Cauchy Schwarz Inequality for Real Numbers Easy Proof

Starting from the definition, we will head into the usage of the Cauchy-Schwarz inequality.

Cauchy-Schwarz inequality states that for all real numbers $a_i$ and $b_i$ , we have

$\left(\sum_{i=1}^n a_i^2\right)\left(\sum_{i=1}^n b_i^2\right)\ge \left(\sum_{i=1}^n a_ib_i\right)^2,$

where equality holds if and only if $\frac{a_i}{b_i}=k$ for some constant $k\in\mathbb{R}^+$ , for all $1\le i\le n$ which have $a_ib_i \neq 0$ .

There are many reformulations of this inequality. There's also a vector form and a complex number version of it. But we only need the above elementary form to tackle Olympiad problems and problems in other areas.

To get a feel of it, let's consider the case of $2$ terms. Let the two sequences be $\{a,b\}$ and $\{c,d\}$ . Then by Cauchy-Schwarz inequality we have

$\left(a^2+b^2\right)\left(c^2+d^2\right)\ge \left(ac+bd\right)^2.$

This easily follows from the Fibonacci-Brahmagupta identity

$\left(a^2+b^2\right)\left(c^2+d^2\right)=\left(ac+bd\right)^2+\left(ad-bc\right)^2\ge \left(ac+bd\right)^2.$

Notice that the equality holds when $ad-bc=0\Rightarrow \dfrac{a}{c}=\dfrac{b}{d}$ if the terms are non-zero. The case of 3 terms states that

$\left( a^2 + b^2 + c^2 \right) \left( d^2 + e^2 + f^2 \right) \geq ( ad + be + cf )^2.$

Let's work through an example followed by a problem to try yourself.

Given $a_1,a_2...a_n\in\mathbb{R}$ such that $a_1+a_2+\cdots+a_n=1$ , prove that $$

$a_1^{2}+a_2^{2}+\cdots+a_n^{2}≥\frac{1}{n}.$

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By Cauchy-Schwarz, we have

$\begin{aligned} (1\cdot a_1+1\cdot a_2+\cdots+1\cdot a_n)^{2}&\le \left(a_1^{2}+a_2^{2}+\cdots+a_n^{2}\right)(1+1+\cdots+1)\\ a_1^{2}+a_2^{2}+\cdots+a_n^{2}&≥\frac{1}{n}. \end{aligned}$

We have equality when $\dfrac{a_i} { 1}$ is a constant for all $i$ , which happens when $a_1 = a_2 = \cdots = a_n = \frac{1}{n}. \ _\square$

Let $a$ , $b$ , $c$ and $d$ be positive reals such that $a + b + c + d = 1$ . Find the minimum value of $\frac {1}{a} + \frac {1}{b} + \frac {1}{c} + \frac {1}{d}$ .

The following is one of the most common examples of the use of Cauchy-Schwarz. We can easily generalize this approach to show that if $x^2 + y^2 + z^2 = 1$ , then the maximum value of $ax + by + cz$ is $\sqrt{ a^2 + b^2 + c^2 }$ .

If $x^2 + y^2 + z^2 = 1$ , what is the maximum value of $x + 2y + 3z?$

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We have $( x + 2y + 3z ) ^ 2 \leq \left( 1^2 + 2^2 + 3^2 \right) \left( x^2 + y^2 + z^2 \right) = 14 .$
Hence, $x + 2y + 3z \leq \sqrt{14}$ with equality holding when $\frac{x}{1} = \frac{y}{2} = \frac{z}{3}$ .
Together with $x^2 + y^2 + z^2 = 1$ , we get

$x = \frac{1}{ \sqrt{14} },\quad y = \frac{2}{ \sqrt{14} },\quad z = \frac{3} { \sqrt{14} }.\ _\square$

Making the right choice of $a_i$ and $b_i$ can simplify a problem. The following problem will make you realize it:

For positive reals $a, b, c > 0$ , show that $$

$abc (a +b + c) \leq a^3 b + b^3 c + c^3 a .$

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At first glance, it is not clear how we can apply Cauchy-Schwarz, as there are no squares that we can use. Furthermore, the RHS is not a perfect square. The power of Cauchy-Schwarz is that it is extremely versatile, and the right choice of $a_i$ and $b_i$ can simplify the problem.

