\(P=\dfrac{1}{4\sqrt{6}}.\left(2\sqrt{24x\left(14x+10y\right)}+2\sqrt{24y\left(14y+10x\right)}\right)\)
\(P\le\dfrac{1}{4\sqrt{6}}\left(24x+14x+10y+24y+14y+10x\right)\)
\(P\le\dfrac{1}{4\sqrt{6}}.48\left(x+y\right)=2\sqrt{6}\left(x+y\right)\le2\sqrt{6}.\sqrt{2\left(x^2+y^2\right)}\le4\sqrt{6}\)
Dấu "=" xảy ra khi \(x=y=1\)
Áp dụng BĐT Bunhiacopski:
\(P^2=\left[\sqrt{x\left(14x+10y\right)}+\sqrt{y\left(14y+10x\right)}\right]^2\\ \Leftrightarrow P^2\le\left(x+y\right)\left(14x+10y+14y+10x\right)\\ \Leftrightarrow P^2\le\left(x+y\right)\left(24x+24y\right)=24\left(x+y\right)^2\)
Mà \(\left(x+y\right)^2\le\left(x^2+y^2\right)\left(1^2+1^2\right)=2\left(x^2+y^2\right)\)
\(\Leftrightarrow P^2\le48\left(x^2+y^2\right)\le48\cdot2=96\\ \Leftrightarrow P\le\sqrt{96}=4\sqrt{6}\)
Dấu \("="\Leftrightarrow x=y=1\)
