DangHuyNgheAn

Cho $a,b,c>0,a+b+c=3$.Tim $Min P=\sum \frac{a(a-2b+c)}{ab+1}$

$x^{2}=a,y=b (a,b\epsilon N*) a^2b+2b/ab+1\Rightarrow a^2b^2+2b^2/ab+1\Rightarrow 2b^2+1/ab+1 do a,b>0 nen 2b^2+1\geq ab+1\Rightarrow 2b-a\geq 0. neu 2b=a,thi 2y=x^2=> x/2, dat x=2k(k\epsilon n*)thi y=2k^2.thu lai thoa man de bai.neu2b-a>0 taco a^2b+2b/ab+1\Leftrightarrow a^2b+a+2b-a/ab+1\Leftrightarrow 2b-a/ab+1=>2b-a\geq ab+1(do 2vedeu>0 )\Leftrightarrow (a-2)(b+1)\leq -3=>a<2=>a=1=>x=1.thay vaodebaitaco3y/y+1\Leftrightarrow 3y+3-3/y+1\Leftrightarrow 3/y+1=>y=2.thu lai thoa man.Vay taco(x,y)la(1,2)(2k,2k^2(k\epsilon n*))$