By Cauchy-Schwarz, we have

$\small \left ( \frac{a}{ \sqrt{c} } \times \sqrt{c} + \frac{ b}{ \sqrt{a} } \times \sqrt{a} + \frac{c}{ \sqrt{b} } \times \sqrt{b} \right) ^2 \leq \left ( \frac{ a^2 }{c} + \frac{ b^2}{a} + \frac{ c^2 } { b} \right) ( c + a + b ).$

As such, $( a + b + c) \leq \left( \frac{ a^2 }{c} + \frac{ b^2}{a} + \frac{ c^2 } { b} \right )$ . Multiplying throughout by $abc$ , we obtain the given inequality. $_ \square$

Proof of Titu's Lemma

For positive reals $a_1, a_2, \ldots, a_n$ and $b_1, b_2, \ldots, b_n$ , prove that

$\frac{ a_1^2 } { b_1 } + \frac{ a_2 ^2 } { b_2 } + \cdots + \frac{ a_n ^2 } { b_n } \geq \frac{ (a_1 + a_2 + \cdots+ a_n ) ^2 } { b_1 + b_2 + \cdots+ b_n }.$

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It is obtained by applying the substitution $a_i= \frac{x_i}{ \sqrt{y_i} }$ and $b_i = \sqrt{y_i}$ into the Cauchy-Schwarz inequality.

It then becomes

$\begin{aligned} \left( \frac{ x_1^2 }{ y_1 } + \frac{ x_2^2 }{ y_2 } + \cdots + \frac{ x_n^2 }{ y_n } \right) (y_1 + y_2 + \cdots + y_n ) &\geq ( x_1 + x_2 + \cdots + x_n )^2 \\ \frac{ x_1^2 }{ y_1 } + \frac{ x_2^2 }{ y_2 } + \cdots + \frac{ x_n^2 }{ y_n } &\geq \frac{ ( x_1 + x_2 + \cdots + x_n )^2 }{ (y_1 + y_2 + \cdots + y_n ) }.\ _\square \end{aligned}$

Let's give it one more try by yourself with this practical use of the Cauchy-Schwarz inequality:

Pink Blue Both sets have the same surface area No definite relation can be made

$a, b, c\ (a<b<c)$ are coprime integers, and there are the following 2 sets of present boxes:

• $3$ pink cubic boxes of respective side lengths $a, b, c$
• $3$ identical blue cuboid boxes of dimensions $a\times b\times c$ .

Which set has a larger total surface area?

An often used consequence of the Cauchy-Schwarz inequality is Titu's lemma. It is so important that we devote an entire page to it. It states that for $a_i,b_i\in\mathbb{R}, b_i>0$ $\frac{a_1^{2}}{b_1}+\frac{a_2^{2}}{b_2}+\cdots+\frac{a_n^{2}}{b_n}≥\frac{(a_1+a_2+\cdots+a_n)^{2}}{b_1+b_2+\cdots+b_n},$ and the equality similarly holds iff ${\dfrac{a_1}{b_1}=\dfrac{a_2}{b_2}=\cdots=\dfrac{a_n}{b_n}}$ .

The red region The blue region They are equal No definite relation can be made

On the left diagram above, the red rectangle is inscribed in a larger outside rectangle.

On the right, in the same outside rectangle, two blue rectangles are formed by the perpendicular lines arising from 2 adjacent vertices of the red rectangle.

Which one has a larger area, the red region or the blue region?

There are various ways to prove this inequality. A short proof is given below.

