Two step question

The powers of the terms are all even, therefore they are all perfect squares. Thus, we can apply the Cauchy-Schwarz Inequality to it. Furthermore, $\frac{x^n}{y^m}$ = $\frac{x^{11n}}{y^{11m}}$ = $\frac{x^{111n}}{y^{111m}}$ , thereby satisfying the condition for equality to hold for the inequality.

Thus, the expression is equal to: $(x^ny^m+x^{11n}y^{11m}+x^{111n}y^{111m})^2$ = $(x^ny^m+(x^{n}y^{m})^{11}+(x^{n}y^{m})^{111})^2$ .

Let o be an odd positive integer and p a positive integer. Notice that $p^o \equiv p\mod 3$ for all p (work through the cases, there's only 3). Thus $x^ny^m \equiv (x^{n}y^{m})^{11} \equiv (x^{n}y^{m})^{111} \ mod 3$ .

Therefore, $x^ny^m+(x^{n}y^{m})^{11}+(x^{n}y^{m})^{111} \equiv0\mod 3$ , which implies $(x^ny^m+(x^{n}y^{m})^{11}+(x^{n}y^{m})^{111})^2 \equiv0\mod 9$ . Thus, 9 is a factor of all expressions of this form.

The task left is to prove that it is indeed the highest common factor. It is sufficient to provide a case that has 9 as its largest factor. This case is when $x=y=1$ and the expression equals 9.

Thus, 9 is indeed the highest common factor