Suppose we have to solve the below problem. For real $a,b,c,abc \neq0$ find maximum of $\frac{ab+ac+bc}{a^2+b^2+c^2}$
I do like this: $$(a-b)^2+(b-c)^2+(c-a)^2\geq 0 \implies 2(a^2+b^2+c^2)\geq 2(ab+ac+bc)\\\div (a^2+b^2+c^2) \\
1\geq \frac{ab+ac+bc}{a^2+b^2+c^2}\implies \max\{\frac{ab+ac+bc}{a^2+b^2+c^2}\}=1$$
now; my question.
(First) Is there any other (suitable for high school) way to find solution ?
(Second) Is there generality for more variables? for example $$\max\{\frac{ab+ac+ad+bc+bd+cd}{a^2+b^2+c^2+d^2} \}=?$$
(Second) For $n$ non-zero (not all zero) reals $\;a_1, \dots, a_n,\;$ we can proceed analogously:
$$0\leq \sum_{i,j=1}^n (a_i-a_j)^2 = 2(n-1)\sum_{i=1}^n (a_i)^2 - 4\sum_{i,j=1}^n a_ia_j,$$ from where $$\frac{\sum_{i,j=1}^n a_ia_j}{\sum_{i=1}^n (a_i)^2}\leq\frac{n-1}{2}.$$ Consequently is $$\max{\frac{\sum_{i,j=1}^n a_ia_j}{\sum_{i=1}^n (a_i)^2}}\leq \frac{n-1}{2}.$$ The value $\;\frac{n-1}{2}\;$ is obtained if all terms $a_i$ are equal.
If the high school students have difficulties to understand due to the formalism, you can do it for a reasonable $n.$
(First) Mathematical induction is suitable (though not easy) at high school level.