Problem 1072

9. 7. Let the weights of the first pile of $n$ coins be represented as $x_{1} \geqslant x_{2} \geqslant \cdots \geqslant x_{n}$, and the weights of the second pile of $m$ coins be represented as $y_{1} \geqslant y_{2} \geqslant \cdots \geqslant y_{m}$.
Also, let $x_{1} \geqslant \cdots \geqslant x_{s} \geqslant x \geqslant x_{s+1} \geqslant \cdots \geqslant x_{n}$,
$$y_{1} \geqslant \cdots \geqslant y_{i} \geqslant x \geqslant y_{t+1} \geqslant \cdots \geqslant y_{m} .$$
(If there are no coins not lighter than $x$, the conclusion is obviously true.) Thus, what needs to be proven is:
$$x s+x_{s+1}+\cdots+x_{n} \geqslant x t+y_{t+1}+\cdots+y_{m} \text {. }$$

Let $x_{1}+x_{2}+\cdots+x_{n}=y_{1}+y_{2}+\cdots+y_{m}=A$,
i.e., to prove $x s+\left[A-\left(x_{1}+\cdots+x_{*}\right)\right]$
$$\geqslant x t+\left[A-\left(y_{1}+\cdots+y_{t}\right)\right],$$

i.e., to prove $x_{1}+\cdots+x_{s}+x(t-s) \leqslant y_{1}+\cdots+y_{t}$.
We will prove this inequality in two cases:
If $t \geqslant s$, then
$$\begin{array}{l} x_{1}+\cdots+x_{s}+x(t-s) \\ =\left(x_{1}+\cdots+x_{s}\right)+(2+\cdots+x) \\ \leqslant\left(y_{1}+\cdots+y_{s}\right)+\left(y_{s+1}+\cdots+y_{t}\right) . \end{array}$$
(Because $x_{1}+\cdots+x_{2} \leqslant y_{1}+\cdots+y_{s}$ can be derived from the given, and $y_{s+1} \geqslant x, \cdots, y_{t} \geqslant x$.)

If $t<s$, then $x_{1}+\cdots+x_{s}+x(t-s) \leqslant y_{1}+\cdots+y_{t}$ is equivalent to
$$x_{1}+\cdots+x_{s} \leqslant y_{1}+\cdots+y_{t}+(\underbrace{x+\cdots+x}_{(t-s) \uparrow}) .$$

This inequality can be derived from the following:
$$\begin{aligned} x_{1}+\cdots+x_{s} & \leqslant y_{1}+\cdots+y_{s} \\ & =\left(y_{1}+\cdots+y_{t}\right)+\left(y_{t+1}+\cdots+y_{s}\right), \end{aligned}$$

and $y_{t+1} \leqslant x, \cdots, y_{s} \leqslant x$.