Global upper limit for function of two variables on unbounded domain I have the following function:

$f(a,b) = \frac{(a + a^2 + 6ab + 3b(1 + b))}{(a + a^2 + 2ab + 3b(1 + b))}, a \geq 0, b \geq 0$

I need to show, that the function has an upper limit / converges to a "global maximum" and also, what that upper limit is.

I am pretty certain from Mathematica use, that the upper limit is $\sqrt{3}$ (it could also be higher though, so I cannot presume, it is that value). Now I am looking for a way to prove it.

There is no point, for which the gradient of the function is zero (no point, for which all partial derivatives are 0 at the so time). So, formally, there is no local maximum.

Looking at the edges: The function for $a \rightarrow 0, a \rightarrow \infty$, a \rightarrow 0 and $b \rightarrow \infty$ yields function values of 1 respectively (doesn't matter, what value the other variables take). $a \rightarrow \infty ~ and ~ b \rightarrow \infty$ is indeterminate $(\frac{\infty}{\infty})$.

So, how I am getting to a solution though, is the following:

I set the partial derivative $\frac{\partial f}{\partial a} = \frac{4b(-a^2 + 3b(1 + b))}{a + a^2 + 2ab + 3b(1 + b))^2} = 0$, which gives me $a = \sqrt{3}\sqrt{b + b^2}$. 

Now I take the partial derivative $\frac{\partial f}{\partial b} = \frac{(4a(a + a^2 - 3b^2))}{(a + a^2 + 2ab + 3b(1 + b))^2}$ and set $a = \sqrt{3}\sqrt{b + b^2}$. For $b \rightarrow \infty$ the value of the partial derivative converges to 0 (so the criterion of the gradient being zero is fulfilled for a local maximum, not checking second derivatives here).

If I set $a = \sqrt{3}\sqrt{b + b^2}$, the initial function becomes: $g(b) = \frac{6b^2 + \sqrt{3}\sqrt{b(1 + b)} + 6b(1 + \sqrt{3}\sqrt{b(1 + b)})}{(6b(1 + b) + \sqrt{3}\sqrt{b(1 + b)}*(1 + 2b)}$.

$g$ has no local maximum (first derivative test). $g(0)=0$, and for $b \rightarrow \infty$ it is $\sqrt 3$, which is my solution.

Now my question is: Does it work, the way I approached this? And if yes, why is that so? I actually cannot give the reasoning myself, why this actually gives us an upper limit. Basically what I'm doing is: I am looking at all points, for which partial derivates of a is zero (might be a local maximum) and look for the maximum of the function of those points. But why is this an upper limit?

Note: this also works the other way round (set partial derivative of b to 0, let a go to infinity).
Hint: It is $$f(a,b)\le \sqrt{3}$$
You will need to show, that $$\sqrt{3} a^2-a^2+2 \sqrt{3} a b-6 a b+\sqrt{3} a-a+3
   \sqrt{3} b^2-3 b^2+3 \sqrt{3} b-3 b\geq 0$$ for $a,b$ with $a\geq 0,b\geq 0$
Solving $$\frac{\partial f(a,b)}{\partial a}=0$$ and $$\frac{\partial f(a,b)}{\partial b}=0$$ we get 
$$a=-\frac{3}{2},b=\frac{1}{2}$$
It is $$f(a,b)\geq 1$$ and the equal sign holds if $a=1,b=0$
Hint: It is $$f(a,b)\le \sqrt{3}$$
You will need to show, that $$\sqrt{3} a^2-a^2+2 \sqrt{3} a b-6 a b+\sqrt{3} a-a+3
   \sqrt{3} b^2-3 b^2+3 \sqrt{3} b-3 b\geq 0$$ for $a,b$ with $a\geq 0,b\geq 0$
Solving $$\frac{\partial f(a,b)}{\partial a}=0$$ and $$\frac{\partial f(a,b)}{\partial b}=0$$ we get 
$$a=-\frac{3}{2},b=\frac{1}{2}$$
It is $$f(a,b)\geq 1$$ and the equal sign holds if $a=1,b=0$