How prove this $(x,y)H\binom{x}{y}\ge p(x^2+y^2)$ let $H_{2\times 2}$ is positive definite matrix,
show that:

There exist  $p>0$, such
  $$(x,y)H\binom{x}{y}\ge p(x^2+y^2)$$

My try: let 
$$H=\begin{bmatrix}
a_{1}&a_{2}\\
b_{1}&b_{2}
\end{bmatrix}$$
and such $$|H|=a_{1}b_{2}-a_{2}b_{1}>0$$
so
$$(x,y)H\binom{x}{y}=(a_{1}x+b_{1}y,a_{2}x+b_{2}y)\binom{x}{y}=a_{1}x^2+b_{2}y^2+(a_{2}+b_{1})xy$$
then How find this $p$,such
$$a_{1}x^2+b_{2}y^2+(a_{2}+b_{1})xy\ge p(x^2+y^2)$$
so I let $t=\dfrac{x}{y}$,then
$$p\le\dfrac{a_{1}t^2+b_{2}+(a_{2}+b_{1})t}{t^2+1}=y$$
then we have
$$(a_{1}-y)t^2+(a_{2}+b_{1})t+b_{2}-y=0$$
so
$$\Delta=(a_{2}+b_{1})^2-4(a_{1}-y)(b_{2}-y)\ge 0$$
$$\Longrightarrow \dfrac{a_{1}+b_{2}}{2}-\sqrt{\dfrac{(a_{2}+b_{1})^2}{4}+\dfrac{(a_{1}-b_{2})^2}{4}}\le y\le  \dfrac{a_{1}+b_{2}}{2}+\sqrt{\dfrac{(a_{2}+b_{1})^2}{4}+\dfrac{(a_{1}-b_{2})^2}{4}}$$
so,There are exist $p \in[I_{1},I_{2}]$
where $$I_{1}= \dfrac{a_{1}+b_{2}}{2}-\sqrt{\dfrac{(a_{2}+b_{1})^2}{4}+\dfrac{(a_{1}-b_{2})^2}{4}},I_{2}= \dfrac{a_{1}+b_{2}}{2}+\sqrt{\dfrac{(a_{2}+b_{1})^2}{4}+\dfrac{(a_{1}-b_{2})^2}{4}}$$
 A: Let $v=[x,y]^T$ and $H$ has the eigen decomposition $H=P^TDP$ where $D$ is the $2\times 2$ diagonal matrix with eigenvalues $\lambda_1,\lambda_2$ as its diagonal entries. Let $\lambda_1\geq \lambda_2$. Define $y=Pv=[y_1,y_2]^T$. Then 
\begin{align}
v^THv=v^TP^TDPv=y^TDy=\lambda_1y_1^2+\lambda_2y_2^2\geq \lambda_2(y_1^2+y_2^2)=\lambda_2(x^2+y^2)
\end{align}
Convince yourself that $y_1^2+y_2^2=x^2+y^2$. Also check if you derive something in the same lines for any $N\times N$ positive definite matrix. 
A: The hypothesis that $H$ is positive definite means that for all nonzero $v= (x, y)^T  \in \Bbb R^2$ we have $\langle v, Hv \rangle > 0$.  In particular, $\langle v, Hv \rangle > 0$ for all $v$ satisfying $\Vert v \Vert = \sqrt{x^2 + y^2} = 1$, i.e. the function $\langle v, Hv \rangle > 0$ is positive on the unit circle in $\Bbb R^2$.  Since this circle is compact and the function $v \to \langle v, Hv \rangle > 0$ is continuous, $v \to \langle v, Hv \rangle > 0$ actually attains its minimum value (on the circle).  Let $p$ denote this value.  Clearly, $p > 0$ or else the unit vector $v_0$ such that $\langle v_0, Hv_0 \rangle = p$ would satisfy $\langle v_0, Hv_0 \rangle = 0$,
contradicting the positive definiteness of $H$.  Thus for every unit vector $u$, $\langle u, Hu \rangle \ge p$.  If $v$ is any nonzero vector, set $u = v / \Vert v \Vert$.  Then $\langle v / \Vert v \Vert, H v / \Vert v \Vert \rangle \ge p$.  But this implies $\langle v, Hv \rangle \ge p \Vert v \Vert^2$, as was required.  QED
Hope this helps.  Cheerio,
and as always,
Fiat Lux!!!
