Let $a : \mathbb{R}^n * \mathbb{R}^n \rightarrow \mathbb{R}$ be a bilinear form: $$a(x,y) = \sum_{i,j=1}^n a_{ij}x_{i}y_{j} \ \ \ \ \ \ \ a_{ij} \in \mathbb{R},\ \ x = \begin{pmatrix}x_1\\.\\.\\.\\x_n\\\end{pmatrix} ,\ \ y = \begin{pmatrix}y_1\\.\\.\\.\\y_n\\\end{pmatrix}$$
Show that if "a" is a positive definite bilinear form $( \ a(x,x) > 0 \ \ \ \forall x \in \mathbb{R}^{n } \backslash\{0\} \ ) $ then there exists $\alpha > 0$ so that $a(x,x) \ge \alpha||x||_{2}^2$
Let $$\alpha = \min_{\|x\|_2=1}a(x,x).$$ Such $\alpha$ exists because $a$ is continuous and unit sphere is compact. From positive definitness of $a$ we have $\alpha\geq 0$. Case $\alpha=0$ is impossible. Otherwise, there would exist $x_0$, $\|x_0\|_2=1$, such that $a(x_0,x_0)=\alpha=0$, contradicting positive definitness of $a$. Therfore, $\alpha>0$.
Now, for every $x\neq0$ we have $a\left(\frac{x}{\|x\|_2},\frac{x}{\|x\|_2}\right)\geq\alpha$. Multiplying this inequality by $\|x\|_2^2$ and using bilinearity of $a$ we obtain $$a(x,x)\geq \alpha\|x\|_2^2.$$ Inequality is trivially true for $x=0$.