# Method of Estimating of the Sign of Difference

The essence of this method is in the following: to prove that f(x,y,z)>g(x,y,z) we need to prove that f(x,y,z)-g(x,y,z)>0.

Example. Prove that if x>=0,y>=0 then (x+y)/2>=sqrt(xy) (arithmetic mean of two non-negative numbers is not less than their geometric mean; this inequality is called Cauchy's Inequality).

Let's make difference (x+y)/2-sqrt(xy).

Next we have that (x+y)/2-sqrt(xy)=(x+y-2sqrt(xy))/2=((sqrt(x)-sqrt(y))^2)/2.

Inequality (sqrt(x)-sqrt(y))^2>=0 is true for all non-negative x and y. Therefore, also ((sqrt(x)-sqrt(y))^2)/2>=0 or (x+y)/2-sqrt(xy).

From this we have that (x+y)/2>=sqrt(xy). Equality holds only when x=y.

From Cauchy's inequality if we take x=x and y=1/x, we have that (x+1/x)/2>=sqrt(x*1/x) or x+1/x>=2. This inequality is true for all x>0.