Đặt \(\left(6a;3b;2c\right)=\left(x;y;z\right)\Rightarrow x+y+z=6\)
\(P=\dfrac{xy}{6\sqrt{x^2+y^2+2z^2}}+\dfrac{yz}{6\sqrt{2x^2+y^2+z^2}}+\dfrac{zx}{6\sqrt{x^2+2y^2+z^2}}\)
Ta có:
\(x^2+z^2+y^2+z^2\ge\dfrac{1}{2}\left(x+z\right)^2+\dfrac{1}{2}\left(y+z\right)^2\ge\left(x+z\right)\left(y+z\right)\)
\(\Rightarrow\dfrac{xy}{6\sqrt{x^2+y^2+2z^2}}\le\dfrac{xy}{6\sqrt{\left(x+z\right)\left(y+z\right)}}\le\dfrac{1}{12}\left(\dfrac{xy}{x+z}+\dfrac{xy}{y+z}\right)\)
Tương tự:
\(\dfrac{yz}{6\sqrt{2x^2+y^2+z^2}}\le\dfrac{1}{12}\left(\dfrac{yz}{x+y}+\dfrac{yz}{x+z}\right)\)
\(\dfrac{zx}{6\sqrt{x^2+2y^2+z^2}}\le\dfrac{1}{12}\left(\dfrac{xz}{x+y}+\dfrac{zx}{y+z}\right)\)
Cộng vế:
\(P\le\dfrac{1}{12}\left(\dfrac{yz+xz}{x+y}+\dfrac{zx+xy}{y+z}+\dfrac{yz+xy}{x+z}\right)=\dfrac{1}{12}\left(x+y+z\right)=\dfrac{1}{2}\)
\(P_{max}=\dfrac{1}{2}\) khi \(x=y=z=2\) hay \(\left(a;b;c\right)=\left(\dfrac{1}{3};\dfrac{3}{2};1\right)\)