Mongolian Mathematical Olympiad

Let us rewrite the inequality:

$$m^2+2amn+a{n}^2\ge m+an.$$

Bring all terms to one side:

$$m^2+2amn+a{n}^2-m-an\ge 0.$$

Group terms:

$$m^2-m+2amn+a{n}^2-an=(m^2-m)+(2amn+a{n}^2-an).$$

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:

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

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

First, check $a=0$:

$$m^2-m\ge 0⟹m(m-1)\ge 0.$$

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, $(m+n{)}^2-(m+n)\ge 0⟹(m+n)(m+n-1)\ge 0$.

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 $m=0$:

$$0^2+2a\cdot 0\cdot n+a{n}^2\ge 0+an$$ $$a{n}^2\ge an⟹an(n-1)\ge 0.$$

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 $m=1$:

$$1+2an+a{n}^2\ge 1+an$$ $$2an+a{n}^2\ge an$$ $$an+a{n}^2\ge 0⟹an(n+1)\ge 0.$$

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:

$$m^2+2amn+a{n}^2-m-an=m^2-m+2amn+a{n}^2-an$$ $$=m^2-m+a{n}^2+2amn-an$$ $$=m^2-m+a(n^2+2mn-n)$$ $$=m^2-m+a((n+m{)}^2-m^2-n)$$

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

$$m^2+2anm+(a{n}^2-m-an)$$ But this is messy.

Alternatively, consider the difference:

Let $f(m,n)=m^2+2amn+a{n}^2-m-an$

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

Try $m=0$:

$0^2+0+a{n}^2\ge 0+an⟹a{n}^2\ge an⟹n^2\ge n$ for $a>0$.

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

Try $m=1$:

$1+2an+a{n}^2\ge 1+an⟹2an+a{n}^2\ge an⟹an+a{n}^2\ge 0⟹n(n+1)\ge 0$ for $a>0$.

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

Try $n=0$:

$m^2\ge m⟹m(m-1)\ge 0$.

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

Try $n=1$:

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

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

For $a=1$, $m^2+m\ge 0⟹m(m+1)\ge 0⟹m\ge 0$ 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 $m+n=1$.