Ta có: (x+y)2>=4xy
\(\Leftrightarrow\)(x+y)2-4xy>=0
\(\Leftrightarrow\)x2+2xy+y2-4xy>=0
\(\Leftrightarrow\)x2-2xy+y2>=0
\(\Leftrightarrow\)(x-y)2>=0 (luôn đúng với mọi x)
Chung minh rang x2+5xy+2x-4xy-10y+14 >0 voi moi x,y
c/m voi moi so thuc x,y,z ta co
x2+y2+z2+3> hoac = 2(x+y+z)
c/m \(\frac{1}{x}+\frac{1}{y}>=\frac{4}{x+y}\)
voi moi x,y>0
c/m (a2+b2) (x2+y2)>=(ax+by)2 voi moi a,b,x,y thuoc R
c/m voi moi x,y>0 ta co
x5+y5>hoac= x4y+xy4
2/ Chung to: x^2+6x+13>0 voi moi x
3/ Chung to : -x^2+6x-11<0 voi moi x
4/ cho x-y=5va x.y=24 .Tinh x^3+y^3
4x(x+y)(x+y+z)(x+z)+y^2z^2 chung minh luon luon >= 0 voi moi x,y,z
4x(x+y)(x+y+z)(x+z)+y^2z^2 chung minh luon luon >= voi moi x,y,z
x-2xy+y2 +1>0 voi moi so thuc x va y
