[toán 10]Bài BĐT thi HSG tỉnh

  • Thread starter bachocanhxtanh_450
  • Ngày gửi
  • Replies 1
  • Views 832

V

vodichhocmai

Cho x,y,z dương. Chứng minh rằng:

[TEX]\frac{2xy}{(x+z)(y+z)} + \frac{2yz}{(x+y)(x+z)} + \frac{3xz}{(x+y)(y+z)} \geq \frac{5}{3}[/TEX]

Các bạn giúp mình với. Bài này hok búa wá

[TEX]LHS:=\frac{2xy\(x+y\)+ 2yz\(y+z\) + 3xz\(x+z\)}{(x+y)(y+z)(z+x)}[/TEX]

[TEX]LHS-\frac{5}{3}:=\frac{6xy\(x+y\)+ 6yz\(y+z\)+9 xz\(x+z\) -5\[xy\(x+y\)+ yz\(y+z\)+xz\(x+z\)+2xyz\]}{(x+y)(y+z)(z+x)}=\frac{MS}{TS}[/TEX]

[TEX]MS:=xy\(x+y\)+ yz\(y+z\)+4 xz\(x+z\) -10xyz\ge 2y^2\sqrt{xz}+8xz\sqrt{zx}-8xyz=2\sqrt{xz}\(y-2\sqrt{xz}\)^2\ge 0[/TEX]

[TEX]\righ Done!![/TEX]
 
Last edited by a moderator:
Top Bottom