(1) First, it is proved that for any two positive numbers x, y, there is1/4x+1/4y >; = 1/(x+y), the proof method is very simple, that is,1/4x+1/4y = (x+y)/4xy, 4xy.

(2) let A = X+ 1 and B = Y+ 1 and substitute it into the original formula: x2+2x+1+y2+2y+1> = xy+x+y+1+y.

It can be discussed in different situations.

When at least one of x and y is 0, the inequality is obviously established;

When X and Y are different symbols, the inequality xy>0 holds;

When the signs of X and Y are the same, X 2+Y 2 >: = 2xy, X 2-xy+Y 2 > = xy>0, the inequality holds.

Note: In the second question, you can also set the binary function F (a, b) = A 2-AB+B 2-A-B+ 1, the partial derivative of A f 1=2a-b- 1, and the partial derivative of B F2 = 2B-A-.