I need to prove (P). For prove that, the hint is to use (D). My question is, how to use (D) to prove (P)?
Let $T$ a triangle or tetrahedron and $\mathbb{P}_k(T)$ the set of polynomials of degree less or equal to $k\geq 2$.
Find the constant $C>0$ such that
$\left(\displaystyle\int_T(\Delta v)^2dx\right)^{1/2}\leq C\,\left(\displaystyle\int_T\nabla v\cdot\nabla v\,dx\right)^{1/2}\quad\textrm{ for all }v\in \mathbb{P}_T(T)\tag{P}$
The hint is to find the largest eigenvalue $\lambda$ of the following generalized eigenvalue problem: find $u\in\mathbb{P}_k(T)\setminus\mathbb{R}$ and $\lambda$ such that
$\displaystyle\int_T\Delta u\,\Delta v\,dx=\lambda\,\int_T\nabla u\cdot\nabla v\,dx\quad\textrm{ for all }v\in\mathbb{P}_k(T)\setminus\mathbb{R} \tag{D}$
I suppose that $C=\lambda>0$, but I can't see it. Obviously, $C$ could depends on $T$ and $k$.
You have to require $k\ge2$. For $k=1$ the left-hand side is identically zero, and all eigenvalues are zero as well.
If $k\ge2$ then (D) is eqivalent to a finite-dimensional eigenvalue problem of the type $$ Ax = \lambda Bx $$ with $A,B$ symmetric, positive semidefinite, $A\ne0$, $B$ positive definite. Multiplying from the left by $B^{-1/2}$ and substituting $x=B^{-1/2}y$, it is symmetric eigenvalue problem $$ B^{-1/2}AB^{-1/2}y = \lambda y. $$ All eigenvalues are $\ge0$. If all eigenvalues are zero, then $A=0$, which is a contradiction.
Then it is easy to see that (D) implies (P): Choose a orthonormal basis of eigenvectors $(y_i)$ of eigenvectors of $B^{-1/2}AB^{-1/2}$ to eigenvalues $(\lambda_i)$, then $x_i:=B^{-1/2}y_i$ are orthonormal with respect to the scalar product induced by $B$. Let $x$ be arbitrary. It can be written as $x=\sum_{i=1}^m x^TBx_i \cdot x_i$. Now $$ x^TAx=\sum_{i=1}^m x^TBx_i \cdot x^TAx_i=\sum_{i=1}^m \lambda_i (x^TBx_i)^2 \le (\max_i \lambda_i )\sum_{i=1}^m \lambda_i (x^TBx_i)^2 = (\max_i \lambda_i ) x^TBx. $$ Redoing the translation into a finite-dimensional problem yields (P) with $C=\sqrt{\max_i \lambda_i }>0$.
This is variant of the proof that the norm of a symmetric positive definite matrix is bounded by (in fact, equal to) the largest eigenvalue. Here, the proof is using a non-standard inner product.