jmc

Let $k=2x^2+3xy+2y^2.$ Then $$2x^2+3xy+2y^2=k=k(4x^2+8xy+5y^2)=4k{x}^2+8kxy+5k{y}^2=0,$$so $(4k-2)x^2+(8k-3)xy+(5k-2)y^2=0.$

If $y=0,$ then $4x^2=1,$ so $$2x^2+3xy+2y^2=\frac{1}{2}.$$Otherwise, we can divide both sides of $(4k-2)x^2+(8k-3)xy+(5k-2)y^2=0$ by $y^2,$ to get $$(4k-2){\left(\frac{x}{y}\right)}^2+(8k-3)\frac{x}{y}+(5k-2)=0.$$This is a quadratic in $\frac{x}{y},$ so and its discriminant must be nonnegative: $$(8k-3{)}^2-4(4k-2)(5k-2)\ge 0.$$This simplifies to $-16k^2+24k-7\ge 0,$ or $16k^2-24k+7\le 0.$ The roots of the quadratic $16k^2-24k+7=0$ are $\frac{3\pm \sqrt{2}}{4},$ so the solution to $16k^2-24k+7\le 0$ is $$\frac{3-\sqrt{2}}{4}\le k\le \frac{3+\sqrt{2}}{4}.$$For any value of $k$ in this interval, we can take $x=ky,$ then substitute into $4x^2+8xy+5y^2=1,$ and obtain solutions in $x$ and $y.$ Thus, $m=\frac{3-\sqrt{2}}{4}$ and $M=\frac{3+\sqrt{2}}{4},$ so $mM=\boxed{\frac{7}{16}}.$