# Proving of Inequalities by Contradiction

The essence of this method is in the following: suppose we need to prove inequality `f(x,y,z)>g(x,y,z)`. We assume contrary proposition, i.e. that for some x,y and z `f(x,y,z)<=g(x,y,z)`.

Using properties of inequalities we perform transformations of inequality. If as result, we obtain incorrect inequality than this means that our assumption that `f(x,y,z)<=g(x,y,z)` was wrong, therefore, `f(x,y,z)>g(x,y,z)`.

**Example**. Prove that `cos(alpha+beta)cos(alpha-beta)<=cos^2(alpha)`.

Assume contrary proposition, i.e. assume that exist some `alpha` and `beta` such that `cos(alpha+beta)cos(alpha-beta)>cos^2(alpha)`.

Using formulas `cos(alpha+beta)cos(alpha-beta)=(cos(2beta)+cos(2alpha))/2` and `cos^2(alpha)=(1+cos(2alpha))/2`, we obtain that `(cos(2beta)+cos(2alpha))/2>(1+cos(2alpha))/2`. From this we have that `cos(2beta)>1`.

Since for any `beta` `cos(2beta)<=1` then we obtained contradiction. Thus, our asumption is incorrect, therefore, `cos(alpha+beta)cos(alpha-beta)<=cos^2(alpha)`.