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Cauchy Inequality

A special case of the Hölder Sum Inequality with $p=q=2$,

\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} (1)

where equality holds for $a_k=cb_k$. In 2-D, it becomes
(a^2+b^2)(c^2+d^2)\geq (ac+bd)^2.
\end{displaymath} (2)

It can be proven by writing
\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} (3)

If $b_i/a_i$ is a constant $c$, then $x=-c$. If it is not a constant, then all terms cannot simultaneously vanish for Real $x$, so the solution is Complex and can be found using the Quadratic Equation
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} (4)

In order for this to be Complex, it must be true that
\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} (5)

with equality when $b_i/a_i$ is a constant. The Vector derivation is much simpler,
({\bf a}\cdot{\bf b})^2 =a^2b^2\cos^2\theta\leq a^2 b^2,
\end{displaymath} (6)

a^2\equiv {{\bf a}\cdot{\bf a}}=\sum_i {a_i}^2,
\end{displaymath} (7)

and similarly for $b$.

See also Chebyshev Inequality, Hölder Sum Inequality


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