Proof of $ d(n)\leq \sqrt {3n}$. Show that
$$ d(n)\leq \sqrt {3n}$$
and  the equality is true if only if $n=12$,
where $d(n)$ is the number of positive divisors of $n$..
Here is a proof Proof of $ d(n)\leq \sqrt {3n}$.
Let $$ n=\prod_{k=1}^m p_k^{\alpha_k},$$
the proof given above considers the case when $\alpha_k\geq 2$,
What I want to know is how to deal with when some $\alpha_k=1$ and some $\alpha_k\geq 2$.
Any help and hint will welcome, or some other method is provided.
Thanks a lot!
 A: Just for fun, here is an alternative proof not invoking Bernoulli's inequality.
After checking the base case $d(1)=1\le\sqrt{3\cdot1}$, let's assume the inequality holds for all integers less than $n$, let $p$ be the largest prime divisor of $n$, and write $n=p^rm$ with $p\not\mid m$, so that $d(n)=(r+1)d(m)$. We want to prove that $(r+1)^2d(m)^2\le3p^rm$.
Now if $p\ge5$, it's easy to prove (by its own induction argument) that $(r+1)^2\le5^r\le p^r$ for all $r\ge0$, in which case the inductive hypothesis $d(m)\le\sqrt{3m}$ (since $m\lt n$) tells us $(r+1)^2d(m)^2\le3p^rm$.
On the other hand, if $p\le3$, then $n$ is of the form $2^a3^b$, in which case $d(n)=(a+1)(b+1)$ and we need only show that
$$(a+1)^2(b+1)^2\le2^a3^{b+1}$$
for all $a,b\in\mathbb{N}$.  This, again, comes down to its own fairly simple induction argument. (The inequality $(b+1)^2\le3^{b+1}$ holds for all $b$, and $(a+1)^2\le2^a$ holds for all $a\ge6$, so it's enough to check things for $0\le a\le5$, which is a little tedious but not difficult. It seems there ought to be some slicker way of doing this part of the induction, but I can't think of one. Perhaps someone can suggest a better approach in comments or another answer.)
A: We will use the following $\textbf{three facts}$:
(a) $$\max_{n\in\mathbb{N}^+}\frac{(n+1)^2}{2^n}=\frac94<3,$$
with equality if and only if $n=2$;
(b) Let $p\geq 3$ and $\alpha\geq 2$, then, by Bernoulli's inequality,
$$p^{\frac{\alpha}{2}}=(1+(p-1))^{\frac{\alpha}{2}}
     \ge 1+\frac{p-1}{2}\alpha\ge1+\alpha,$$
with equality if and only if $\alpha=2,p=3$;
(c) Let $$n=\prod_{k=1}^m p_k^{\alpha_k},$$
then $$d(n)=\prod_{k=1}^{m}(1+\alpha_k).$$
The discussion is divided into the following cases:
$(1)$ If $n=p^\alpha$, where $p$ is a prime,
then $d(n)=\alpha+1$.
It is easy to see that
$$d(n)=\sqrt{(\alpha+1)^2}\leq\sqrt{\frac94\cdot2^\alpha}
<\sqrt{3\cdot p^\alpha}=\sqrt{3n}.$$
$(2)$ If $n=p_1p_2\cdots p_m$ with $2\leq p_1<p_2<\cdots<p_m,m\geq 2$,
then $d(n)=2^m$. It is easy to see that
$$4^m<3p_1p_2\cdot4^{m-2}\iff16<3p_1p_2.$$
So
$$d(n)=\sqrt{4^m}<\sqrt{3p_1p_2\cdots p_m}=\sqrt{3n}.$$
$(3)$ If $n=2^2 p_1p_2\cdots p_m$
with $3\leq p_1<p_2<\cdots<p_m$,
then
\begin{align*}
d(n)
&=3\cdot 2^m\\
&=\sqrt{3\cdot 3\cdot 4^m}\\
&\leq\sqrt{3\cdot 2^2p_1p_2\cdots p_m}\\
&=\sqrt{3n}.\ (\mbox{equality is true}\ \iff n=12)
\end{align*}
$(4)$ If $n=2^{\alpha} p_1p_2\cdots p_m$
with $3\leq p_1<p_2<\cdots<p_m,\alpha\geq 3$,
then
\begin{align*}
d(n)
&=(1+\alpha)\cdot 2^m\\
&=\sqrt{(1+\alpha)^2\cdot 4^m}\\
&<\sqrt{\frac94\cdot 2^{\alpha}\cdot4^m}\\
&=\sqrt{3\cdot 2^{\alpha}\cdot 3\cdot4^{m-1}}\\
&\leq\sqrt{3\cdot 2^{\alpha}p_1p_2\cdots p_m}\\
&=\sqrt{3n}.
\end{align*}
$(5)$ If $n=2^{\alpha}p_1^{\alpha_1}p_2^{\alpha_2}\cdots p_m^{\alpha_m}$
with $3\leq p_1<p_2<\cdots<p_m,\alpha\geq 3,\alpha_k\geq 2$,
then
\begin{align*}
d(n)
&=(1+\alpha)\prod_{k=1}^{m}(1+\alpha_k)\\
&<\sqrt{3\cdot 2^{\alpha}}\prod_{k=1}^{m}p_k^{\frac{\alpha_k}2}\\
&=\sqrt{3n}.
\end{align*}
$(6)$ If $n=2^{\alpha}p_1\cdots p_l\cdot p_{l+1}^{\alpha_{l+1}}\cdots p_m^{\alpha_m}$
with $3\leq p_1<\cdots<p_l,3\leq p_{l+1}<\cdots<p_m,\alpha_k\geq 2$,
then
\begin{align*}
d(n)
&=(1+\alpha)\cdot2^l\prod_{k=l+1}^{m}(1+\alpha_k)\\
&\leq\sqrt{\frac94\cdot 2^{\alpha}\cdot 4^l}\prod_{k=1}^{m}p_k^{\frac{\alpha_k}2}\\
&=\left(\frac{3\cdot 4^{l-1}}{p_1\cdots p_l}\right)^{\frac12}
  \cdot\sqrt{3\cdot2^{\alpha}p_1\cdots p_l\cdot
  p_{l+1}^{\alpha_{l+1}}\cdots p_m^{\alpha_m}}\\
&<\sqrt{3n}.
\end{align*}
