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...

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## Comments

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

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

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

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x = r

sin(phi)cos(theta)y = r

sin(phi)sin(theta)z = r*cos(phi)

Here, phi and theta are two different angles........

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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.......

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