Square roots and calculus

Plugging in $x=1$ in the inequality $f(x) \leq \frac{\sqrt{x}}{2a}$ shows us that $\sqrt{2} \leq \frac{1}{2a}$ , from which follows that $a \leq \frac{1}{2\sqrt{2}}.$ This choice $x=1$ might be pretty random at first glance, but remember that $\ln(1)=0$ , which makes the inequality easier to deal with.

We will now show that the upperbound, $a = \frac{1}{2\sqrt{2}}$ , will satisfy the conditions given. We will do this using the following theorem: Let $b\in\mathbb{R}$ , and let $f, g: [b, \infty) \to \mathbb{R}$ be differentiable functions on their domain. If $f(b) \leq g(b)$ , $\lim_{x\to\infty} \left(g(x) - f(x)\right) \geq 0$ and there exists $x_0\in [b, \infty)$ , such that $f(x_0)=g(x_0)$ and furthermore $f'(x) \geq g'(x)$ for $x \in [b, x_0]$ and $f'(x) \leq g'(x)$ for $x \in [x_0, \infty)$ , then $f(x) \leq g(x)$ for all $x \in [b, \infty)$ .

We will come to this fact via a proof by contradiction. Suppose that there exists $x_1 \in [b, \infty)$ such that $f(x_1) > g(x_1)$ . Without loss of generality, assume that $x_1 \in [b, x_0]$ . Then, because $f'(x) \geq g'(x)$ for all $x \in [b, x_0]$ , we notice that $f(x_0) > g(x_0)$ , leading to a contradiction, as $f(x_0) = g(x_0)$ . $\blacksquare$

We will be using this theorem with $b = \frac{1}{e^2}$ , $f(x) = \frac{\ln(x)}{2\sqrt{2}} + \sqrt{x+1}$ and $g(x) = \frac{\sqrt{x}}{2\left(\frac{1}{2\sqrt{2}}\right)} = \sqrt{2x}$ , furthermore, we use $x_0 = 1$ . However, we will have to check if the conditions are satisfied.

Note that because $e > 2$ we have that $f(\frac{1}{e^2}) = \sqrt{\frac{1}{e^2}+1} - \frac{\sqrt{2}}{2} < \sqrt{\frac{5}{4}} - \frac{\sqrt{2}}{2} = \frac{\sqrt{5}-\sqrt{2}}{2}$ . Now, notice that $\left(\frac{9}{4}\right)^2 = \frac{81}{16} > 5$ and $1.4^2 < 2$ , so that $\frac{\sqrt{5}-\sqrt{2}}{2} < \frac{9}{8}-0.7 = \frac{9\cdot 12.5 - 70}{100} < \frac{43}{100}$ . Furthermore, notice that because $e < 3$ , $g(\frac{1}{e^2}) = \frac{\sqrt{2}}{e} > \frac{\sqrt{2}}{3} > \frac{1.4}{3} = \frac{1.4\cdot 33}{99} > \frac{46}{100} > \frac{43}{100}$ , so that we can conclude that $f(\frac{1}{e^2}) \leq g(\frac{1}{e^2})$ .

Now we will show that $\lim_{x\to\infty} \left(g(x) - f(x)\right) = \lim_{x\to\infty} \left(\sqrt{2x} - \frac{\ln(x)}{2\sqrt{2}} - \sqrt{x+1}\right) \geq 0$ . Notice that $\lim_{x\to\infty} \left(\sqrt{2x} - \frac{\ln(x)}{2\sqrt{2}} - \sqrt{x+1}\right) = \lim_{x\to\infty} \left((\sqrt{2}-1)\sqrt{x} - \frac{\ln(x)}{2\sqrt{2}}\right)$ . With l'Hôpital's rule we can show that $\lim_{x\to\infty} \frac{\ln(x)}{\sqrt{x}} = 0$ , so that in the limit $\ln(x)$ can be neglected and $\lim_{x\to\infty} \left(g(x) - f(x)\right) \geq 0$ .

Note that $f(1) = g(1)$ per definition of $a$ , so we can skip this. However, we still have to show that $f'(x) \geq g'(x)$ for $x \in [\frac{1}{e^2}, 1]$ and $f'(x) \leq g'(x)$ for $x \in [1, \infty)$ . Note that $f'(x) = \frac{1}{2x\sqrt{2}} + \frac{1}{2\sqrt{x+1}}$ and $g'(x) = \frac{1}{\sqrt{2x}}$ .

Now that we have this out of the way, let us show that for $x \in [\frac{1}{e^2}, 1] : f'(x) \geq g'(x)$ . Because $x \in [\frac{1}{e^2}, 1]$ we have that $x^2 \leq 1, \text{ from which follows that } x^2+2x+1 = (x+1)^2 \leq 2(x+1) \text{ so that } 0 \leq |x+1| = x+1 \leq \sqrt{2(x+1)}.$ If we go further, we notice that $2x(x+1) = 2x^2+2x \leq 2x\sqrt{2(x+1)}, \text{ so that } 2x^2 + 3x \leq 1 + 2x\sqrt{2(x+1)} \leq x + 2x\sqrt{2(x+1)}.$ The last step follows from $1 \leq x$ . By adding $2x^2 +2x$ to both sides, we notice that $4x^2+4x \leq (\sqrt{x+1})^2+2(x\sqrt{2})(\sqrt{x+1})+(x\sqrt{2})^2 = (\sqrt{x+1}+x\sqrt{2})^2$ Taking the square root on both sides gives us $0 \leq 2\sqrt{x(x+1)} - \sqrt{x+1} \leq x\sqrt{2}, \text{ from which follows that } f'(x) \geq g'(x).$ The last step requires a couple of relatively easy steps, which I leave for the reader. Likewise can be shown that for $x \in [1, \infty): f'(x) \leq g'(x)$ , as then $x^2 \geq 1$ . With this, we can use the specified theorem to notice that $a = \frac{1}{2\sqrt{2}}$ satisfies. Remember that $a \leq \frac{1}{2\sqrt{2}}$ , so that $A = \boxed{\frac{1}{2\sqrt{2}}}$ .

We know that aln(x) + \sqrt{x + 1} is less or equal to \frac{\sqrt{x}}{2a} for x greater or equal to e^{-2}, and,in particular, for x = 1. Substituting x = 1 in the inequality we get aln(1) + \sqrt{2} less or equal to \frac{\sqrt{1}}{2a}. Solving for a we get that a is less or equal to \frac{1}{2\sqrt{2}} . So A = \frac{1}{2\sqrt{2}} and the solution is \boxed{3535}