Mongolian Mathematical Olympiad

Let us rewrite the inequality:

$$m^2 + 2amn + an^2 \ge m + an.$$

Bring all terms to one side:

$$m^2 + 2amn + an^2 - m - an \ge 0.$$

Group terms:

$$m^2 - m + 2amn + an^2 - an = (m^2 - m) + (2amn + an^2 - an).$$

Factor $a$ in the last three terms:

$$(m^2 - m) + a(2mn + n^2 - n).$$

Note that $2mn + n^2 - n = n(2m + n - 1)$, so:

$$(m^2 - m) + a n (2m + n - 1).$$

Now, since $0 \le a \le 1$, the minimum occurs at $a = 0$ or $a = 1$.

First, check $a = 0$:

$$m^2 - m \ge 0 \implies m(m-1) \ge 0.$$

This is true for $m \le 0$ or $m \ge 1$.

Now, check $a = 1$:

$$m^2 - m + n(2m + n - 1) = m^2 - m + 2mn + n^2 - n.$$

Group terms:

$$m^2 + 2mn + n^2 - m - n = (m + n)^2 - (m + n).$$

So, $(m + n)^2 - (m + n) \ge 0 \implies (m + n)(m + n - 1) \ge 0$.

This is true for $m + n \le 0$ or $m + n \ge 1$.

Now, for $0 < a < 1$, the expression is a convex combination of the two cases above, so the minimum is achieved at the endpoints $a = 0$ or $a = 1$.

Therefore, the inequality holds for all $m, n$ if and only if $m(m-1) \ge 0$ and $(m + n)(m + n - 1) \ge 0$.

But the problem asks for all $m, n$ (no restriction), so let's check the minimum value.

Consider $m = 0$:

$$0^2 + 2a \cdot 0 \cdot n + a n^2 \ge 0 + a n$$
$$a n^2 \ge a n \implies a n(n-1) \ge 0.$$

Since $a \ge 0$, this is true for $n \le 0$ or $n \ge 1$.

But for $a = 0$, the inequality is $0 \ge 0$, always true.

For $a > 0$, $n(n-1) \ge 0$.

Similarly, for $m = 1$:

$$1 + 2a n + a n^2 \ge 1 + a n$$
$$2a n + a n^2 \ge a n$$
$$a n + a n^2 \ge 0 \implies a n(n+1) \ge 0.$$

Since $a \ge 0$, this is true for $n \ge 0$ or $n \le -1$.

But for $a = 0$, always true.

Therefore, the inequality holds for all $m, n$ if $a = 0$ (trivially), and for $a > 0$, for all $m, n$ if and only if $m(m-1) \ge 0$ and $(m + n)(m + n - 1) \ge 0$.

But the problem says $0 \le a \le 1$ and asks for all $m, n$.

But for $a = 0$, the inequality is $m^2 \ge m$, i.e., $m(m-1) \ge 0$.

For $a = 1$, the inequality is $(m + n)^2 \ge m + n$, i.e., $(m + n)(m + n - 1) \ge 0$.

So, for $0 < a < 1$, the minimum is at $a = 0$ or $a = 1$.

Therefore, the inequality holds for all $m, n$ if and only if $m(m-1) \ge 0$ and $(m + n)(m + n - 1) \ge 0$ for all $m, n$, which is only possible if $m$ is an integer and $m \le 0$ or $m \ge 1$, and $m + n \le 0$ or $m + n \ge 1$.

But the problem says "for all integers $m$ and $n$". Let's check the original expression:

Let us complete the square:

$$m^2 + 2amn + a n^2 - m - a n = m^2 - m + 2a m n + a n^2 - a n$$
$$= m^2 - m + a n^2 + 2a m n - a n$$
$$= m^2 - m + a (n^2 + 2 m n - n)$$
$$= m^2 - m + a ((n + m)^2 - m^2 - n)$$

But perhaps a better approach is to consider the expression as a quadratic in $m$:

$$m^2 + 2a n m + (a n^2 - m - a n)$$
But this is messy.

Alternatively, consider the difference:

Let $f(m, n) = m^2 + 2a m n + a n^2 - m - a n$

Set $m = k$, $n$ arbitrary.

Try $m = 0$:

$0^2 + 0 + a n^2 \ge 0 + a n \implies a n^2 \ge a n \implies n^2 \ge n$ for $a > 0$.

This is true for $n \le 0$ or $n \ge 1$.

Try $m = 1$:

$1 + 2a n + a n^2 \ge 1 + a n \implies 2a n + a n^2 \ge a n \implies a n + a n^2 \ge 0 \implies n(n+1) \ge 0$ for $a > 0$.

This is true for $n \ge 0$ or $n \le -1$.

Try $n = 0$:

$m^2 \ge m \implies m(m-1) \ge 0$.

So, for $m \le 0$ or $m \ge 1$.

Try $n = 1$:

$m^2 + 2a m + a \ge m + a \implies m^2 + 2a m \ge m \implies m^2 + (2a - 1)m \ge 0$

For $a = 0$, $m^2 - m \ge 0 \implies m \le 0$ or $m \ge 1$.

For $a = 1$, $m^2 + m \ge 0 \implies m(m+1) \ge 0 \implies m \ge 0$ or $m \le -1$.

So, in all cases, the inequality holds for $m \le 0$ or $m \ge 1$, $n \le 0$ or $n \ge 1$, and $m + n \le 0$ or $m + n \ge 1$.

Equality holds when $a = 0$ and $m = 0$ or $m = 1$, or $a = 1$ and $m + n = 0$ or $m + n = 1$.

In summary:

- The inequality holds for all integers $m, n$ and $0 \le a \le 1$.
- Equality holds if and only if either $a = 0$ and $m = 0$ or $m = 1$, or $a = 1$ and $m + n = 0$ or $m + n = 1$.