# Cauchy-Schwarz Inequality By Engr. Roy Roque Rivera Jr. Part 1

Our star of the day is another gifted man. I am actually his number 1 fan. He graduated as valedictorian during high school at Negros Occidental High School. He took up B.S. Chemical Engineering at the University of St. La Salle, Bacolod City. December 3, 2013, he took up the ChE board examination and ranked 6th out of 777 takers and 426 passers. What made him different from other takers was that he just did a SELF-REVIEW. One of our friends told me that he is the guy who always has an answer. He is currently working as a Chemical Engineer at Coral Bay Nickel Corporation.

As a quizzer since grade school, here are just a few glimpse of his achievements.

College:

• 6th Placer Chemical Engineering Board Examinations
• Champion, 2011 Visayas – Mindanao Wide MTAP – Tertiary Level (Team and Individual Category)
• 2nd Place, Philippine Statistics Quiz (Regional Level)
• Champion, 2010 Council of Engineering and Architectural Schools (CEAS) – Visayas Math and Science Quiz Region VI (Team and Individual Category)
• 1st Runner Up, 2011 Council of Engineering and Architectural Schools (CEAS) – Visayas Math and Science Quiz Region VI (2012)

High School

• National Finalist, 2007 Metrobank MTAP DepEd Mathematics Challenge (Team and Individual Category)
• Champion, 2007 Metrobank MTAP DepEd Mathematics Challenge Regional Level (Team and Individual Category)

Here is the first part of the article he shared. This is the conventional Cauchy – Schwarz Inequality. The second part of the article is the application of the same inequality in Engel Form which has a huge application in proving hard problems in inequalities.

1. The Cauchy – Schwarz Inequality

1.1 Cauchy – Schwarz Inequality

Theorem: Let $x_i$ and $y_i$ be real numbers for $i=1,2,\dots , n,$ then the following inequality holds

$(\sum_{i=1}^n x_i^2)^{\frac{1}{2}}(\sum_{i=1}^n y_i^2)^{\frac{1}{2}}\geq \sum_{i=1}^n x_iy_i$

Proof: Let $A=(\sum_{i=1}^n x_i^2)^{\frac{1}{2}}$ and $B=(\sum_{i=1}^n y_i^2)^{\frac{1}{2}}$. Consider

$\sum_{i=1}^n x_iy_i=AB\sum_{i=1}^n \displaystyle\frac{x_iy_i}{AB}$

From AM-GM Inequality, we know that (1)

$\displaystyle\frac{1}{2}(\displaystyle\frac{x_1^2}{A^2}+\displaystyle\frac{y_1^2}{B^2})\geq \displaystyle\frac{x_i^2y_i^2}{A^2B^2}$

Or

$\displaystyle\frac{x_i^2y_i^2}{A^2B^2}\leq \displaystyle\frac{1}{2}(\displaystyle\frac{x_1^2}{A^2}+\displaystyle\frac{y_1^2}{B^2})$

(1)AM-GM Inequality states that for any positive real numbers  $a_i,(i=1,2,\ldots ,n),$ the inequality

$\displaystyle\frac{a_1+a_2+\ldots+a_n}{n}\geq \root n\of{a_1a_2\ldots a_n}$ always holds. Equality is attained if and only if $a_1=a_2=\ldots=a_n$

Therefore, (1) becomes

$AB\sum_{i=1}^n \displaystyle\frac{x_iy_i}{AB}\leq AB\sum_{i=1}^n \displaystyle\frac{1}{2}(\displaystyle\frac{x_i^2}{A^2}+\displaystyle\frac{y_i^2}{B^2})$

$=\displaystyle\frac{AB}{2}\sum_{i=1}^n (\displaystyle\frac{x_i^2}{A^2}+\displaystyle\frac{y_i^2}{B^2})$

$=\displaystyle\frac{AB}{2}(\sum_{i=1}^n \displaystyle\frac{x_i^2}{A^2}+\sum_{i=1}^n \displaystyle\frac{y_1^2}{B^2})$

