<s>[INST] 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.
 [/INST] I think, it means that $a+b>0$, otherwise your inequality is wrong for $a=b=0$.
Let $a+b=2x$.
Thus, by C-S and AM-GM we obtain:
$$I(a,b)=\frac{1}{a+b}\left(\frac{a^2}{1+a}+\frac{b^2}{1+b}\right)+\frac{1}{1+a+b+ab}\geq$$
$$\geq\frac{1}{2x}\cdot\frac{(a+b)^2}{1+a+1+b}+\frac{1}{1+2x+\left(\frac{a+b}{2}\right)^2}=$$
$$=\frac{x}{1+x}+\frac{1}{(x+1)^2}=\frac{3}{4}+\frac{1}{4}-\frac{1}{1+x}+\frac{1}{(1+x)^2}=$$
$$=\frac{3}{4}+\left(\frac{1}{2}-\frac{1}{1+x}\right)^2\geq\frac{3}{4}.$$
Also, $I(a,b)\leq1$ it's $$a^2(1+b)+b^2(1+a)+a+b\leq(1+a)(1+b)(a+b)$$ or $$ab\geq0$$ and we are done!
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