Let $v$ be some function in $C^1_c(\Bbb{R})$.
Assume we already has the inequality $$|v(x)|^p\le p\|v\|_p^{p-1}\|v'\|_p \tag{1}$$
where $v'$ is the weak derivative of $v$.
Now we want to deduce the Sobolev inequality $$\|v\|_{\infty}\le C\|v\|_{W^{1,p}} \tag{2}$$
where $C$ independent of $p$ and $v$.
To do this ,it's sufficient to prove that $$p\|v\|_p^{p-1}\|v'\|_p \le C^p(\|v\|_p^p+\|v'\|_p^p) \tag{*}$$
Let's write it in the symble that is needs to find some $C$ with :
$$C^p(a^p+b^p) - pa^{p-1}b \ge 0$$ for $a,b\ge 0$.
Just taking derivative we find $C = (p-1)^{\frac{p-1}{p^2}}$ is the constant which makes the derivative for $F(a,b)$ zero,and subsititude the optimal condition that $(p-1)b = a C^p$ into it we find that minimal value is exactly zero.
Such that $C \sim (p-1)^{1/(p-1)}\le e^{1/e}$ is the constant independent of $p$.
I think my proof is almost correct.Is there some better proof from $(1)$ to $(2)$ without the heavy computation?
Based on the solution below,I make the proof precise as follows:
$$p^{1/p}\|v\|_p^{\frac{p-1}{p}}\|v'\|_p^{1/p} \le p^{1/p}(\frac{p-1}{p}\|v\|_p + \frac{1}{p}\|v'\|_p) \le C(\|v\|_p +\|v'\|_p)\sim \|v\|_{W^{1,p}}$$
(Where the last estimate since in finite dimension $\ell^p$ and $\ell^1$ are equivalent with the constant only depend on $k$ of $W^{k,p}$.)
There is an easier proof (or at least it’s just one application of a standard tool).
You have $a\le b^\theta c^{1-\theta}$ for some $\theta\in[0,1]$. Just apply Young’s inequality $$ BC \le \frac{B^p}p+\frac{C^q}q$$ where $$B=b^\theta, \quad C=c^{1-\theta}, \quad\frac1p+\frac1q=1,\quad p=\frac1\theta$$