Any better way to find the minima/maxima?

The teacher asked us to minimize an expression:x2+y2+z2xy+yz\frac{x^2+y^2+z^2}{xy+yz}, where x,y,z>0x, y, z>0

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

x2+12y2+12y2+z2xy+yz\frac{x^2+\frac{1}{2}y^2+\frac{1}{2}y^2+z^2}{xy+yz}

Use the inequality a+b2ab(a,b>0)a+b\ge2\sqrt{ab} (a,b>0) and we get:

x2+12y2+12y2+z2xy+yz2x212y2+212y2z2xy+yz=2 \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 10x2+10y2+z2xy+xz+yz \frac { 10x^{ 2 }+10y^{ 2 }+z^{ 2 } }{ xy+xz+yz } :

10x2+10y2+z2xy+xz+yz=[ax2+ay2]+[(10a)x2+bz2]+[(10a)y2+(1b)z2]xy+xz+yz2ax2ay2+2(10a)x2bz2+2(10a)y2(1b)z2xy+xz+yz=2axy+2(10a)bxz+2(10a)(1b)yzxy+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=(10a)b=(10a)(1b) a=\sqrt{(10-a)b} = \sqrt{(10-a)(1-b)} , that is a=2 and b=12 a = 2 \ and\ b = \frac{1}{2}

Then

4xy+4xz+4yzxy+xz+yz=4 \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
10 months, 2 weeks ago

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

Aaghaz Mahajan - 10 months, 2 weeks ago

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

Pi Han Goh - 10 months, 2 weeks ago

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

Aaghaz Mahajan - 10 months, 2 weeks ago

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

Pi Han Goh - 10 months, 2 weeks ago

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@Pi Han Goh 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........

Aaghaz Mahajan - 10 months, 2 weeks ago

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

Aaghaz Mahajan - 10 months, 2 weeks ago

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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.......

Aaghaz Mahajan - 10 months, 2 weeks ago

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