As I said in the title, the problem states:
Solve the following inequation in $\mathbb{N}^2$: $$ 2\sqrt{u} + 4\sqrt{v - u} - 7 \geq v $$
Source: Mathematical challenges for $8^{th}$ grade.
Approach: By very long calculations, replacing $v - 7 = w$, I obtained an inequation like: $$400u^2 + [40(w^2 - 16w + 112) - 16w^2]u + (w^2 - 16w + 112)^2 \leq 0$$
So, the function admits a maximum, therefore the $a$ coefficient ($f(x) = ax^2 + bx + x$) is negative. But, $400u^2 \geq 0$, so the inequation is, in fact, an equation!
Back into the first inequation, we can replace $\geq$ by $=$.
Now, we can begin solving like a Diophantine Equation: $$2\sqrt{u} + 4\sqrt{v - u} - 4 = v$$
Because $v \in \mathbb{N}$, $u$ is a perfect square and $v - u$ is also a perfect square.
Noting $u = \alpha^2$ and $v = \beta^2$. The equation looks like: $$2\alpha + 4\beta - 7 \geq \alpha^2 + \beta^2$$
Equivalent to: $$(\alpha - 1)^2 + (\beta - 2)^2 + 2 = 0$$
Impossible in $\mathbb{N}$.
Question: My solution finds that no $(u, v)$ natural verify the inequation. Is there any mistake in my solution? Could you find a better approach?
You have a little problem with swapping signs.
If $a= \sqrt{u} $ and $b=\sqrt{v-u}$ then you have to solve $$2a+4b-7\geq a^2+b^2$$ so $$-2 \geq (a-1)^2+(b-2)^2$$ and thus no solution.