$=\displaystyle\frac{AB}{2}(\displaystyle\frac{x_1^2+x_2^2+\ldots+x_n^2}{A^2}+\displaystyle\frac{y_1^2+y_2^2+\ldots +y_n^2}{B^2}$

But $A^2=\sum_{i=1}^n x_i^2$ and $B^2=\sum_{i=1}^n y_i^2$; hence,

$\displaystyle\frac{AB}{2}(\displaystyle\frac{x_1^2+x_2^2+\ldots+x_n^2}{A^2}+\displaystyle\frac{y_1^2+y_2^2+\ldots+y_n^2}{B^2}\leq \displaystyle\frac{AB}{2}(\displaystyle\frac{x_1^2+x_2^2+\ldots+x_n^2}{ x_1^2+x_2^2+\ldots+x_n^2}+\displaystyle\frac{y_1^2+y_2^2+\ldots+y_n^2}{y_1^2+y_2^2+\ldots +y_n^2})$

$=\displaystyle\frac{AB}{2}(1+1)$

$=AB$

It follows that

$\sum_{i=1}^n x_iy_i\leq AB$

Or

$(\sum_{i=1}^n x_i^2)^{\frac{1}{2}}(\sum_{i=1}^n y_i^2)^{\frac{1}{2}}\geq \sum_{i=1}^n x_iy_i$

Which was to be demonstrated

Equality occurs if and only if

$\displaystyle\frac{x_1}{y_1}=\displaystyle\frac{x_2}{y_2}=\ldots =\displaystyle\frac{x_n}{y_n}=\lambda$

For some constant $\lambda$

1.2 Cauchy- Schwarz Inequality in Engel Form

Theorem ( Cauchy-Schwarz Inequality in Engel Form) : Suppose that $x_i,y_i,a_i \in \Re,$ with $x_i,y_i \ne 0$ for $i=1,2,\ldots ,n$ then

$\displaystyle\frac{a_1^2}{x_1}+\displaystyle\frac{a_2^2}{x_2}+\ldots+\displaystyle\frac{a_n^2}{x_n}\geq \displaystyle\frac{(a_1+a_2+\ldots+a_n)^2}{x_1+x_2+\ldots+x_n}$ (2)

Proof:  The proof of (2) directly follows from the Cauchy-Schwarz Inequality. From Cauchy- Schwarz Inequality, we have

$(\displaystyle\frac{a_1^2}{(\sqrt{x_1})^2}+\ldots+\displaystyle\frac{a_n^2}{(\sqrt{x_n})^2}((\sqrt{x_1})^2+\ldots+(\sqrt{x_n})^2)\geq (\displaystyle\frac{a_1}{\sqrt{x_1}}\sqrt{x_1}+ \displaystyle\frac{a_2}{\sqrt{x_2}}\sqrt{x_2}+\ldots+\displaystyle\frac{a_n}{\sqrt{x_n}}\sqrt{x_n})^2$

Hence,

$\displaystyle\frac{a_1^2}{x_1}+\displaystyle\frac{a_2^2}{x_2}+\ldots+\displaystyle\frac{a_n^2}{x_n}\geq \displaystyle\frac{(a_1+a_2+\ldots+a_n)^2}{x_1+x_2+\ldots+x_n}$

Which was to be demonstrated

Equality occurs if and only if

$\displaystyle\frac{x_1}{y_1}=\displaystyle\frac{x_2}{y_2}=\ldots =\displaystyle\frac{x_n}{y_n}=\lambda$

For some constant $\lambda$

2. Illustrations and Examples

2.1 Elementary Applications

Problem 1: What is the minimum value of $x^2+9y^2$ given that $4x+9y=36$

Solution 1: (Method of Lagrange Multipliers): The method of Lagrange Multipliers 2 involves finding the values of $x,y$ and $\lambda$ such that the system of equations hold

