 2ca + 2{a^2}} }} \le \frac{1}{9}\left( {\frac{2}{c} + \frac{1}{a}} \right)\):

\(P \le \frac{1}{9}\left( {\frac{2}{a} + \frac{1}{b} + \frac{2}{b} + \frac{1}{c} + \frac{2}{c} + \frac{1}{a}} \right) = \frac{1}{3}\left( {\frac{1}{a} + \frac{1}{b} + \frac{1}{c}} \right) \Rightarrow P \le \frac{1}{3} \cdot \sqrt {3\left( {\frac{1}{{{a^2}}} + \frac{1}{{{b^2}}} + \frac{1}{{{c^2}}}} \right)} = \frac{1}{3} \cdot \sqrt 3 = \frac{{\sqrt 3 }}{3}\)

Dấu “=” xảy ra \( \Leftrightarrow \left\{ \begin{array}{l}a = b = c\\\frac{1}{{{a^2}}} + \frac{1}{{{b^2}}} + \frac{1}{{{c^2}}} = 1\end{array} \right. \Leftrightarrow a = b = c = \sqrt 3 \)

Vậy \(\max P = \frac{{\sqrt 3 }}{3}\) khi \(a = b = c = \sqrt 3 \).: