Let $$p(X)=\sum_{r}a_{r_1r_2...r_n}{x_1}^{r_1}{x_2}^{r_2}...{x_n}^{r_n}$$ be a homogeneous polynomial of degree $r$ and suppose that $p(X)>0$ for all non-zero $X\in\mathbb{R}^n$.
Then $p$ attains a non-zero minimum on the unit n-sphere.
I have not gotten very far nor do I have a clear strategy in mind but this is my attempt so far:
Let $S=\{X\hspace{1mm}|\hspace{1mm}|X|=1\}$ and $T=\{p(X)\hspace{1mm}|\hspace{1mm}X\in S\}$.
It is quite easy to show that $T$ is bounded and it is also non-empty, thus inf$(T)=\alpha$ exists. Here I though the right way to go was to assume that $\alpha=0$ and show that it leads to a contradiction and then use the fact that $\alpha$ is the infimum of $T$ and non-zero to get to the desired conclusion. However, I can only get as far as showing that if $\varepsilon$ is given there is some $X_0\in S$ such that $\alpha<P(X_0)<\alpha+\varepsilon$, and from here I cant see a way to a contradiction if I assume $\alpha=0$.
If anyone can lead me in the right direction or give me a solution, that would be great.
The unit $n$-sphere $S^n$ is compact, therefore every continuous real-valued function on $S^n$ has a minimum and a maximum. Therefore the value $\alpha=\inf T$ is actually a minimum, meaning there is $x_0\in S^n$ such that $\alpha=p(x_0)>0$.
Your proof is also on the right track. Continuing from $\alpha<p(x_0)<\alpha+\varepsilon$ and assuming $\alpha=0$, then there is, for each $n$, a point $x_n$ such that $p(x_n)<\frac{1}{n}$. Again, using compactness of $S^n$, there is a point $x_\infty$ such that for each $\delta>0,N\in\mathbb N$, there is $m>N$ such that $|x_\infty-x_m|<\delta$. From here use that $p(x)$ is a continuous function and try to show that $p(x_\infty)=\alpha=0$.