Show that a positive definite ternary quadratic form must be anisotropic over $\mathbb{Q}_p$ for some prime $p$.
This is my answer. Please give any advise.
Let $V=\langle a_1,a_2,a_3\rangle$ with $a_i>0$. We know that $V_p$ is isotropic if and only if $S_p(V)=(-1,-1)_p$, where $S_p(V)$ is the Hasse Symbol of $V_p$. For $2<p<\infty$ we have:
$$(-1,-1)_p=1$$
$$(-1,-1)_2=-1$$
$$(-1,-1)_{\infty}=-1$$
By Hilbert reciprocity law we have that
$${\Large\Pi}_{p\leq \infty}(-1,-1)_p=1.$$
But $S_{\infty}(V)=(a_1,a_1)_{\infty}(a_2,a_1a_2)_{\infty}(a_3,a_1a_2a_3)_{\infty}=1\neq (-1,-1)_{\infty}$, hence must be a prime $p>2$ such that, $S_p(V)=-(-1,-1)_p$ and so is anisotropic.