I have not been able to conclude, but I think the following is a good start.
Because the diameter is at least $4$, there exist $x,y$ with $d(x,y)\geq4$. In particular for all $z$ , $d(x,z)+d(y,z)\geq4$ We recall that $$ \lambda=\frac{1}{2}\max_{\sum|u(i)|^{2}=1}\sum_{i\sim j}(u(i)-u(j))^{2} $$ and therefore is monotone increasing in the edges. To consider the maximum of $\lambda$, it is enough to study the maximal graphs $G=(V,E)$ with $d(x,y)=4$. We can then suppose that for all $z$: $d(x,z)+d(y,z)=4$. (Indeed if for example $d(x,z)\geq3$ (resp. $d(y,z)$) let $z'$ such that $d(x,z')=2$, we can add the edges $(z,y)$ and $(z,z')$ to our graph).
Let us call $$ A=\text{\{}z:d(x,z)=1\text{\}} $$
$$ B=\text{\{}z:d(x,z)=2\text{\}} $$
$$ C=\text{\{}z:d(x,z)=3\text{\}} $$ Because of the maximality of $G,$ we have $$ V=A\cup B\cup C\cup\text{\{}x,y\text{\}} $$ $$ E=\text{\{}(x,z):z\in A\text{\}}\cup\text{\{}(z_{1},z_{2}):z_{1},z_{2}\in A\cup B\text{\}}\cup\text{\{}(z_{1},z_{2}):z_{1},z_{2}\in B\cup C\text{\}}\cup\text{\{}(y,z):z\in C\text{\}}. $$
Let us now consider the following orthonormal family : $1_{\text{\{}x\text{\}}},1_{\text{\{}y\text{\}}},u_{A}=\frac{1_{A}}{\sqrt{|A|}},u_{B}=\frac{1_{B}}{\sqrt{|B|}},u_{C}=\frac{1_{C}}{\sqrt{|C|}}$. We have
$$ \begin{cases} L(1_{\text{\{}x\text{\}}})=|A|1_{\text{\{}x\text{\}}}-1_{A}\\ L(1_{A})=-|A|1_{\text{\{}x\text{\}}}-|A|1_{B}+(1+|B|)1_{A}\\ L(1_{B})=-|B|1_{A}-|B|1_{C}+(|A|+|C|)1_{B}\\ L(1_{C})=-|C|1_{\text{\{}y\text{\}}}-|C|1_{B}+(1+|B|)1_{C}\\ L(1_{\text{\{}y\text{\}}})=|C|1_{\text{\{}y\text{\}}}-1_{C}. \end{cases} $$
This gives in the family $(1_{\text{\{}x\text{\}}},u_{A},u_{B},u_{C},1_{\text{\{}y\text{\}}})$ the following matrix
$$ M=\begin{pmatrix}|A| & -\sqrt{|A|} & & 0 & 0\\ -\sqrt{|A|} & |B|+1 & -\sqrt{|A||B|} & & 0\\ & -\sqrt{|A||B|} & |A|+|C| & -\sqrt{|B||C|}\\ 0 & & -\sqrt{|B||C|} & |B|+1 & -\sqrt{|C|}\\ 0 & 0 & & -\sqrt{|C|} & |C| \end{pmatrix} $$ Actually we can write $L$ as the block matrix $$ L=\begin{pmatrix}M & 0\\ 0 & L' \end{pmatrix} $$ and because of min-max principal the larger eigenvalue of $L'$ is smaller than the laplacian on the graph $G-\text{\{}x,y\text{\}}$ which is smaller than $n-1$. So we only have to deal with $M$. The problem can be state as follow :Do we have $\|M\|\leq|A|+|B|+|C|+1$?
Here is where I get stuck. It should be possible to do a complete analysis of $M$ and calculate explicitely the largest eigenvalue but it seems a bit tedious. We can also make some numerical simulations : no counter example appear for $|A|+|B|+|C|\leq 100$