Chủ đề này có 3 trả lời

[nguyennamphu1810]

$CMR:x^2+y^2+z^2\geq \sqrt{2}.(xy+yz)$

[Thu Huyen 21]

$CMR:x^2+y^2+z^2\geq \sqrt{2}.(xy+yz)$

$VT= (\frac{x}{\sqrt{2}})^{2}+y^{2}+(\frac{x}{\sqrt{2}})^{2}+z^{2}\geq 2.\frac{x}{\sqrt{2}}.y+2.\frac{x}{\sqrt{2}}.z=\sqrt{2}.(xy+xz)$ (ĐPCM)
Dấu bằng xảy ra khi $x=\sqrt{2}.y=\sqrt{2}.z$

[Issac Newton of Ngoc Tao]

$CMR:x^2+y^2+z^2\geq \sqrt{2}.(xy+yz)$

ta có:$VT\geq y^{2}+\frac{(x+z)^{2}}{2}\geq \sqrt{2}.y(x+z)$. đpcm

[phamngochung9a]

$CMR:x^2+y^2+z^2\geq \sqrt{2}.(xy+yz)$

$x^{2}+y^{2}+z^{2}= (x^{2}+z^{2})+y^{2}\geq \frac{(x+z)^{2}}{2}+y^{2}\geq 2\sqrt{\frac{(x+z)^{2}}{2}.y^{2}}= 2\sqrt{2}(xy+xz)$