Question

Cho các số thực dương thỏa mãn \(x^2+y^2+z^2=2\)

CMR: \(x+y+z\le xyz+2\)

Answer 1

\(BDT\Leftrightarrow x+y+z-xyz\le2\)

Áp dụng BĐT Cauchy-Schwarz ta có:

\(\left(x\left(1-yz\right)+\left(y+z\right)\right)^2\le\left(x^2+\left(y+z\right)^2\right)\left(\left(1-yz\right)+1\right)\)

\(=\left(x^2+y^2+z^2+2yz\right)\left(2-2yz+y^2z^2\right)\)

\(=2\left(1+yz\right)\left(2-2yz+y^2z^2\right)\)do \(x^2+y^2+z^2=2\)

\(=4\left(1-y^2z^2\right)+2\left(1+yz\right)y^2z^2\)

\(=4+2y^2z^2\left(yz-1\right)\le4\) do \(yz\le\frac{y^2+z^2}{2}\le\frac{x^2+y^2+z^2}{2}=1\)

\(\left(x\left(1-yz\right)+\left(y+z\right)\right)^2\le4\Rightarrow x\left(1-yz\right)+\left(y+z\right)\le2\)

Hay ta có ĐPCM

Answer 2

câu 3 đề hsg HN 2016-2017