Consider $f(x) = ax +\frac{1}{x+1}$. Let $ $L = the max of $ S$ = the min of $f(x)$ for $x ∈[0, 1]$. Show that $L-S >\frac{1}{12}$ for any $a > 0$. Consider the function $f(x) = ax +\frac{1}{x+1}$, where $a$ is a positive constant. Let $L$ = the largest value of $f(x)$ and $S$ = the smallest value of $f(x)$ for $x ∈[0, 1]$. Show that $L-S >\frac{1}{12}$ for any $a > 0$. 
 A: $f'(x)= a - \dfrac{1}{(x+1)^2}= 0 \iff x = \dfrac{1}{\sqrt{a}} - 1$. So if $a = 1$, then $f'(x) = 0 \iff x = 0$, Thus $L = f(1) = \dfrac{3}{2}, S = f(0) = 1 \implies L-S = \dfrac{3}{2} > \dfrac{1}{2}$. If $a > 1 \implies L = f(1) = a+\dfrac{1}{2}, S = f(0) = 1\implies L-S = a-\dfrac{1}{2} > \dfrac{1}{2} > \dfrac{1}{12}$. If $a < 1$. We have two cases: $a) 0 < a < \dfrac{1}{4} \implies \dfrac{1}{\sqrt{a}} - 1 > 1\implies S = f(1) = a+ \dfrac{1}{2}, L = f(0) = 1\implies L-S = 1- (a - \dfrac{1}{2}) > \dfrac{3}{2} - a > \dfrac{3}{2} - \dfrac{1}{4} = \dfrac{5}{4} > \dfrac{1}{12}$.
$b) \dfrac{1}{4} \le a < 1$. Thus $0 < x_c = \dfrac{1}{\sqrt{a}} - 1 \le 1$. For if $a = \dfrac{1}{4} \implies x_c = 1\implies L = f(0) = 1, S = f(1) = a+\dfrac{1}{2} = \dfrac{1}{4} + \dfrac{1}{2} = \dfrac{3}{4}\implies L - S = 1-\dfrac{3}{4} = \dfrac{1}{4} > \dfrac{1}{12}$. Thus for the final case that $\dfrac{1}{4} < a < 1 $. We have: $f(0) = 1, f(1) = a+\dfrac{1}{2}, f\left(\dfrac{1}{\sqrt{a}}-1\right)= a\left(\dfrac{1}{\sqrt{a}} - 1\right) + \sqrt{a}=2\sqrt{a} - a$. We have: $2\sqrt{a}-a -\left(a+\dfrac{1}{2}\right)= 2\sqrt{a} - 2a-\dfrac{1}{2}= -\dfrac{\left(2\sqrt{a}-1\right)^2}{2} < 0$, and $2\sqrt{a} - a - 1 = -\left(\sqrt{a}-1\right)^2 < 0$. This implies that $L = 1, S = 2\sqrt{a} - a$ if $\dfrac{1}{4} < a \le \dfrac{1}{2}\implies L-S = \left(1-\sqrt{a}\right)^2 \ge \left(1-\dfrac{1}{\sqrt{2}}\right)^2= \dfrac{3}{2} - \sqrt{2} > \dfrac{1}{12}$ as you can check it yourself. Lastly, $L = a+\dfrac{1}{2}, S = 2\sqrt{a} - a$ if $\dfrac{1}{2} \le a < 1$. For this case, $L - S = 2a - 2\sqrt{a} + \dfrac{1}{2}= \dfrac{(2\sqrt{a}-1)^2}{2} \ge \dfrac{(\sqrt{2}-1)^2}{2} > \dfrac{1}{12}$ ( verify thisfact ! ). Done.
