The following problem comes from the 43rd Polish Mathematical Olympiad, linked here: https://imomath.com/othercomp/Pol/PolMO392.pdf
Show that the inequality $$\sum_{n=1}^{r} \sum_{m=1}^{r} \frac{a_ma_n}{m+n} \geq 0$$ holds for any real numbers $a_1,a_2,...,a_r$. Additionally, find the conditions for which there's equality.
I was able to prove this result* using some tools of Calculus (some fairly basic ones, but from Calculus nevertheless). On the other hand, I've been struggling to produce a proof that is Calculus-free.
The reason I even seek such a proof under these restrictions to begin with is because I encountered this problem within the context of a book which doesn't presume the reader's knowledge of Calculus. Granted, the book does occasionally state some theorems whose proofs may rely on Calculus, like Jensen's Inequality, or the Power Mean Inequality, which follows from it.
So, I'd prefer to see proofs which either avoid the use of Calculus altogether, or otherwise use it implicitly (e.g. by using a result which is possibly derived from it).
Nevertheless, if anyone has solutions which explicitly use Calculus, but which are enlightening in some way or are otherwise different from any listed solution, I'd be happy to see that too.
*: I want to verify that my proof is valid, so I'll detail it here: Let $A(x):= a_1x + a_2x^2 + ... + a_rx^r$. Then $A(x)^2 = (\sum_{n=1}^{r}a_nx^n)(\sum_{m=1}^{r}a_mx^m)=\sum_{n=1}^{r} \sum_{m=1}^{r}a_ma_nx^{m+n}$.
Thus, $$\frac{A(x)^2}{x} = \sum_{n=1}^{r} \sum_{m=1}^{r} a_n a_m x^{m+n-1}$$ for which it's not hard to see that $D(x):= \sum_{n=1}^{r} \sum_{m=1}^{r} \frac{1}{m+n}a_n a_m x^{m+n}$ is an anti-derivative. Note that $D(1)$ is exactly the expression which we wish to show is nonnegative, and that $D(0) = 0$.
Hence, by the Fundamental Theorem of Calculus $$D(1) = D(1) - D(0) = \int_{0}^{1} \frac{A(x)^2}{x}dx$$
Since the integrand $\frac{A(x)^2}{x}$ is a nonnegative function over the interval $[0,1]$, it follows that the resulting definite integral is nonnegative. Thus, $D(1) = \sum_{n=1}^{r} \sum_{m=1}^{r} \frac{a_ma_n}{m+n} \geq 0$, as needed.