Consider the function

$f(x)=\left(a_1x-b_1\right)^2+\left(a_2 x-b_2\right)^2+\cdots +\left(a_nx-b_n\right)^2.$

Being a sum of squares, $f(x)$ is always non-negative. Now we expand out $f(x)$ and collect terms. Then it becomes

$f(x)=\left(\sum_{i=1}^n a_i^2\right)x^2-2\left(\sum_{i=1}^n a_i b_i\right)x+\left(\sum_{i=1}^n b_i^2\right),$

which is a quadratic in $x$ . The graph of this quadratic is an up parabola. Since $f\ge 0$ , the graph either touches the $x$ -axis or stays in the upper half of it. That means either it's a perfect square (has a repeated root) or doesn't have any real root. So the discriminant is $\le 0$ . Computing the discriminant, we have

$4\left(\sum_{i=1}^n a_i b_i\right)^2-4\left(\sum_{i=1}^n a_i^2\right)\left(\sum_{i=1}^n b_i^2\right)\le 0.$

Dividing by $4$ and rearranging yields the Cauchy-Schwarz inequality.

Notice that equality holds when $f$ has a real root (repeated, of course). From the first form of $f(x)$ and using the fact that the sum of squares equals to $0$ only when each square equals to $0$ , we have $a_i x-b_i=0 \implies \frac{b_i}{a_i}=x$ for all $1\le i\le n$ . That is when $\frac{a_i}{b_i}$ is constant for all $i$ . $_\square$

We can also derive the Cauchy-Schwarz inequality from the more general Hölder's inequality. Simply put $m = 2$ and $r = 2$ , and we arrive at Cauchy Schwarz. As such, we say that Holders inequality generalizes Cauchy-Schwarz.

This section is devoted to defining and proving the vector form of Cauchy-Schwarz inequality.

For all vectors $x$ and $y$ of a real inner product space,

$|\langle x,y\rangle| ^2 \leq \langle x,x\rangle \cdot \langle y,y\rangle,$

where $\langle\cdot,\cdot\rangle$ is the inner product.

Equivalently, $|\langle x,y\rangle| \leq \|x\| \cdot \|y\|.$

Equality hold if and only if the 2 vectors are linearly dependent.

For a complex vector space, we have

$\left| \sum_{i=1}^n x_i \bar{y}_i \right|^2 \leq \sum_{j=1}^n |x_j|^2 \sum_{k=1}^n |y_k|^2.$

Equality holds if and only if $x$ and $y$ are linearly dependent, that is, one is a scalar multiple of the other (which includes the case when one or both are zero).

Now we will prove the above claim by the vector form of Cauchy-Schwarz.

If $v = 0,$ it is clear that we have equality, and in this case $u$ and $v$ are also trivially linearly dependent. We henceforth assume that $v$ is non-zero. We also assume that $\langle u, v \rangle \ne 0$ because otherwise the inequality is obviously true.

Let

$z= u-\frac {\langle u, v \rangle} {\langle v, v \rangle} v.$

From the theory of vectors, we know that $z$ is the projection of $u$ onto the plane orthogonal to $v$ . We will show this is true, by showing that $z$ is orthogonal to the vector $v$ , or equivalently that $\langle z , v \rangle = 0$ . By the linearity of the inner product in its first argument, one has

$\langle z, v \rangle = \left\langle u -\frac {\langle u, v \rangle} {\langle v, v \rangle} v, v\right\rangle = \langle u, v \rangle - \frac {\langle u, v \rangle} {\langle v, v \rangle} \langle v, v \rangle = 0.$

Since these vectors are orthogonal, we can apply the Pythagorean theorem to

$u =\dfrac {\langle u, v \rangle} {\langle v, v \rangle} v+z,$

which gives

$\left\|u\right\|^2 = \left|\frac{\langle u, v \rangle}{\langle v, v \rangle}\right|^2 \left\|v\right\|^2 + \left\|z\right\|^2 = \frac{|\langle u, v \rangle|^2}{\left\|v\right\|^2} + \left\|z\right\|^2 \geq \frac{|\langle u, v \rangle|^2}{\left\|v\right\|^2}.$

After multiplying by $||v||^ 2$ , we obtain the Cauchy–Schwarz inequality.

If the relation in the above expression is actually an equality, then $||z||^2 = 0$ and hence $z = 0$ . From the definition of $z$ as the projection vector, this implies that there is no part of $u$ that is perpendicular to $v$ , and hence these two vectors are parallel. This establishes the theorem. $_ \square$

Another approach to prove it, in an algebraic way, is given below.

Consider two sequences $A=\{a_{i}\}$ and $B=\{b_{j}\}$ and a matrix, whose composing rule for each element is $M_{ij}={a_{i}}^{2}.{b_{j}}^{2}$ . For sequences with equal number of terms, we'll have

$\left| \begin{matrix} {a_{1}}^{2}.{b_{1}}^{2} & {a_{1}}^{2}.{b_{2}}^{2} & \ldots & {a_{1}}^{2}.{b_{n}}^{2}\\ {a_{2}}^{2}.{b_{1}}^{2} & {a_{2}}^{2}.{b_{2}}^{2} & \ldots & {a_{2}}^{2}.{b_{n}}^{2}\\ \vdots & \vdots & \ddots & \vdots\\ {a_{n}}^{2}.{b_{1}}^{2} & {a_{n}}^{2}.{b_{2}}^{2} & \ldots & {a_{n}}^{2}.{b_{n}}^{2} \end{matrix} \right| .$

If we attempt to sum all the terms on the matrix, by the distributive property, this would be given as

$S_{M}=\left( \displaystyle \sum_{i=1}^n a_i^2\right) \left( \displaystyle \sum_{j=1}^n b_j^2\right) .$

Now consider two related sequences $A'=\{a_{i}b_i\}$ and $B'=\{a_jb_{j}\}$ and matrix, whose composing rule for each element is $M'_{ij}=a_{i}b_ia_jb_{j}$ , as follows

$\left| \begin{matrix} {a_1^2b_1^2} & {a_1b_1a_2b_2} & \ldots & {a_1b_1a_nb_n}\\ {a_2b_2a_1b_1} & {a_2^2b_2^2} & \ldots & {a_2b_2a_nb_n}\\ \vdots & \vdots & \ddots & \vdots\\ {a_1b_1a_nb_n} & {a_2b_2a_nb_n} & \ldots & {a_n^2b_n^2} \end{matrix} \right| .$

Summing all terms gives

$S_{M'}=\left( \displaystyle \sum_{k=1}^n a_kb_k\right)^2 = \sum_{i=1}^n a_ib_i \sum_{j=1}^n a_jb_j.$

Subtracting one sum from the other cancels the diagonal terms and yields

$S_M - S_{M'} = \sum_{1 \le i < j}^{j < i \le n} \sum_{j=1}^n a_ia_ib_jb_j - a_ib_ia_jb_j.$

Because for every pair $i < j$ , there exists two terms $a_ia_ib_jb_j - a_ib_ia_jb_j$ and $a_ja_jb_ib_i - a_jb_ja_ib_i$ , the above difference may be re-written as

$S_M - S_{M'} = \sum_{i = 1}^{j - 1} \sum_{j=1}^n a_ia_ib_jb_j - a_ib_ia_jb_j + a_ja_jb_ib_i - a_jb_ja_ib_i = \sum_{i = 1}^{j - 1} \sum_{j=1}^n (a_ib_j - a_jb_i)^2.$

Because squares are non-negative, thus $S_M \geq S_{M'}$ , and therefore

$\left( \displaystyle \sum_{i=1}^n a_i^2\right) \left( \displaystyle \sum_{j=1}^n b_j^2\right) \geq \left( \displaystyle \sum_{i=1}^n a_ib_i\right)^2. \ _\square$

The following example is motivated by the vector form of Cauchy-Schwarz.

What is the maximum of the function $a \sin \theta + b \cos \theta?$

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By Cauchy-Schwarz,

$( a \sin \theta + b \cos \theta ) ^2 \leq \big( a^2 + b^2 \big) \big( \sin ^2 \theta + \cos ^2 \theta \big) = a^2 + b^2 .$

Hence, the maximum is $\sqrt{ a^ 2 + b ^2 }$ , which is achieved when $\tan \theta = \frac{ a}{b}$ . $_ \square$

In this section, we will learn the applications of Cauchy-Schwarz inequality through some examples and problems.

