$$16=(2x+2)(y+2)\le \frac{(2x+2+y+2)^2}{4}=\frac{(10-xy)^2}{4}$$

MariusStanean wrote: $$16=(2x+2)(y+2)\le \frac{(2x+2+y+2)^2}{4}=\frac{(10-xy)^2}{4}$$ Thanks. Proof of Zhangyunhua:

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From \[8=(x+1)(y+2)\geqslant (\sqrt{xy}+\sqrt 2)^2 \implies xy\leqslant 2,\]so \[(xy-10)^2=(10-xy)^2\geqslant 8^2=64.\]

Keith50 wrote: From \[8=(x+1)(y+2)\geqslant (\sqrt{xy}+\sqrt 2)^2 \implies xy\leqslant 2,\]so \[(xy-10)^2=(10-xy)^2\geqslant 8^2=64.\] Note that $x,y$ are real numbers, so $xy$ might be negative. For instance, we could have $y=-1$, $x=7$. Also it is not necessarily true that $xy \leq 2$, it could be that, e.g., $x=-9$, $y=-3$, $xy=27$.