Cho x,y,x dương
CM $x^{2}+y^{2}+z^{2}\geq \sqrt{2}(xy+xz)$
CM $x^{2}+y^{2}+z^{2}\geq \sqrt{2}(xy+xz)$
Started By nhunghongmnsd, 11-12-2016 - 22:24
#2
Posted 11-12-2016 - 22:33
Baoriven
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- Điều hành viên OLYMPIC
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- 1426 posts
Thượng úy
Ta có: $x^2+y^2+z^2\geq x^2+\frac{(y+z)^2}{2}\geq 2x.\frac{y+z}{\sqrt{2}}=\sqrt{2}(xy+xz)$.
Dấu bằng xảy ra khi $y=z;x=\sqrt{2}y$.
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