## From Cauchy-Schwartz Inequality to Triangle Inequality, and Back Again

In class (as is the case also in Anton’s book) the strategy for establishing the triangle inequality ||u + v|| < ||u|| + ||v|| was to establish first the Cauchy-Schwartz Inequality and then to argue as follows, given vectors u and v in some Rn (or Rn ):

1. Expand and then estimate ||u + v||2 :
||u + v||2 = (u + v) • (u + v) = (uu) + 2(uv) + (vv)
< (uu) + 2 | uv | + (vv) = ||u||2 + 2 | uv | + ||v||2
< ||u||2 + 2 ||u|| ||v|| + ||v||2 = (||u|| + ||v||)2 ; and then
2. take square roots on both sides:
||u + v|| < ||u|| + ||v|| .

Amusingly enough, one can also reverse this inference: that is, it is possible to derive the Cauchy-Schwartz inequality from the triangle inequality; here’s how.
Recall (from class, or from Anton, Chapter 4, section 1, item 4.1.6) the formula
uv = ¼[(||u + v||)2 – (||uv||)2] ,
though it will only get used at the very end, after we’ve prepared suitable estimates on the two norms (squared) within the RHS.
On the one hand, applying the triangle inequality to uv and v tells us ||u|| = ||(uv) + v|| < ||uv|| + ||v|| , which rebalances to
||u|| – ||v|| < ||uv|| ;
and applying it instead to vu and u tells us ||v|| = ||(vu) + u|| < ||vu|| + ||u|| , which rebalances and rewrites to
||v|| – ||u|| < ||vu|| = ||uv|| .
But these two inequalities taken together can be summarized as ||uv|| > | ||u|| – ||v|| | , whence
||uv||2 > (||u|| – ||v||)2 .
On the other hand, the triangle inequality applied to u and v tells us ||u + v|| < ||u|| + ||v|| , whence
||u + v||2 < (||u|| + ||v||)2 .
From these last two inequalities it follows that
||u + v||2 – ||uv||2 < (||u|| + ||v||)2 – (||u|| – ||v||)2 = ... = 4 ||u|| ||v|| (why?),
and if we apply the thinking above not to u and v but to u and – v , we find (since || – v|| = ||v|| ):
||uv||2 – ||u + v||2 < (||u|| + ||v||)2 – (||u|| – ||v||)2 = 4 ||u|| ||v|| .
Finally, these two inequalities taken together tell us that
| ||u + v||2 – ||uv||2 | < 4 ||u|| ||v|| ,
and so, at last exploiting the formula for uv displayed at the outset, we have
| uv | = | ¼[||u + v||2 – ||uv||2] | = ¼ | ||u + v||2 – ||uv||2 | < ¼ (4 ||u|| ||v||) = ||u|| ||v|| ,
as was to be proved.
 First posted: 2 Apr 2006. Math 221 Home  Contents © 2006 by Fred E.J. Linton