# The Cauchy–Schwarz Inequality II

I have previously written about the Cauchy-Schwarz Inequality.

$\displaystyle\sqrt{\sum_{i=1}^n{a_i^2}}\sqrt{\sum_{i=1}^n{b_i^2}} \geq \sum_{i=1}^n{a_ib_i}$

Here is another proof of this fun inequality.  We start at a similar point to the previous proof by applying the fact that the square of any number is always non-negative.  Suppose $t$ is an arbitrary real number, then for $i=1,2,\ldots, n$

$(a_i-t b_i)^2 \geq 0$.

This would be true for whatever real numbers we used, but there is a reason for using these particular real numbers in this particular way. This is because we can construct a very useful sum,

$\displaystyle\sum_{i=1}^n{a^2_i-2ta_ib_i+t^2b^2_i }\geq 0$,

which with some rearrangement can be seen to be a polynomial in $t$,

$\displaystyle\sum_{i=1}^n{a^2_i}+t^2\sum_{i=1}^n{b^2_i}-2t\sum_{i=1}^n{a_ib_i} \geq 0$

This is a quadratic equation.  For a quadratic expression,

$ax^2+bx+c$,

the minimum value over all possible real values of $x$ occurs when

$\displaystyle x= -\frac{b}{2a}$.

If we apply this information to the quadratic expression, which we have previously constructed, we get that the minimum should occur at

$\displaystyle\frac{2\sum_{i=1}^n{a_ib_i}}{2\sum_{i=1}^n{b^2_i}} = \frac{\sum_{i=1}^n{a_ib_i}}{\sum_{i=1}^n{b^2_i}}$.

If we substitute this value into the quadratic expression, we further see that

$\displaystyle\sum_{i=1}^n{a^2_i}+\left(\frac{\sum_{i=1}^n{a_ib_i}}{\sum_{i=1}^n{b^2_i}}\right)^2\sum_{i=1}^n{b^2_i}-2\left(\frac{\sum_{i=1}^n{a_ib_i}}{\sum_{i=1}^n{b^2_i}}\right)\sum_{i=1}^n{a_ib_i} \geq 0$,

and with some algebraic manipulation that,

$\displaystyle\sum_{i=1}^n{a^2_i}\sum_{i=1}^n{b^2_i} \geq \left(\sum_{i=1}^n{a_ib_i}\right)^2.$

This is the Cauchy-Schwarz Inequality.  However, it may bother some readers that it does not look exactly like the form of the Cauchy-Schwarz equality that I have given.  If so, note that if we take the square root of both sides,

$\displaystyle\sqrt{\sum_{i=1}^n{a_i^2}}\sqrt{\sum_{i=1}^n{b_i^2}} \geq \left|\sum_{i=1}^n{a_ib_i}\right|$

But,

$\displaystyle\left|\sum_{i=1}^n{a_ib_i}\right| \geq \sum_{i=1}^n{a_ib_i}$

and so,

$\displaystyle\sqrt{\sum_{i=1}^n{a_i^2}}\sqrt{\sum_{i=1}^n{b_i^2}} \geq \sum_{i=1}^n{a_ib_i}$

## 2 thoughts on “The Cauchy–Schwarz Inequality II”

1. […] · Leave a Comment I have previously written about the Cauchy-Schwarz Inequality here and here.  To recap, the Cauchy-Schwarz […]

2. […] to discuss a fourth proof of the Cauchy-Schwarz inequality. I have previously presented a first, second and third proof.  The fourth proof involves some algebraic pyrotechnics where we measure the […]