I have only found proof of Lagrange multiplier with one constraint only. I could not foound proof of multiple constraints. So, I generalized the proof. If you find any mistakes in my proof (as if there would be anyone reading my posts XD), please let me know.
One Constraint
Let \(P = (q_1(t_0),...,q_n(t_0),\dot{q}_0(t_0),...,\dot{q}_n(t_0))\) be the point where the Lagragian \(L\) is extremized at the constraint \(f_1\). Let \(\vec{r}(t) = (q_1(t),...,q_n(t),\dot{q}_0(t),...,\dot{q}_n(t))\) be a parametrized function on \(f_1\). Define
$$\nabla L = (\frac{\partial L}{\partial q},...,\frac{\partial L}{\partial \dot{q}})$$
Let \(h(t) = L(q_1(t),...\dot{q}(t))\). Then,
\begin{align}
\frac{d}{dt}h(t) &= \sum_k \frac{\partial L}{\partial q_k}\dot{q}_k + \frac{\partial L}{\partial \dot{q}_k}\frac{d\dot{q}_k}{dt}\\
&= \nabla L \cdot \frac{d\vec{r}}{dt}.
\end{align}
At extremum \(P\),
$$\frac{dh}{dt} = \nabla L \cdot \frac{d\vec{r}}{dt} = 0.$$
As \(\vec{r}\) is an arbitrary parametrized function of \(f_1\), \(L\) is orthogonal to any parametrized function of \(f_1\). Then \(L\) is parallel to \(f_1\)
$$\nabla L = \lambda_1 \nabla f_1,$$
for some constant \(\lambda\).
More Constraints
As we have seen \(L\) is extremized at \(f_1\) when
$$\nabla L = \lambda_1 \nabla f_1,$$
let
$$h(t) = L - \lambda_1 f_1.$$
Then, using argument from above, we know that extremum occurs when
$$\nabla L - \lambda_1\nabla f_1 = \lambda_2 \nabla f_2.$$
Similarly, for \(m\) constraints,
$$\nabla L = \sum^m_{\alpha=1}\lambda_\alpha \nabla f_\alpha.$$
Aftermath
As we know \(\delta S=0\) can be written as
\begin{align}
\int^{t_2}_{t_1}\left(\delta L - \sum^m_{\alpha=1}\lambda_\alpha\delta f_\alpha\right)dt &= 0 \\
\int^{t_2}_{t_1}\sum_k\left( \frac{\partial L}{\partial q_k}\delta q_k + \left(\frac{\partial L}{\partial \dot{q_k}}\right)\delta\dot{q}_k - \sum^m_{\alpha=1}\lambda_\alpha\left(\frac{\partial f_\alpha}{\partial q_k}\delta q_k + \left(\frac{\partial f_\alpha}{\partial \dot{q}_k}\right)\delta\dot{q}_k \right)\right)dt &=0\\
\end{align}
Integrating by part, we have
$$\int\left(\frac{\partial L}{\partial \dot{q}}\right)\frac{d\delta q}{dt}dt = \frac{\partial L}{\partial \dot{q}}\delta q - \int\left(\delta q\frac{d}{dt}\frac{\partial L}{\partial \dot{q}}\right)dt$$
and
$$\int\left(\frac{\partial f_\alpha}{\partial \dot{q}}\right)\frac{d\delta q}{dt}dt = \frac{\partial f_\alpha}{\partial \dot{q}}\delta q - \int\left(\delta q\frac{d}{dt}\frac{\partial f_\alpha}{\partial \dot{q}}\right)dt$$
So,
\begin{align}
\sum_k\left[\delta q_k\frac{\partial L}{\partial \dot{q_k}}\right]^{t_2}_{t_1} + \int^{t_2}_{t_1}\sum_k\delta q_k\left(\frac{\partial L}{\partial q_k}-\frac{d}{dt}\frac{\partial L}{\partial \dot{q}_k}\right) & \\
+\sum^m_{\alpha=1}\left[\lambda_\alpha\left(\frac{\partial f_\alpha}{\partial q_k}-\frac{d}{dt}\frac{\partial f_\alpha}{\partial \dot{q}_k}\right)\delta q_k + \lambda_\alpha\frac{d}{dt}\left(\frac{\partial f_\alpha}{\partial \dot{q}_\alpha}\delta q_k\right)\right]dt &= 0.
\end{align}
As \(\delta q_k(t_2) = \delta q_k(t_1)\),
$$\int^{t_2}_{t_1}\sum_k\delta q_k\left(\frac{\partial L}{\partial q_k}-\frac{d}{dt}\frac{\partial L}{\partial \dot{q}_k}\right) +\sum^m_{\alpha=1}\lambda_\alpha\left[\left(\frac{\partial f_\alpha}{\partial q_k}-\frac{d}{dt}\frac{\partial f_\alpha}{\partial \dot{q}_k}\right)\delta q_k + \frac{d}{dt}\left(\frac{\partial f_\alpha}{\partial \dot{q}_\alpha}\delta q_k\right)\right]dt = 0.$$
Similarly, integrating by part again, we have
$$\lambda_\alpha\frac{d}{dt}\left(\frac{\partial f}{\partial \dot{q}}\delta q\right) = \frac{d}{dt}\left(\lambda_\alpha\frac{\partial f_\alpha}{\partial \dot{q}}\delta q\right) - \frac{d\lambda_\alpha}{dt}\left(\frac{\partial f}{\partial \dot{q}}\delta q\right).$$
Then,
$$\left[\lambda_\alpha\frac{\partial f_\alpha}{\partial \dot{q}_k}\delta q_k\right]^{t_2}_{t_1}+\int^{t_2}_{t_1}\left[\frac{\partial L}{\partial q_k}-\frac{d}{dt}\frac{\partial L}{\partial \dot{q}_k}+\sum^m_{\alpha=1}\lambda_\alpha\left(\frac{\partial f_\alpha}{\partial q_k}-\frac{d}{dt}\frac{\partial f_\alpha}{\partial \dot{q}_k}\right)-\sum^m_{\alpha=1}\left(\frac{d\lambda_\alpha}{dt}\right)\left(\frac{\partial f}{\partial \dot{q}_k}\right)\right]\delta q_kdt = 0.$$
Define
$$Q_k=\sum^m_{\alpha=1}\left[\lambda_{\alpha}\left(\frac{\partial f_{\alpha}}{\partial q_k}-\frac{d}{dt}\frac{\partial f_\alpha}{\partial \dot{q}_k}\right)-\frac{d\lambda_{\alpha}}{dt}\frac{\partial f_{\alpha}}{\partial \dot{q}_k}\right].$$
As \(\delta q_k(t_2) = \delta q_k(t_1)\),
$$\int^{t_2}_{t_1}\sum_k\left[\frac{\partial L}{\partial q_k}-\frac{d}{dt}\frac{\partial L}{\partial \dot{q}_k}-Q_k\right]\delta q_kdt = 0.$$
From above section, we know that
$$\nabla L = \sum^m_{\alpha=1}\lambda_\alpha \nabla f_\alpha.$$
So, this time we know that for each coordinate
$$ \bbox[5px,border:2px solid #666]
{
\frac{\partial L}{\partial q_k}-\frac{d}{dt}\frac{\partial L}{\partial \dot{q}_k}=Q_k
}
$$
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