Theorem 3.3 (Cauchy-Schwarz Inequality) If u and v are any two vectors in an innner product space V, then

| < u,v > | || u || || v ||

Discussion of the ideas behind the proof: This is an alternate proof that I hope you will think is better motivated than the one in the text. The only inequality that shows up in the definition of an inner product space appears in the condition that 0 < x,x > for any vector x. Since this seems to be the only possible tool, we need to rewrite the Cauchy-Schwarz inequality until it looks like the inner product of a vector with itself. The first thing to do is to square both sides and then rewrite the lengths in term of the inner product.

| < u,v > | || u || || v ||
< u,v >2 || u ||2 || v ||2
< u,v >2 < u,u > < v,v >
0 < u,u > < v,v > - < u,v > < u,v >

If u is the zero vector, then the Cauchy-Schwarz inequality clearly holds, so we can assume that u is nonzero. The next step is to divide through by < u,u >.

0 < v,v > - c < u,v >, where c = < u,v > / < u,u >.
0 < v - cu , v >

Now we have a new form of the Cauchy-Schwarz inequality to work with. We can complete the proof if we show that

< v - cu , v > = < v - cu , v - cu >,

since the inner product of any vector with itself must be nonnegative. To show this, we only need to check that

< v - cu , u > = 0,

since

< v - cu , v - cu > = < v - cu , v > - < v - cu , -cu > = < v - cu , v > + c < v - cu , u >.

Expanding < v - cu , u > gives < v,u > - c< u,u >, and this is equal to zero because c =

< u,v > / < u,u >.

In summary, we finally have the inequality

0 < v - cu , v - cu > = < v - cu , v > = < v,v > - c < u,v >.

As we have already shown, this inequality is the same as the Cauchy-Schwarz inequality

< u,v >2 || u ||2 || v ||2.

Formal proof of the Cauchy-Schwarz inequality:

If u = 0, then the inequality certainly holds, so we can assume that u is nonzero. Then < u,u > is nonzero, and so we can define

c = < u,v > / < u,u >.

It follows from the definition of an inner product that for the vector v - cu we have

0 < v - cu , v - cu >.

Computing this inner product gives

< v - cu , v - cu > = < v , v - cu > - c < u , v - cu > = < v , v - cu >

because c = < u,v > / < u,u > and therefore

< u , v - cu > = < u,v > - c < u,u > = 0.

Thus we have

0 < v - cu , v > = < v,v > - c < u,v >,

and adding c < u,v > to both sides gives

c < u,v > < v,v >.

Finally, multiplying both sides of the inequality by < u,u > gives the identity

< u,v >2 < u,u > < v,v >,

which is the same as the Cauchy-Schwarz inequality.