[Toán 9] Bất đẳng thức.

C

congchuaanhsang

Xét $x^2+y^2-\sqrt{2}xy+z^2-\sqrt{2}yz$

=$(x^2-2x.\dfrac{\sqrt{2}}{2}y+\dfrac{y^2}{2})+\dfrac{1}{2}(y^2-2y\sqrt{2}{z}+2z^2)$

=$(x-\dfrac{\sqrt{2}}{2}y)^2+\dfrac{1}{2}(y-\sqrt{2}z)^2$\geq0

\Rightarrowđpcm
 
E

eye_smile

Cách khác:
$x^2+y^2+z^2$ \geq $y^2$+ $\dfrac{(x+z)^2}{2}$ \geq 2 $\sqrt{y^2.\dfrac{(x+z)^2}{2}}$ \geq $2y$.$\dfrac{x+z}{\sqrt{2}}$ = $\sqrt{2}.(yx+yz)$
 
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