Is $\lambda(n) + \max\limits_{p\mid n} v_p(n)\leqslant n$?

Question

Given an integer $n = \prod\limits_{n\mid p}p^{v_p(n)}$, is $$\lambda(n) + \max\limits_{p\mid n} v_p(n)\leqslant n$$ where $\lambda(n)$ is the Carmichael function?

Answer 1

For a single-prime factor, i.e. $n=p^v$, this is true, since $$\begin{align} \lambda(p^v) &\leqslant \phi(p^v) = p^v-p^{v-1} \stackrel{v\geqslant1}\leqslant p^v - v \end{align}$$ (since $p^{v-1}-v$ is strictly increasing in $v$) and therefore $$\lambda(p^v)+v\leqslant\phi(p^v)+v\leqslant p^v.$$ Note how for $v\geqslant\begin{cases}3 & p=2 \\ 2 & p > 2\end{cases}$ this is even $\phi(p^v)+v < p^v$.

For $n=m\cdot p^v$ where $m\geqslant2$ and $p$ are coprime, observe that $$\begin{align} \lambda(n) &= \lambda(m\cdot p^v) \\ &\leqslant \phi(m\cdot p^v) = \overbrace{\phi(m)}^{< m}\overbrace{\phi(p^v)}^{\leqslant p^v - v} \\ &< m(p^v-v) = (mp^v) - \underbrace{\overbrace{m}^{\geqslant v_\max(m)}v}_{\geqslant vv_\max(m)\geqslant v_\max(n)} \\ &\leqslant n - v_\max(n) \end{align}$$ where $v_\max(m) := \max\limits_{p\mid m}v_p(m)$.

So in summary the even stronger relation $$\phi(n) + v_\max(n) < n \quad n\in\mathbb N\backslash(\mathbb P\cup\{4\})$$ holds for all composite $n\neq4$.