Let $H$ be a symmetric positive-definite and symmetric matrix of order $n$ with real entry. Then there exists a real positive constant $a$ such that: $$\forall x \in \Bbb R^n, x^THx \geq a||x||^2$$
Proof: Let $x \in \Bbb R^n$ be an arbitrary non zero vector and diagonalize $H$ as $H = O^TDO$ where $O$ is orthogonal and $D$ is a diagonal matrix with positive entries. Then: $$(x|Hx) \geq a(x|x)$$ $$(x|(H-aI)x) \geq 0$$ $$(x|(O^TDO-aI)x) \geq 0$$ $$(Ox|(DO-aO)x) \geq 0$$ $$(Ox|(D-aI)Ox) \geq 0$$ and since $Ox$ can not be the zero vector, if we choose $a$ such that each diagonal element of $D$ is strictly greater than $a$, then we may conclude that the inequality hold.
I would like to know if such a claim and the proposed proof are correct.
As always, any comment or answer is welcome and let me know if I can explain myself clearer!
The proof is basically correct. I would like to give a second proof that might be even easier (you can decide about that...).
Since we know $H$ is diagonalizable by an orthogonal matrix we have an orthonormal basis $(v_1, \dots, v_n)$ consisting of eigenvectors for $H$ w.r.t. eigenvalues $\lambda_i>0$. Let $\tilde{\lambda}:= \min_{i=1, \dots, n} \lambda_i >0$. Then for any $x$ we have a representation $x= \sum_{i=1}^n a_iv_i$ which gives $$x^THx =(x, Hx)= \left ( \sum_{i=1}^n a_iv_i, \sum_{j=1}^n a_j\lambda_j v_j \right )= \sum_{i,j=1}^n a_ia_j \lambda_j (v_i, v_j) = \sum_{j=1}^na_j^2 \lambda_j \ge \tilde{\lambda} (x,x)$$ where we used the fact that $(v_i, v_j) =0$ for $i \ne j$ and $(v_i, v_i ) =1$ (properties of an orthonormal basis). This gives us the chance to explicitely name the constant (that you called $a$ in the statement) with which the result holds, namely $\tilde{\lambda}$.