Thread: AM GM Inequality Application

Can someone tell me why if a and b are positive reals, that sqrt((a+1)(b+1)) >= sqrt(ab) + 1. I know that this statement is true and have been told is so because it comes from the Arithmetic Mean-Geometric Mean Inequality: (a+b)/2 >= sqrt(ab). I have been trying and just can't see how statement is derived from the AM GM Inequality. I apologize if the answer is simple and should've been apparent to me. The AM GM idea is new to me.

Take the square of both sides.

I see it now. Squaring both sides gives ab + a + b + 1 >= ab + 2sqrt(ab) + 1. Canceling like terms from each side, leaves
a + b >= 2sqrt(ab) or (a + b)/2 >= sqrt(ab), which is the general form of the AM GM Inequality. Thanks. I had tried squaring many times before, but I must have made a mistake somewhere. I probably just needed to step away from it for a while.