Áp dụng BĐT cauchy-schwarz
Ta có:[tex]\left(9a^3+3b^2+c\right)\left(\dfrac{1}{9a}+\dfrac{1}{3b}+c\right)\ge\left(a+b+c\right)^2[/tex]
[tex]\Leftrightarrow\dfrac{a}{\left(9a^3+3b^2+c\right)}\le\dfrac{a\left(\dfrac{1}{9a}+\dfrac{1}{3}+c\right)}{\left(a+b+c\right)^2}=\dfrac{\dfrac{1}{9}+\dfrac{a}{3}+ac}{\left(a+b+c\right)^2}[/tex]
Tương tự:
[tex]\Rightarrow P\le\dfrac{1}{9}\cdot3+\dfrac{a+b+c}{3}+ab+bc+ca[/tex]
[tex]\Rightarrow P\le\dfrac{1}{9}\cdot3+\dfrac{a+b+c}{3}+\dfrac{\left(a+b+c\right)^2}{3}=1[/tex]
Dấu "=" xảy ra [tex]\Leftrightarrow[/tex] a=b=c=[tex]\frac{1}{3}[/tex] (ĐPC/M)