Ta có:
\(bc\sqrt{1\left(a-1\right)}\le bc.\frac{1+a-1}{2}=\frac{abc}{2}\)
\(ca\sqrt{b-4}=\frac{1}{2}ca\sqrt{4\left(b-4\right)}\le\frac{1}{2}ca.\frac{4+b-4}{2}=\frac{abc}{4}\)
\(ab\sqrt{c-9}=\frac{1}{3}ab.\sqrt{9\left(c-9\right)}\le\frac{1}{3}ab.\frac{9+c-9}{2}=\frac{abc}{6}\)
Từ đó suy ra \(P\le\frac{abc\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}\right)}{abc}=\frac{11}{12}\)
Đẳng thức xảy ra khi a = 2; b = 8; c = 18
Is that true?