Question

Cho a,b,c >0. và x^2+y^2+z^2=1. Tìm max P=x(y+z)

Answer 1

\(x,y,z>0\Rightarrow x^2,y^2,z^2>0\)  áp dụng bđt cosi ta có :

\(x^2+y^2+z^2=x^2+\left(y^2+z^2\right)>=2\sqrt{x^2\left(y^2+z^2\right)}>=2\sqrt{\frac{x^2\left(y+z\right)^2}{2}}=\frac{2x\left(y+z\right)}{\sqrt{2}}\)

\(\Rightarrow1>=\frac{2x\left(y+z\right)}{\sqrt{2}}\Rightarrow\sqrt{2}>=2x\left(y+z\right)\Rightarrow\frac{\sqrt{2}}{2}=\frac{1}{\sqrt{2}}>=x\left(y+z\right)\)

dấu = xảy ra khi \(x=\sqrt{\frac{1}{2}};y=z=\frac{1}{2}\)

vậy max P là \(\frac{1}{\sqrt{2}}\)khi \(x=\sqrt{\frac{1}{2}};y=z=\frac{1}{2}\)