# 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`.