Question

如何比较下列不等式的大小?1.当X>1时,X⒊与X⒉-X+12.X⒉+Y⒉+1与2(X+Y+1)

Answer 1

1.当X>1时,X⒊与X⒉-X+1

x^3-(x^2-x+1)

=x(x^2-x+1)+1

=x(x^2-x+1/4+3/4)+1

=x[(x-1/2)^2+3/4]+1>0

所以当X>1时,X⒊>（X⒉-X+1）

2.(X⒉+Y⒉+1)-2(X+Y+1)

=(x-y)^2>=0

所以(X⒉+Y⒉+1)>=[2(X+Y+1)]