# Proving Inequalities Using Monotony of the Function

Monotony of the function can be used to prove some not obvious inequalities.

Example 1. Prove that sin(x)>2/pi x when 0<x<pi/2.

Consider function f(x)=(sin(x))/x defined on interval (0,pi/2].

We have that f'(x)=(xcos(x)-sin(x))/x^2=(cos(x)(x-tan(x)))/x^2. On interval (0,pi/2) cos(x)>0 and x<tan(x), so f'(x)<0. This means that function is decreasing on interval (0,pi/2).

So, for any x from (0,pi/2) we have that f(x)=(sin(x))/x>f(pi/2)=2/pi or equivalently sin(x)>2/pi x.

Example 2. Prove that cos(x)>1-1/2x^2 for x>0.

Consider function f(x)=cos(x)-1+1/2x^2. We have that f(0)=cos(0)-1+1/2 0^2=0.

derivative of function f'(x)=-sin(x)+x>0 because sin(x)<x.

This means that for x>=0 function is increasing and for x>0 we have that f(x)>f(0)=0, i.e. cos(x)-1+1/2 x^2>0. This is equivalent to cos(x)>1-1/2x^2.

Example 3. Prove that ln(x)<=x-1 for x>0.

Consider function f(x)=ln(x)-x. Its derivative is f'(x)=1/x-1. Clearly f'(x)>0 when 0<x<1 and f'(x)<0 when x>1.

So, function is increasing on interval (0,1) and decreasing on interval (1,oo). So, at point x=1 it takes its largest value f(1)=ln(1)-1=-1.

That's why ln(x)-x<=-1 for x>0. This is equivalent to ln(x)<=x-1 for x>0.