Now,

$f(x,y)=x^2+9y^2$

And

$g(x,y)=4x+9y\longleftrightarrow g(x,y)=36$

We have

$f_x=\displaystyle\frac{\partial}{\partial x}(x^2+9y^2)=2x$

$f_x=\displaystyle\frac{\partial}{\partial x}(x^2+9y^2)=18y$

$g_x=\displaystyle\frac{\partial}{\partial x}(4x+9y)=4$

$g_x=\displaystyle\frac{\partial}{\partial x}(4x+9y)=9$

Hence, we have the following system of equations to solve

Now,

$x=2\lambda$   and   $y=\displaystyle\frac{\lambda}{2}$

Thus,

$4x+9y=36\longrightarrow 8\lambda+\displaystyle\frac{9}{2}\lambda=36$

2The method of Lagrange Multipliers involves finding the extreme values of the function $f$ given the constrant $g(x,y)=k$

Solving for ʎ gives $\lambda=\displaystyle\frac{72}{25}$. Now, we have

$x^2+y^2=4\lambda^2+\displaystyle\frac{9}{4}\lambda^2$

$=\displaystyle\frac{25}{4}\lambda^2$

$=\displaystyle\frac{25}{4}(\displaystyle\frac{72}{25})^2$

$=\displaystyle\frac{36^2}{25}$

$=\displaystyle\frac{1296}{25}$

Thus, min $\{x^2+9y^2\}=\boxed{\displaystyle\frac{1296}{25}}$

Solution 2 (Cauchy-Schwarz Inequality): Observe that $4x+9y=36$ can be written as $4x+3(3y)=36$. Now, by Cauchy –Schwarz Inequality, we have

$\sqrt{4^2+3^2}\sqrt{x^2+(3y)^2}\geq 4x+3(3y)$

or

$\sqrt{4^2+3^2}\sqrt{x^2+(3y)^2}\geq 36$

Solving for $x^2+9y^2$ gives

$x^2+9y^2=\displaystyle\frac{36^2}{25}$

Hence, min $\{x^2+9y^2\}=\boxed{\displaystyle\frac{1296}{25}}$

Problem 2: What is the minimum distance between the ellipse $\displaystyle\frac{x^2}{a^2}+\displaystyle\frac{y^2}{b^2}=1$ and the line $Ax+By=1$. The ellipse and the plane do not intersect

Solution: Let $(p,q)$ be a point on the ellipse $\displaystyle\frac{x^2}{a^2}+\displaystyle\frac{y^2}{b^2}=1$, and $(r,s)$ be a point on the line $Ax+By=1$. Also, let the distance between $(p,q)$ and $(r,s)$ be equal to $d$; hence,

$d^2=(p-q)^2+(q-s)^2$

Now, consider the product

$((p-r)^2+(q-s)^2)(A^2+B^2)$

By Cauchy- Schwarz Inequality, we have

$((p-q)^2+(q-s)^2)(A^2+B^2)\geq (A(p-r)+B(q-s))^2$

$=(Ap+Bq-(Ar+Bs))^2$

$=(Ap+Bq-1)^2$

Or equivalently

$d=\displaystyle\frac{1-(Ap+Bq)}{\sqrt{A^2+B^2}}$

Note that because $(r,s)$ is one the line $Ax+By=1$, it follows that $Ar+Bs=1$. Also since the ellipse and the plane do not intersect

$1-(Ap+Bq)>0\longrightarrow Ap+Bq<1$

Now, consider the product

$(\displaystyle\frac{x^2}{a^2}+\displaystyle\frac{y^2}{b^2})(aA^2+bB^2)$

By Cauchy-Schwarz Inequality, the product becomes

$(\displaystyle\frac{p^2}{a^2}+\displaystyle\frac{q^2}{b^2})(a^2A^2+b^2B^2)\geq (Ap+Bq)^2$

