Prove inequalities $\frac 34 \le I(a,b) \le 1$ Given the expression,
$$ I(a,b) = \frac{a^2}{(1+a)(a+b)} + \frac{b^2}{(1+b)(a+b)} + \frac{1}{(1+a)(1+b)}$$
where $a\ge 0$ and $b\ge 0$, prove the following inequalities:
$$\frac 34 \le I(a,b) \le 1$$
I had trouble figuring it out. Tried some inequity techniques I am aware of and had no lack so far. I am not sure if it is wise to go brute force by examining the derivatives of $I(a,b)$, which I feel would be pretty messy due to its dual dimension. 
Appreciate if anyone could offer a viable approach for the proof.

Putting everything over a common denominator shows that
$$I(a,b) = \frac{a^2+b^2+a^2b+ab^2+a+b}{a^2+b^2+a^2b+ab^2+a+b+2ab} = \frac{(1+a)(1+b)(a+b)-2ab}{(1+a)(1+b)(a+b)} \leq 1$$
since $a,b\geq 0$. Simplifying this expression, we get
$$I(a,b) = 1 - \frac{2ab}{(1+a)(1+b)(a+b)}$$
Now take a look at the second term and consider it rewritten in the following way:
$$\frac{2ab}{(1+a)(1+b)(a+b)} = \frac{\frac{ab}{4}}{\frac{a^2+b^2+a^2b+ab^2+a+b+ab+ab}{8}}$$
then by AM-GM inequality
$$\frac{a^2+b^2+a^2b+ab^2+a+b+ab+ab}{8} \geq (a^8b^8)^{\frac{1}{8}} = ab$$
$$\implies \frac{2ab}{(1+a)(1+b)(a+b)} \leq \frac{\frac{ab}{4}}{ab} = \frac{1}{4}$$
which means that 
$$I(a,b) \geq 1 -\frac{1}{4} = \frac{3}{4}$$
