When a>0 and a≠ 1, for any real number x, y, x >；; y，x+y & gt； 0, proving: a x+1/a x > a y+1/a y.

Syndrome: ax+1/ax-(ay+1/ay) = (ax-ay) (1-1(a (x+y)).

When 0 0;

When a> is at 1, ax-ay >; 0,1-1/a (x+y) > 0, then ax+1/ax-(ay+1/ay) > 0.

So ax+1/ax >; a^y + 1/a^y.