# Synthetic Method of Proving Inequalities

The essence of this method is in following: with the help of the sequence of transformations we infer inequality from well-known inequalities.

Well-known inequalities are:

1. (x+y)/2>=sqrt(xy), where x>=0,y>=0 (Cauchy's inequality);
2. x+1/x>=2, where x>0;
3. |x+a|<=|x|+|a| (triangle inequality);
4. -1<=sin(alpha)<=1;
5. -1<=cos(alpha)<=1.

Example. Show that (a+b+c+d)/4>=root(4)(abcd).

If we take x=(a+b)/2,y=(c+d)/2 and apply Cauchy's inequality then we will obtain that ((a+b)/2+(c+d)/2)/2>=sqrt((a+b)/2*(c+d)/2). This can be rewritten as (a+b+c+d)/4>=sqrt((a+b)/2*(c+d)/2).

Now, apply Cauchy's inequality to numbers a and b; and c and d: (a+b)/2>=sqrt(ab) and (c+d)/2>=sqrt(cd). Taking square roots of both sides yields sqrt((a+b)/2)>=root(4)(ab) and sqrt((c+d)/2)>=root(4)(cd).

Therefore, (a+b+c+d)/4>=sqrt((a+b)/2*(c+d)/2)=sqrt((a+b)/2)sqrt((c+d)/2)>=root(4)(ab)*root(4)(cd)=root(4)(abcd) .

Thus, (a+b+c+d)/4>=root(4)(abcd).

Equality is possible only if a=b=c=d.