Ta có \(P=\sum\dfrac{1}{\sqrt{2a^2+5ab+2b^2}}\le\sum\dfrac{1}{\sqrt{9ab}}=\dfrac{1}{3}\sum\dfrac{1}{\sqrt{ab}}\le\dfrac{1}{6}\sum\left(\dfrac{1}{a}+\dfrac{1}{b}\right)=\dfrac{1}{3}\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)=\dfrac{2}{3}\).
Dấu "=" xảy ra khi và chỉ khi \(a=b=c=\dfrac{3}{2}\)