Question

cm x^2/a+y^2/b+z^2/c >= (x+y+z)^2/a+b+c

Answer 1

Can them dieu kien a;b;c>0 nhe

Theo BDT Cauchy-Schwarz ta co

\(\left(a+b+c\right)\left(\dfrac{x^2}{a}+\dfrac{y^2}{b}+\dfrac{z^2}{c}\right)\ge\left(x+y+z\right)^2\)

\(\Leftrightarrow\dfrac{x^2}{a}+\dfrac{y^2}{b}+\dfrac{z^2}{c}\ge\dfrac{\left(x+y+z\right)^2}{a+b+c}\)

Dau "=" xay ra khi \(\dfrac{x^2}{a^2}=\dfrac{y^2}{b^2}=\dfrac{z^2}{c^2}\Leftrightarrow\dfrac{x}{a}=\dfrac{y}{b}=\dfrac{z}{c}\)

Answer 2

$a+b+c)\left(\frac{x^2}{a}+\frac{y^2}{b}+\frac{z^2}{c}\right)\ge {\left(x+y+z\right)}^2$

$\iff \frac{x^2}{a}+\frac{y^2}{b}+\frac{z^2}{c}\ge \frac{{\left(x+y+z\right)}^2}{a+b+c}$

Dấu "=" xay ra khi $\frac{x^2}{a^2}=\frac{y^2}{b^2}=\frac{z^2}{c^2}\iff \frac{x}{a}=\frac{y}{b}=\frac{z}{c}$