Cho $x,y>0$ và $x+y+1=3xy.$ Tìm GTLN:
$P=\frac{3x}{y(x+1)}+\frac{3y}{x(y+1)}-\frac{1}{x^2}-\frac{1}{y^2}$
ủa sao lại vote down vậy chamhocdethihsgtoan? –  dolaemon 17-04-16 05:26 PM
up luôn a chiêm ngưỡng 1 thể –  dolaemon 14-04-16 09:16 PM
bữa làm bài này em có cách làm hay hay :)) –  ๖ۣۜDevilღ 14-04-16 07:41 PM
dòng thứ 2 thiếu y kìa! -_- –  Confusion 11-04-16 12:54 PM
-_-....... –  Confusion 11-04-16 12:51 PM
hôm nay ko đi học à? –  Confusion 11-04-16 12:51 PM
dòng max cuối cùng kia là sao vậy? –  Confusion 11-04-16 01:06 PM
sao e toàn viết thiếu thế! -_-! –  Confusion 11-04-16 01:06 PM
Đúng click V và voteup dùm e với ạ –  111 11-04-16 08:53 AM
e có cách bđổi khác, có lẽ dễ nhìn hơn! ~~
Gthiết suy ra: $(1+x)(1+y)=4xy$ 
A/d Cauchy suy ra $xy\geq 1$
Ta có : Quy đồng $P$
$P=\frac{3x^3(1+y)+3y^2(1+x)}{xy(1+x)(1+y)}-\frac{x^2+y^2}{x^2y^2}$
$\Leftrightarrow P=\frac{3x^2+3y^2+3xy(x+y)}{xy.4xy}-\frac{x^2+y^2}{x^2y^2}$
$\Leftrightarrow P=\frac{3xy(3xy-1)-(x^2+y^2)}{4x^2y^2}$
$\Leftrightarrow P=\frac{5xy-1}{4x^2y^2}$
$\Rightarrow .....$
đến đây dùng típ cách của 111 ( xét hiệu $P-1$)! ^_^
Đặt:
$\frac{1}{x}=a;\frac{1}{y}=b\Rightarrow 3=a+b+ab\geq 2\sqrt{ab}+ab\geq 3\sqrt[3]{a^2b^2}\Rightarrow ab\leq 1$
$P=\frac{ab}{a+1}+\frac{ab}{b+1}=\frac{ab(5-ab)}{4}=\frac{-(ab-1)^2+3ab+1}{4}\leq 1$
dấu = xảy ra khi $a=b=1$

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