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...