Diễn Đàn MathScope - Xem bài viết đơn - [VMO 2014] Bài 6

daicahuyvn · 04-01-2014, 09:41 PM

$\frac{x^3y^4z^3}{(x^4+y^4)(xy+z^2)^3}\le \frac{x^3y^4z^3}{8xy(x^2+y^2)z^3xy\sqrt{xy}}=\frac {y\sqrt{xy}}{8(x^2+y^2)}$
$a=\sqrt{\frac{x}{y}}$.abc=1
$8T\le \sum \frac{a}{a^4+1}\le \sum \frac{1}{\sqrt{2(a^4+1)}}\le\frac{3}{2}$

04-01-2014, 09:41 PM	#14
daicahuyvn +Thành Viên+ Tham gia ngày: May 2011 Bài gởi: 9 Thanks: 6 Thanked 12 Times in 3 Posts	$\frac{x^3y^4z^3}{(x^4+y^4)(xy+z^2)^3}\le \frac{x^3y^4z^3}{8xy(x^2+y^2)z^3xy\sqrt{xy}}=\frac {y\sqrt{xy}}{8(x^2+y^2)}$ $a=\sqrt{\frac{x}{y}}$.abc=1 $8T\le \sum \frac{a}{a^4+1}\le \sum \frac{1}{\sqrt{2(a^4+1)}}\le\frac{3}{2}$