HỦY DIỆT THE WORLD
04/12/2017 at 21:05-
KEITA FC 8C 04/12/2017 at 21:14
Áp dụng bất đẳng thức Bunhia ta có :
\(\left(1x+1y+1z\right)^2\)
\(\le\left(1^2+1^2+1^1\right)\left(x^2+y^2+z^2\right)=3\)
\(\Rightarrow x+y+z\le\sqrt{3}\)
\(\left(xy+yz+zx\right)^2\)
\(\le\left(x^2+y^2+z^2\right)\left(y^2+z^2+x^2\right)=1\)
\(\Rightarrow xy+yz+zx\le1\)
\(\Rightarrow p\le1+\sqrt{3}\)
Mặt khác \(x=y=z=\dfrac{1}{\sqrt{3}}\) thì \(p=1+\sqrt{3}\)
Vậy maxP \(=1+\sqrt{3}\)
HỦY DIỆT THE WORLD selected this answer. -
Mons Trần 04/12/2017 at 21:29
You don't speak Vietnamese here.You must speak English
You are violating rules.
You should redress.Are you okayy..?