A special case of the Hölder Sum Inequality with
,
![\begin{displaymath}
\left({\,\sum_{k=1}^n a_kb_k}\right)^2 \leq \left({\,\sum_{k=1}^n {a_k}^2}\right)\left({\,\sum_{k=1}^n {b_k}^2}\right),
\end{displaymath}](c1_640.gif) |
(1) |
where equality holds for
. In 2-D, it becomes
![\begin{displaymath}
(a^2+b^2)(c^2+d^2)\geq (ac+bd)^2.
\end{displaymath}](c1_642.gif) |
(2) |
It can be proven by writing
![\begin{displaymath}
\sum_{i=1}^n (a_ix+b_i)^2=\sum_{i=1}^n {a_i}^2\left({x+{b_i\over a_i}}\right)^2=0.
\end{displaymath}](c1_643.gif) |
(3) |
If
is a constant
, then
. If it is not a constant, then all terms cannot simultaneously vanish for
Real
, so the solution is Complex and can be found using the
Quadratic Equation
![\begin{displaymath}
x={-2\sum a_ib_i\pm\sqrt{4\left({\sum a_ib_i}\right)^2-4\sum {a_i}^2\sum{b_i}^2}\over 2\sum {a_i}^2}.
\end{displaymath}](c1_646.gif) |
(4) |
In order for this to be Complex, it must be true that
![\begin{displaymath}
\left({\sum_i a_ib_i}\right)^2\leq \left({\sum_i {a_i}^2}\right)\left({\sum_i {b_i}^2}\right),
\end{displaymath}](c1_647.gif) |
(5) |
with equality when
is a constant. The Vector derivation is much simpler,
![\begin{displaymath}
({\bf a}\cdot{\bf b})^2 =a^2b^2\cos^2\theta\leq a^2 b^2,
\end{displaymath}](c1_648.gif) |
(6) |
where
![\begin{displaymath}
a^2\equiv {{\bf a}\cdot{\bf a}}=\sum_i {a_i}^2,
\end{displaymath}](c1_649.gif) |
(7) |
and similarly for
.
See also Chebyshev Inequality, Hölder Sum Inequality
References
Abramowitz, M. and Stegun, C. A. (Eds.).
Handbook of Mathematical Functions with Formulas, Graphs, and Mathematical Tables, 9th printing.
New York: Dover, p. 11, 1972.
© 1996-9 Eric W. Weisstein
1999-05-26