Chứng minh BĐT Xvacơ $\frac{x^2}{a}+\frac{y^2}{b}+\frac{z^2}{c}$ $\geq$ $\frac{(x+y+z)^2}{a+b+c}$

N

noinhobinhyen

theo bđt Bunhia-cốpxki ta có:

$(\dfrac{x^2}{a}++\dfrac{y^2}{b}+\dfrac{z^2}{c})(a+b+c) \geq (x+y+z)^2$

$\Rightarrow \dfrac{x^2}{a}++\dfrac{y^2}{b}+\dfrac{z^2}{c} \geq \dfrac{(x+y+z)^2}{a+b+c}$
 
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