Vì \(\hept{\begin{cases}x;y;z\ge0\\x+y+z=1\end{cases}\Rightarrow0\le x;y;z\le1}\)
\(\Rightarrow\hept{\begin{cases}x\left(1-x\right)\ge0\\y\left(1-y\right)\ge0\\z\left(1-z\right)\ge0\end{cases}}\)\(\Leftrightarrow\hept{\begin{cases}x-x^2\ge0\\y-y^2\ge0\\z-z^2\ge0\end{cases}}\Leftrightarrow\hept{\begin{cases}x^2\le x\\y^2\le y\\z^2\le z\end{cases}}\)
Ta có \(S=\sqrt{3x^2+1}+\sqrt{3y^2+1}+\sqrt{3z^2+1}\)
\(=\sqrt{x^2+2x^2+1}+\sqrt{y^2+2y^2+1}+\sqrt{z^2+2z^2+1}\)
\(\le\sqrt{x^2+2x+1}+\sqrt{y^2+2y+1}+\sqrt{z^2+2z+1}\)
\(=\sqrt{\left(x+1\right)^2}+\sqrt{\left(y+1\right)^2}+\sqrt{\left(z+1\right)^2}\)
\(=x+1+y+1+z+1\)
\(=x+y+z+3=4\)
Dấu "=" xảy ra khi x = y = 0 ; z = 1 và các hoán vị
xét :\(\sqrt{3a^2+1}=< a+1\)
=>\(3a^2+1=< a^2+2a+1\)
=>\(2a\left(a-1\right)=< 0\)luon dung
ap dụng bđt vừa chứng minh ta có :S>=x+y+z+3=1
xay ra dấu = khi x=y=0,z=1(hoán vị)
