P=\(6\sqrt{x-3}+8\sqrt{5-x}\)(P>0)
<=> \(P^2=\left(6\sqrt{x-3}+8\sqrt{5-x}\right)^2\le\left(6^2+8^2\right)\left(x-3+5-x\right)\)(bđt bunhiacopski)
=> \(P^2\le200\)
=> \(P\le\sqrt{200}=10\sqrt{2}\)(do P>0)
Dấu "=" xảy ra <=>\(6\sqrt{5-x}=8\sqrt{x-3}\)
<=> x =3,75(t/m)
Vậy max\(P=10\sqrt{2}\) tại x=3,75