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.

Roy Pic

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

lagrange1Now,

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

lagrange3Now,

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

lagrange2

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

lagrange4

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}}

Proceed to Part 2.

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