AM/GM Inequality and Cauchy-Schwarz

Section 4.4 AM/GM Inequality and Cauchy-Schwarz

In this section we cover some very useful inequalities that allow us to separate terms in a product. As a warm-up, let's begin with the following.

Lemma 4.4.1.

Let \(u,v\in\RR\text{.}\) Then

\begin{equation*} uv \leq \frac{u^2+v^2}{2}\text{.} \end{equation*}

Notice that on one side of the above inequality, the terms \(u\) and \(v\) are multiplied together, and on the other side, they are separate (but now squared).

Proof.

Since\((u-v)^2\geq 0\text{,}\) we get

\begin{equation*} 0\leq (u-v)^2 =u^2-2uv + v^2\text{.} \end{equation*}

So we can add \(2uv\) to both sides to get

\begin{equation*} 2uv \leq u^2+v^2 \end{equation*}

and then we can multiply both sides by \(\frac{1}{2}\) to get \(uv \leq \frac{u^2+v^2}{2}\) as desired.

Exercise 4.4.2.

When does equality hold in the previous lemma? That is, when do we have \(uv = \frac{u^2+v^2}{2}\text{?}\)

Solution.

Let's retrace the proof of Lemma 4.4.1 backwards trying to use only equalities. If \(2uv=u^2+v^2\text{,}\) then

\begin{equation*} 0 = 2+v^2-2uv = (u-v)^{2}\text{,} \end{equation*}

and this last equality holds if and only if \(u=v\text{.}\)

From this, we deduce the famous Arithmetic Mean-Geometric Mean inequality:

Theorem 4.4.3. AM/GM Inequality.

Let \(n\in\NN\text{,}\) and let \(x_{1},\dots,x_{n}\geq 0\) be nonnegative real numbers. Then

\begin{equation*} (x_{1}x_{2}\cdots x_{n})^{\frac{1}{n}}\leq \frac{x_{1}+\cdots + x_{n}}{n}\text{,} \end{equation*}

with equality if and only if \(x_1=x_2=\cdots=x_n\text{.}\)

That is, the geometric mean on the left is at most the arithmetic mean on the right.

Proof.

Write \(a = (x_1 + \cdots + x_n)/n\) for the arithmetic mean. First, let us consider the case in which we have \(x_1=\cdots=x_n=a\text{.}\) In that case, we have equality, because both the arithmetic and geometric means are equal to \(a\text{.}\) In particular, the claim is true for \(n=1\text{.}\)

Now for an induction argument, we may assume that the claim holds for \(n=k\text{,}\) where \(k\in\NN\text{,}\) and we aim to prove it for \(n=k+1\text{.}\) Furthermore, we may assume that at least two of the \(x_i\) are unequal to \(a\text{;}\) in particular \(a>0\text{.}\) Since one of these must be larger than \(a\) and the other must be smaller, we are free to assume that \(x_k\lt a\) and \(x_{k+1}\gt a\text{.}\) We aim to show that

\begin{equation*} x_1x_2\cdots x_{k+1} \lt a^{k+1}\text{.} \end{equation*}

If any of the \(x_i\)s are zero, then the geometric mean vanishes, so we have our strict inequality.

Now set \(\xi_k = x_k + x_{k+1}-a\text{;}\) note that \(\xi_k > 0\text{.}\) Now \(a = (x_1+x_2+\cdots +x_{k-1}+\xi_k)/k\text{.}\) So by hypothesis, we have

\begin{equation*} x_1x_2\cdots x_{k-1}\xi_k \leq a^k\text{,} \end{equation*}

with equality if and only if \(x_1=x_2=\cdots=x_{k-1}=\xi_k\text{.}\) Now note that

\begin{equation*} 0 \lt (a-x_k)(x_{k+1}-a) = (x_k + x_{k+1}-a)a - x_kx_{k+1} = \xi_ka-x_kx_{k+1}\text{,} \end{equation*}

so we deduce that

\begin{equation*} x_1x_2\cdots x_{k-1}x_kx_{k+1} \lt x_1x_2\cdots x_{k-1}\xi_ka \leq a^ka = a^{k+1}\text{,} \end{equation*}

as desired.

The next useful inequality is the Cauchy-Schwarz inequality:

Theorem 4.4.4. Cauchy-Schwarz Inequality.

Let \(n\in\NN\text{,}\) and let \(x_{1},\dots,x_{n},y_{1},\dots,y_{n}\in\RR\text{.}\) Then

\begin{equation*} x_{1}y_{1}+\cdots + x_{n}y_{n} \leq \sqrt{x_{1}^2+\cdots + x_{n}^{2}} \sqrt{y_{1}^2+\cdots + y_{n}^{2}}\text{.} \end{equation*}

Proof.

Let \(a=\sqrt{x_1^2+\cdots+x_n^2}\text{,}\) and let \(b=\sqrt{y_1^2+\cdots+y_n^2}\text{.}\) We aim to show that

\begin{equation*} \frac{x_1y_1}{ab}+\cdots+\frac{x_ny_n}{ab} \leq 1 \end{equation*}

Now by the AM/GM Inequality, we have, for each \(i\text{,}\)

\begin{equation*} \frac{x_iy_i}{ab} \leq \frac{1}{2}\left(\frac{x_i^2}{a^2}+\frac{y_i^2}{b^2}\right)\text{.} \end{equation*}

Now if we sum these over all \(i\text{,}\) the right hand side becomes

\begin{equation*} \frac{1}{2}\left(\frac{x_1^2}{a^2}+\frac{y_1^2}{b^2}\right)+\cdots+\frac{1}{2}\left(\frac{x_n^2}{a^2}+\frac{y_n^2}{b^2}\right) = \frac{1}{2}\left(\frac{x_1^2+\cdots+x_n^2}{a^2}+\frac{y_1^2+\cdots+y_n^2}{b^2}\right) = 1\text{,} \end{equation*}

and so the proof is complete.

Why is this useful? On the left we have a mixed product of \(x\)-terms and \(y\)-terms, and we are able to relate it to a product of two terms where now the \(x\)'s and \(y\)'s are separated.

Example 4.4.5.

Show that

\begin{equation*} \frac{a+b}{2} \leq \left(\frac{a^2+b^2}{2}\right)^{\frac{1}{2}}\text{.} \end{equation*}

Which of our above inequalities should we try using? Lemma 4.4.1 and AM-GM don't seem relevant since we don't have something like \(ab\) on the left, so let's see if we can use the Cauchy-Schwarz inequality: Rewrite \(\frac{a+b}{2}=a\cdot \frac{1}{2} +b\cdot\frac{1}{2}\text{,}\) then the Cauchy Schwarz inequality implies:

\begin{equation*} \frac{a+b}{2} = a\cdot\frac{1}{2} + b\cdot\frac{1}{2} \leq \sqrt{a^2+b^2}\sqrt{\frac{1}{2^{2}}+\frac{1}{2^{2}}} =\left(\frac{a^2+b^2}{2}\right)^{\frac{1}{2}}\text{.} \end{equation*}