# Any better way to find the minima/maxima?

The teacher asked us to minimize an expression:$\frac{x^2+y^2+z^2}{xy+yz}$, where $x, y, z>0$

Since the denominator and numerator are homogeneous, we try to divide the y^2 into two halves:

$\frac{x^2+\frac{1}{2}y^2+\frac{1}{2}y^2+z^2}{xy+yz}$

Use the inequality $a+b\ge2\sqrt{ab} (a,b>0)$ and we get:

$\frac { x^{ 2 }+\frac { 1 }{ 2 } y^{ 2 }+\frac { 1 }{ 2 } y^{ 2 }+z^{ 2 } }{ xy+yz } \ge \frac { 2\sqrt { x^{ 2 }\cdot \frac { 1 }{ 2 } y^{ 2 } } +2\sqrt { \frac { 1 }{ 2 } y^{ 2 }\cdot z^{ 2 } } }{ xy+yz } =\boxed{2}$

also, the same way to find the minima of $\frac { 10x^{ 2 }+10y^{ 2 }+z^{ 2 } }{ xy+xz+yz }$ :

$\frac { 10x^{ 2 }+10y^{ 2 }+z^{ 2 } }{ xy+xz+yz } \\ =\frac { [ax^{ 2 }+ay^{ 2 }]+[(10-a)x^{ 2 }+bz^{ 2 }]+[(10-a)y^{ 2 }+(1-b)z^{ 2 }] }{ xy+xz+yz } \\ \ge \frac { 2\sqrt { ax^{ 2 }\cdot ay^{ 2 } } +2\sqrt { (10-a)x^{ 2 }\cdot bz^{ 2 } } +2\sqrt { (10-a)y^{ 2 }\cdot (1-b)z^{ 2 } } }{ xy+xz+yz } \\ =\frac { 2a\cdot xy+2\sqrt{(10-a)b}\cdot xz+2\sqrt{(10-a)(1-b)}\cdot yz }{ xy+xz+yz }$

Now we let $a=\sqrt{(10-a)b} = \sqrt{(10-a)(1-b)}$ , that is $a = 2 \ and\ b = \frac{1}{2}$

Then

$\frac { 4xy+4xz+4yz }{ xy+xz+yz } \\ = \boxed{4}$

But this method is quite troublesome... right?

Is there any other way to solve the minima? I also tried partial derivative, but it's not easy to solve that equation either... Note by John Lee
1 year, 1 month ago

This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.

When posting on Brilliant:

• Use the emojis to react to an explanation, whether you're congratulating a job well done , or just really confused .
• Ask specific questions about the challenge or the steps in somebody's explanation. Well-posed questions can add a lot to the discussion, but posting "I don't understand!" doesn't help anyone.
• Try to contribute something new to the discussion, whether it is an extension, generalization or other idea related to the challenge.

MarkdownAppears as
*italics* or _italics_ italics
**bold** or __bold__ bold
- bulleted- list
• bulleted
• list
1. numbered2. list
1. numbered
2. list
Note: you must add a full line of space before and after lists for them to show up correctly
paragraph 1paragraph 2

paragraph 1

paragraph 2

[example link](https://brilliant.org)example link
> This is a quote
This is a quote
    # I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
# I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
MathAppears as
Remember to wrap math in $$ ... $$ or $ ... $ to ensure proper formatting.
2 \times 3 $2 \times 3$
2^{34} $2^{34}$
a_{i-1} $a_{i-1}$
\frac{2}{3} $\frac{2}{3}$
\sqrt{2} $\sqrt{2}$
\sum_{i=1}^3 $\sum_{i=1}^3$
\sin \theta $\sin \theta$
\boxed{123} $\boxed{123}$

Sort by:

Yes there is another approach................Try to use spherical co-ordinates...........That makes things simpler and transforms everything into two variables.........!!

- 1 year, 1 month ago

Care to elaborate on this? What substitution works here? Are you sure the constraints are easily manageable?

- 1 year, 1 month ago

Umm yes.........well, we can use the standard substitutions used when transforming from xyz co-ordinates to spherical co-ordinates.....!!

- 1 year, 1 month ago

I don't understand. What standard substitutions are you referring to?

- 1 year, 1 month ago

Well.........here you go, Sir........
x = rsin(phi)cos(theta)
y = rsin(phi)sin(theta)
z = r*cos(phi)
Here, phi and theta are two different angles........

- 1 year, 1 month ago

Also, r is the radius of the sphere.........we are just taking these as dummy variables.........

- 1 year, 1 month ago

Also, after the transformation, we will have a unit fraction to deal with........and hence it would simply be the matter of maximising the denominator.......

- 1 year, 1 month ago