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$x$ and $y$ are reals such that $1 \le y \le 2$ and $2y \le xy + 2$. Calculate the minimum value of $$\large \frac{x^2 + 4}{y^2 + 1}$$

This problem is adapted from a recent competition... It really is. There must be better answers.

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We first need to prove that $x \ge 0$. It is true that $2y \le xy + 2 \iff 0 \le \dfrac{2(y - 1)}{y} \le x$ (because $1 \le y$).

Moreover, $$y \le 2 \implies \left\{ \begin{align} y^2 &\le 2y\\ xy + 2 &\le 2x + 2 \end{align} \right. (x, y \ge 0)$$

We have that $$\frac{x^2 + 4}{y^2 + 1} \ge \frac{2x + 3}{2y + 1} \ge \frac{2x + 3}{xy + 3} \ge \frac{2x + 3}{2x + 3} = 1$$

The equality sign occurs when $x = 1$ and $y = 2$.

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Let $x=1$ and $y=2$.

Thus,we get a value $1$.

We'll prove that it's a minimal value.

Indeed, since $$x\geq\frac{2(y-1)}{y}\geq0,$$ we obtain: $$\frac{x^2+4}{y^2+1}-1=\frac{\left(\frac{2(y-1)}{y}\right)^2+4}{y^2+1}-1=\frac{(2-y)(y^3+2y^2-3y+2)}{y^2(y^2+1)}\geq0$$ and we are done!

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$$ x y \ge 2y-2\Rightarrow x\ge 2-\frac {2}{\max y}\ge 1 $$

because $y > 0$ now

$$ \frac{\min x^2+4}{\max y^2+1}\le \frac{x^2+4}{y^2+1} $$

hence

$$ \frac 55 = 1 \le \frac{x^2+4}{y^2+1} $$

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