Reach for the Summit - M-S1-A4

Given that $f(x) = \sqrt x - \ln (x+a)$ , $\implies f'(x) = \dfrac 1{2\sqrt x} - \dfrac 1{x+a}$ . For $f(x)$ to be monotonic increasing for $x \in (0, \infty)$ , $f'(x) \ge 0$ for $x \in (0, \infty)$ . Then we have

$\begin{aligned} \frac 1{2\sqrt x} - \frac 1{x+a} & \ge 0 \\ \frac 1{2\sqrt x} & \ge \frac 1{x+a} \\ x+a & \ge 2\sqrt x \\ x^2 + 2ax + a^2 & \ge 4x \\ x^2 + (2a-4) x + a^2 & \ge 0 & \small \blue{\text{For the inequality to be true,}} \\ (2a-4)^2 - 4a^2 & \le 0 & \small \blue{\text{the discriminant }\le 0.} \\ -16a + 16 & \le 0 \\ \implies a & \ge \boxed 1 \end{aligned}$

$f(x)=\sqrt x-\ln (x+a)\implies f'(x) =\frac{1}{2\sqrt x}-\frac{1}{x+a}>0$

$\implies a>2\sqrt x-x$

Therefore minimum of $a$ is $\boxed 1$ when $x=1$ .