Bất đẳng thức

H

huynhbachkhoa23

$A=\sum \dfrac{y^2+z^2}{x-1} \ge \sum \dfrac{(y+z)^2}{2x-2} \ge \dfrac{2(x+y+z)^2}{(x+y+z)-3}=\dfrac{2t^2}{t-3}=2m>0$ với $t>3$

$t^2-mt+3m=0$

$\Delta = m^2-12m = m(m-12) \ge 0$

Suy ra $m\ge 12 \rightarrow A \ge 24$

Đẳng thức xảy ra khi $x=y=z=2$
 
E

eye_smile

$A=\dfrac{x^2+y^2}{z-1}+\dfrac{y^2+z^2}{x-1}+\dfrac{x^2+z^2}{y-1} \ge \dfrac{2xy}{z-1}+\dfrac{2yz}{x-1}+\dfrac{2xz}{y-1} \ge \dfrac{8xy}{z^2}+\dfrac{8xz}{y^2}+\dfrac{8yz}{x^2} \ge 3\sqrt[3]{\dfrac{8xy.8yz.8zx}{x^2.y^2.z^2}}=24$

Dấu "=" xảy ra \Leftrightarrow $x=y=z=2$
 
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