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.