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An advanced mathematical solution to a competition problem. Help me.
Of course, the equation can be derived to solve this problem, but it has a very simple method. I noticed that.

The numerator and denominator are both homogeneous twice, so if g(x, y, z) is the numerator, r(x, y, z) is the denominator.

The value of f=g/r satisfies f(kx, ky, kz)=f(x, y, z), so it can be defined as r= 1 to find the maximum value.

This is equivalent to limiting r(x, y, z)= 1, and finding the maximum value of g(x, y, z) can start Lagrange.

The machine is broken.

More simply, note that r= 1 is the unit sphere S. In order to get the maximum value of G on S, the gradient of G is needed.

Grad(g) is perpendicular to S. grad(g) = (gx, gy, gz) = (y, x+2z, 2y).

The normal vector of S at (x, y, z) is (x, y, z).

Therefore, the above conditions are equivalent to

Y/x = (x+2z)/y = 2y/z = undetermined constant k

This is y = kx, z = 2x, and k * k = 5.

Thus, the radical number 5 with f = k/2 = 2 is obtained.

Note that this method is actually consistent with Lagrangian method.