MHB Show that equality holds in Cauchy-Schwarz inequality if and only if....

[Ragnarok7]

This is from section I 4.9 of Apostol's Calculus Volume 1. The book states the Cauchy-Schwarz inequality as follows:

$$\left(\sum_{k=1}^na_kb_k\right)^2\leq\left(\sum_{k=1}^na_k^2\right)\left(\sum_{k=1}^nb_k^2\right)$$

Then it asks you to show that equality holds in the above if and only if there is a real number $$x$$ such that $$a_kx+b_k=0$$ for every $$k=1,2,\ldots,n$$.

The "if" direction is easy, but I'm not sure how to get the "only if" - i.e., that such an $$x$$ implies equality. There are proofs of this online involving vectors and linear algebra, but I am wondering if it can be done without that, as the book has not introduced such things yet. For the record, the way the Cauchy-Schwarz inequality was proved in the book itself is as follows:

We have $$\sum_{k=1}^n(a_kx+b_k)^2\geq0$$ for every real $$x$$ because a sum of squares can never be negative. This may be written in the form $$Ax^2+2Bx+C\geq0$$, where $$A=\sum_{k=1}^na_k^2$$, $$B=\sum_{k=1}^na_kb_k$$, and $$C=\sum_{k=1}^nb_k^2$$. We wish to prove that $$B^2\leq AC$$. If $$A=0$$, then each $$a_k=0$$, so $$B=0$$ and the result is trivial. If $$A\neq 0$$, we may complete the square and write

$$Ax^2+2Bx+C=A\left(x+\frac{B}{A}\right)^2+\frac{AC-B^2}{A}$$.

The right side has its smallest value when $$x=-B/A$$. Putting $$x=-B/A$$ in the above, we obtain $$B^2\leq AC$$.

 

[Deveno]

Suppose there is an $x$ such that $a_kx + b_k = 0$ for each $k$.

Thus $b_k = -a_kx$, and:

$\displaystyle \left( \sum_{k = 1}^n a_kb_k \right)^2 = x^2\left(\sum_{k=1}^n (a_k)^2 \right)^2$

(really we are squaring the sum of terms $-(a_k)^2$, but when we pull out a factor of $(-1)^2$ it goes away).

On the other hand:

$\displaystyle \left( \sum_{k=1}^n (a_k)^2 \right)\left( \sum_{k=1}^n (b_k)^2 \right)$

$\displaystyle = \left( \sum_{k=1}^n (a_k)^2 \right)\left( \sum_{k=1}^n (-a_kx)^2 \right)$

$\displaystyle = \left( \sum_{k=1}^n (a_k)^2 \right)\left( x^2\sum_{k=1}^n (a_k)^2 \right)$

$\displaystyle = x^2\left( \sum_{k=1}^n (a_k)^2 \right)^2$.

This is actually the "if" part, the "only if" part would mean showing there is such an $x$ if equality holds (I get these mixed up, sometimes, too).

 

[Euge]

This is from section I 4.9 of Apostol's Calculus Volume 1. The book states the Cauchy-Schwarz inequality as follows:

$$\left(\sum_{k=1}^na_kb_k\right)^2\leq\left(\sum_{k=1}^na_k^2\right)\left(\sum_{k=1}^nb_k^2\right)$$

Then it asks you to show that equality holds in the above if and only if there is a real number $$x$$ such that $$a_kx+b_k=0$$ for every $$k=1,2,\ldots,n$$.

