Cauchy's Inequality

Theorem

$\displaystyle \sum {r_i^2} \sum {s_i^2} \ge \left({\sum {r_i s_i}}\right)^2$

where all of $r_i, s_i \in \R$.

Proof 1

For any $\lambda \in \R$, we define $f: \R \to \R$ as the function:

$\displaystyle f \left({\lambda}\right) = \sum {\left({r_i + \lambda s_i}\right)^2}$

Now:

$f \left({\lambda}\right) \ge 0$

because it is the sum of squares of real numbers.

Hence:

$\displaystyle \forall \lambda \in \R: f \left(\lambda\right) \equiv \sum {r_i^2} + 2 \lambda \sum {r_i s_i} + \lambda^2 \sum {s_i^2} \ge 0$

This is a simple quadratic in $\lambda$, and we can solve it using Quadratic Equation, where:

$\displaystyle a \lambda^2 + b \lambda + c = 0: a = \sum {s_i^2}, b = 2 \sum {r_i s_i}, c = \sum {r_i^2}$

The discriminant of this equation (i.e. $b^2 - 4 a c$) is:

$\displaystyle 4 \left({\sum {r_i s_i}}\right)^2 - 4 \sum {r_i^2} \sum {s_i^2}$

If this were positive, then $f \left({\lambda}\right) = 0$ would have two distinct real roots, $\lambda_1 < \lambda_2$, say.

If this were the case, then $f$ must be negative somewhere.

To see this, note that $f$ must factor either as $\left({\lambda - \lambda_1}\right) \left({\lambda - \lambda_2}\right)$ or $-\left({\lambda - \lambda_1}\right) \left({\lambda - \lambda_2}\right)$.

When $\lambda^*$ is between $\lambda_1$ and $\lambda_2$, we have $\lambda^* - \lambda_1$ positive and $\lambda^* - \lambda_2$ negative, and so their product is negative.

When $\lambda^*$ is greater than both $\lambda_1$ and $\lambda_2$, both terms are positive and so their product is positive.

This means that the sign of $f$ must change at $\lambda_2$.

But we have:

$\forall \lambda \in \R: f \left({\lambda}\right) \ge 0$

so the discriminant can not be positive.

Thus:

$\displaystyle 4 \left({\sum {r_i s_i}}\right)^2 - 4 \sum {r_i^2} \sum {s_i^2} \le 0$

which is the same thing as saying:

$\displaystyle \sum {r_i^2} \sum {s_i^2} \ge \left({\sum {r_i s_i}}\right)^2$

$\blacksquare$

Proof 2

From the Complex Number form of the Cauchy-Schwarz Inequality, we have:

$\displaystyle \sum \left|{w_i}\right|^2 \left|{z_i}\right|^2 \ge \left|{\sum w_i z_i}\right|^2$

where all of $w_i, z_i \in \C$.

As elements of $\R$ are also elements of $\C$, it follows that:

$\displaystyle \sum \left|{r_i}\right|^2 \left|{s_i}\right|^2 \ge \left|{\sum r_i s_i}\right|^2$

where all of $r_i, s_i \in \R$.

But from the definition of modulus, it follows that:

$\displaystyle \forall r_i \in \R: \left|{r_i}\right|^2 = r_i^2$

Thus:

$\displaystyle \sum {r_i^2} \sum {s_i^2} \ge \left({\sum {r_i s_i}}\right)^2$

where all of $r_i, s_i \in \R$.

$\blacksquare$

Source of Name

This entry was named for Augustin Louis Cauchy.

He first published this result in 1821.

It is a special case of the Cauchy-Schwarz Inequality.

Sources

• K.G. Binmore: Mathematical Analysis: A Straightforward Approach (1977)... (previous)... (next): $\S 1.11$
• For a video presentation of the contents of this page, visit the Khan Academy.