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`.