Áp dụng BĐT Cauchy-Schwarz ta có:
\(\dfrac{ab}{\sqrt{3c+ab}}=\dfrac{ab}{\sqrt{\left(a+b+c\right)c+ab}}=\dfrac{ab}{\sqrt{c^2+ab+bc+ca}}\)
\(=\dfrac{ab}{\sqrt{\left(a+c\right)\left(b+c\right)}}\le\dfrac{1}{2}\left(\dfrac{ab}{a+c}+\dfrac{ab}{b+c}\right)\)
Tương tự cho 2 BĐT còn lại rồi cộng theo vế:
\(P\le\dfrac{1}{2}\left(a+b+c\right)=\dfrac{3}{2}\)
\("="\Leftrightarrow a=b=c=1\)