For $x,y,z\in\mathbb{R^{+}}$ , prove that $$

$\sqrt{x(3x+y)}+\sqrt{y(3y+z)}+\sqrt{z(3z+x)}≤2(x+y+z).$

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By Cauchy-Schwarz, we have

$\begin{aligned} &\sqrt{x(3x+y)}+\sqrt{y(3y+z)}+\sqrt{z(3z+x)} \\ \leq& \sqrt{\left(\big(\sqrt{x}\big)^{2}+\big(\sqrt{y}\big)^{2}+\big(\sqrt{z}\big)^{2}\right)\left(\big(\sqrt{3x+y}\big)^{2}+\big(\sqrt{3y+z}\big)^{2}+\big(\sqrt{3z+x}\big)^{2}\right)}\\ \leq& \sqrt{4(x+y+z)^{2}}\\ =&2(x+y+z). \ _\square \end{aligned}$

It's interesting to know that even triangle inequality in $n$ dimensions leads to Cauchy-Schwarz inequality, which can be proved easily.

For real values $\{ a_i, b_i \}$ , show that

$\small \sqrt{ a_1 ^2 + a_2 ^2 + \cdots+ a_n^2 } + \sqrt{ b_1 ^2 + b_2^2 + \cdots+ b_n^2 } \geq \sqrt{ ( a_1 + b_1)^2 + ( a_2 + b_2) ^2 + \cdots + ( a_n+ b_n)^2}.$

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Squaring both sides and subtracting common terms, we want to show that

$\small 2 \sqrt{ a_1 ^2 + a_2 ^2 + \cdots+ a_n^2 } \times \sqrt{ b_1 ^2 + b_2^2 + \cdots +b_n^2 } \geq 2 \left( a_1 b_1 + a_2 b_2 + \cdots + a_n b_n \right).$

Squaring this and dividing by 4, we get

$\left( a_1 ^2 + a_2 ^2 + \cdots+ a_n^2 \right) \times \left( b_1 ^2 + b_2^2 + \cdots+ b_n^2 \right) \geq \left( a_1 b_1 + a_2 b_2 + \cdots + a_n b_n \right)^2,$

which is Cauchy-Schwarz inequality and we know its proof. $_\square$

Here is a simple problem involving the application of Cauchy-Schwarz inequality:

Given $a+b+c=1$ and ${a,b,c}>0,$ find the minimum value of $a^2+2b^2+c^2.$

Here are some additional problems you can try to prove on your own:

• Prove that if $\{a_i, b_i\}$ are infinite sequences such that $\sum a_i ^2 < \infty$ and $\sum b_i ^2 < \infty$ , then $\sum |a_i b_i | < \infty$ .

• [APMO 1991] For positive reals $\{ a_i, b_i \}$ such that $a_1 + a_2 + \cdots + a_ n = b_1 + b_2 + \cdots+ b_n$ , show that

$\frac{ a_1 ^2 } { a_1 + b_1 } + \frac{ a_2 ^2 } { a_2 + b_2 } + \cdots + \frac{ a_n^2 } { a_n + b_n } \geq \frac{ a_ 1 + a_2 + \cdots + a_n } { 2}.$

• Show that for $x, y, z > 0$ , we have

$x + y + z \leq 2 \left ( \frac{ x^2 } { y+z } + \frac{ y^2 }{ x+z } + \frac{ z^2 } { x+y } \right).$

• Prove that

$\left( \sum a_i b_i c_i \right)^2 \leq \left( \sum a_i ^2 \right ) \left( \sum b_i ^2 \right) \left( \sum c_i ^2 \right ).$

Solve this last problem and you're excellent at Cauchy-Schwarz inequality. Here you go:

Real numbers $a,b,c,d$ satisfy

$a^2+b^2=c^2+d^2.$

If the maximum of the expression

$\dfrac{20bc+14bd+20ad-14ac}{a^2+b^2+c^2+d^2}$

can be expressed as $\sqrt{n}$ for some positive integer $n$ , what is the value of $n?$

• Hölder's inequality

• Rearrangement Inequality

• Chebyshev Inequality

• Jensen's Inequality

• Titu's Lemma

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Source: https://brilliant.org/wiki/cauchy-schwarz-inequality/

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