3 replies to this topic

[tienduc]

Cho x,y > 1. Tìm Min $P=\frac{(x^{3}+y^{3})-(x^{2}+y^{2})}{(x-1)(y-1)}$

[Hoanganh2001]

trong đề thi hsg toán hà nội phải không

[Shin Janny]

Cho x,y > 1. Tìm Min $P=\frac{(x^{3}+y^{3})-(x^{2}+y^{2})}{(x-1)(y-1)}$

$P=\frac{x^{2}(x-1)+y^{2}(y-1)}{(x-1)(y-1)}$

  $=\frac{x^{2}}{y-1}+\frac{y^{2}}{x-1}$

  $\geq 2\sqrt{\frac{x^{2}y^{2}}{(x-1(y-1)}}$ (bđt AM-GM)

  $= \frac{2xy}{\sqrt{(x-1)1}.\sqrt{(y-1)1}}$

  $\geq \frac{2xy}{\frac{(x-1)+1}{2}.\frac{(y-1)+1}{2}}$ (bđt AM-GM)

  $= \frac{8xy}{xy}=8$

Vậy Min P=8, dấu bằng xảy ra khi x=y=2

Edited by Shin Janny, 30-04-2016 - 22:21.

[Hoangtheson2611]

P=$\sum \frac{x^{2}}{(y-1).1}\geqslant \sum \frac{4x^{2}}{y^{2}}\geqslant 2.\sqrt{\frac{16x^{2}y^{2}}{x^{2}.y^{2}}}=8$ . Dấu "=" xảy ra khi a=b=2