Inequality with Max., Min.

[man111]

If $x,y,z>0$ and $x^2+y^2+z^2=3$ Then find

Max. and Min. value of $\displaystyle \left(xy+yz+zx\right)+\frac{5}{\left(x+y+z\right)}$

[ИСН]

Look at your thing as a function of $t=x+y+z$.

[ewert]

man111 в сообщении #474375 писал(а):

If $x,y,z>0$ and $x^2+y^2+z^2=3$ Then find

Max. and Min. value of $\displaystyle \left(xy+yz+zx\right)+\frac{5}{\left(x+y+z\right)}$

$=\frac{(x+y+z)^2-3}{2}+\frac{5}{x+y+z},$ где $x+y+z\in(\sqrt3;3].$