Then

$(a^2A^2+b^2B^2)\geq (Ap+Bq)^2$

or

$Ap+Bq\leq \sqrt{a^2A^2+b^2B^2}$

Observe that in order to minimize the value of d, the value of $Ap+Bq$ must be maximized; hence, take $Ap+Bq=\sqrt{a^2A^2+b^2B^2}$. Therefore, it follows that the minimum distance between the ellipse $\displaystyle\frac{x^2}{a^2}+\displaystyle\frac{y^2}{b^2}=1$ and the line $Ax+By=1$ is equal3

$d=\boxed{\displaystyle\frac{1-\sqrt{a^2A^2+b^2B^2}}{\sqrt{A^2+B^2}}}$

Problem 3: Find the maximum value of $f(x,y,z)=4x+3y+2z$ on the ellipsoid $16x^2+9y^2+4z^2=1$.

Solution: We use Cauchy-Schwarz Inequality

$(1^2+1^2+1^2)(16x^2+9y^2+4z^2)\geq (4x+3y+2z)^2$

or

$(4x+3y+2z)^2\le (1^2+1^2+1^2)(16x^2+9y^2+4z^2)$

Thus,

$(4x+3y+2z)^2\leq 3$

Or

$-\sqrt{3}\leq 4x+3y+2z\leq \sqrt{3}$

Therefore, the maximum value of $latex f(x,y,z)=4x+3y+2z$ is $\boxed{\sqrt{3}}$

Problem 4: Find the maximum value of $f(x,y,z)=2x+3y+8z$ on the ellipsoid $x^2+y^2+4z^2=4$

Solution: By Cauchy – Schwarz Inequality, we have

$(2^2+3^2+4^2)(x^2+y^2+4z^2)\geq (2x+3y+8z)^2$

Or

$(2x+3y+8z)^2\leq (2^2+3^2+4^2)(x^2+y^2+4z^2)$

Thus,

$(2x+3y+8z)^2\leq 116$

or

$-2\sqrt{29}\leq 4x+3y+8z\leq 2\sqrt{29}$

Therefore, the maximum value of $f(x,y,z)=2x+3y+8z$ is $\boxed{2\sqrt{29}}$

One can use the Method of Lagrange Multipliers and get the same result.

Problem 5 (R. Guadalupe): Let $a,b,c \in \Re$ such that

Find the maximum value of a.

Solution: We write the equation as

$a^2+b^2+c^2=2014\longleftrightarrow b^2+c^2=2014-a^2$

and

$a+2b+3c=86\longleftrightarrow 2b+3c=86-a$

Now, by Cauchy – Schwarz Inequlity we have $(b^2+c^2)(2^2+3^2)\geq (2b+3c)^2$. Hence,

$(b^2+c^2)(2^2+3^2)\geq (2b+3c)^2$

$(2014-a^2)(4+9)\geq (86-a^2)^2$

$26182-13a^2\geq a^2-172a+7396$

Which can be simplified into

$7a^2-86a-9393\leq 0$

$(7a-303)(a+31)\leq 0$

It follows that

$-31\leq x\leq \displaystyle\frac{303}{7}$

Thus, the maximum value of a must be $\boxed{\displaystyle\frac{303}{7}}$

Problem 6 (United States of America Mathematical Olympiad): Let $a,b,c,d,e\in \Re$ such that

What is the maximum value of e?

Solution:

By Cauchy-Schwarz Inequality, we know that

$(a^2+b^2+c^2+d^2)(1^2+1^2+1^2+1^2)\geq (a+b+c+d)^2$

Or equivalently

$4(8-e^2)\geq (16-e)^2$

Or

$e(5e-16)\leq 0$

Which implies that    $e\in [0, \displaystyle\frac{16}{5}]$.  Thus,  max  $\{e\}=\boxed{\displaystyle\frac{16}{5}}$

### Dan

Blogger and a Math enthusiast. Has no interest in Mathematics until MMC came. Aside from doing math, he also loves to travel and watch movies.

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