1. The problem is to find the best approximation interval for $\sqrt{91}$.\n\n2. We know that $9^2 = 81$ and $10^2 = 100$. Since $91$ is between $81$ and $100$, its square root must lie between $9$ and $10$.\n\n3. Checking the other ranges: $8^2 = 64$ (less than 91), $7^2 = 49$, $5^2 = 25$, and $4^2 = 16$. So $\sqrt{91}$ cannot be less than 9 or between these lower numbers.\n\n4. Therefore, the correct inequality for the best approximation of $\sqrt{91}$ is:\n\n$$9 < \sqrt{91} < 10$$\n\nFinal answer: $9 < \sqrt{91} < 10$