The "if" direction is easy, but I'm not sure how to get the "only if" - i.e., that such an $$x$$ implies equality. There are proofs of this online involving vectors and linear algebra, but I am wondering if it can be done without that, as the book has not introduced such things yet. For the record, the way the Cauchy-Schwarz inequality was proved in the book itself is as follows:

We have $$\sum_{k=1}^n(a_kx+b_k)^2\geq0$$ for every real $$x$$ because a sum of squares can never be negative. This may be written in the form $$Ax^2+2Bx+C\geq0$$, where $$A=\sum_{k=1}^na_k^2$$, $$B=\sum_{k=1}^na_kb_k$$, and $$C=\sum_{k=1}^nb_k^2$$. We wish to prove that $$B^2\leq AC$$. If $$A=0$$, then each $$a_k=0$$, so $$B=0$$ and the result is trivial. If $$A\neq 0$$, we may complete the square and write

$$Ax^2+2Bx+C=A\left(x+\frac{B}{A}\right)^2+\frac{AC-B^2}{A}$$.

The right side has its smallest value when $$x=-B/A$$. Putting $$x=-B/A$$ in the above, we obtain $$B^2\leq AC$$.

Hi Ragnarok,

Note that for all real $x$,

$\displaystyle (*) \sum_{k = 1}^n (a_k x + b_k)^2 = Ax^2 + 2Bx + C$.

Suppose $B^2 = AC$. If $A = 0$, we must have $B = 0$. So $a_k = b_k = 0$ for all $k$. Consequently, for any real $x$, $a_k x + b_k = 0$ for all $k$. Now assume $A\neq 0$. Then $-\frac{B}{A}$ is the root of the polynomial $Ax^2 + 2Bx + C$, and so the left hand side of $(*)$ is zero for $x = -\frac{B}{A}$. This forces $a_k x + b_k = 0$ for all $k$.

 

[Ragnarok7]

Thank you both so much! Deveno, I just realized that I asked the wrong question (I shouldn't post when tired). What I meant to ask was how equality implies the given condition. That's the harder part, for me at least. So I got the "if" and "only if" right, I just asked it the other way around. Euge, you seem to have realized what I meant anyway. Thanks!

 

[Evgeny.Makarov]

Suppose $B^2 = AC$. If $A = 0$, we must have $B = 0$. So $a_k = b_k = 0$ for all $k$. Consequently, for any real $x$, $a_k x + b_k = 0$ for all $k$.

The fact that $A=\sum_{k=1}^n a_k^2=0$ and $B=\sum_{k=1}^n a_kb_k=0$ does not imply that all $b_k=0$, but it does imply that all $a_k=0$. In fact, the required statement is not entirely correct: $AC=B^2$ holds iff vectors $\vec{a}=(a_1,\dots,a_n)$ and $\vec{b}=(b_1,\dots,b_n)$ are linearly dependent, which means that one of them equals the other multiplied by a scalar (possible 0). So either there exists an $x$ such that $a_kx+b_k=0$ for all $k$ (i.e., $\vec{b}=-x\vec{a}$) or there exists a $y$ such that $a_k+b_ky=0$ (i.e., $\vec{a}=-y\vec{b}$).

 

[Euge]

The fact that $A=\sum_{k=1}^n a_k^2=0$ and $B=\sum_{k=1}^n a_kb_k=0$ does not imply that all $b_k=0$, but it does imply that all $a_k=0$. In fact, the required statement is not entirely correct: $AC=B^2$ holds iff vectors $\vec{a}=(a_1,\dots,a_n)$ and $\vec{b}=(b_1,\dots,b_n)$ are linearly dependent, which means that one of them equals the other multiplied by a scalar (possible 0). So either there exists an $x$ such that $a_kx+b_k=0$ for all $k$ (i.e., $\vec{b}=-x\vec{a}$) or there exists a $y$ such that $a_k+b_ky=0$ (i.e., $\vec{a}=-y\vec{b}$).

Yes I mixed up the definitions of $B$ and $C$ (First time for everything, right? :) )

Since Ragnarok wants the "only if" part without the use of vectors, start with the identity

$\displaystyle AC - B^2 = \sum_{1 \le j < k \le n} (a_j b_k - a_k b_j)^2$.

If $AC - B^2 = 0$, then for all $k < j$, $a_j b_k = a_k b_j $. Hence $a_1, a_2, \ldots a_n$ is in proportion with $b_1, b_2,\ldots, b_n$.