Monotonicity of a fraction. So I want to prove that the following fraction is monotone increasing, as a part of another proof, that's why I stumbled on:
$$\frac{4^{n+1}}{2\sqrt{n+1}} \ge \frac{4^{n}}{2\sqrt{n}}$$
I know it's basic, though how to prove it?
 A: Both sides are positive, hence we may tyr to show the quotient is $\ge 1$:
$$ \frac{4^{n+1}}{2\sqrt{n+1}}>\frac{4^{n}}{2\sqrt{n}}\iff \frac{\frac{4^{n+1}}{2\sqrt{n+1}}}{\frac{4^{n}}{2\sqrt{n}}}=\frac{4\sqrt n}{\sqrt{n+1}}=\sqrt{\frac{16n}{n+1}}> 1 $$
The latter is certainly true for all $n\ge1$ (becaue $16n>n+1$).
A: You have that
$$
\frac{4^{n+1}}{2\sqrt{n+1}}\geq \frac{4^n}{2\sqrt{n}} \Leftrightarrow \frac{4^{n+1}}{4^n}\geq \frac{\sqrt{n+1}}{\sqrt{n}}\Leftrightarrow 4\geq\sqrt{\frac{n+1}{n}}=\sqrt{1+\frac{1}{n}}.
$$
The term inside the square root is decreasing, while the left-hand side is constant. For $n=1$ obviously $4\geq \sqrt{2}$, so your inequality is true for all $n\in \mathbb{N}$.
A: First, multiply both sides by $2 \times 4^{-n} \sqrt{n+1}$. You are left with the following equivalent statement : $$ 4 \geq \sqrt{1 + \frac{1}{n}} $$
The function on the right hand side is strictly decreasing, and takes value $\sqrt{2}$ for $n=1$, which is lower than $4$. Therefore, the inequality holds for all integer $n \geq 1$.
A: May be, you could consider the function $$f(x)=\frac{4^{x}}{2\sqrt{x}}$$ Taking logarithms $$\log\big(f(x)\big)=x\log(4)-\frac 12\log(x)-\log(2)$$ Using  differentiation  $$\frac{f'(x)}{f(x)}=\log(4)-\frac 1 {2x}$$ which is positive as soon as $x \gt \frac{1}{4\log (2)}$. So, the function is always increasing for $x > 1$.
