Determine the minimum term of the sequence

$$
\sqrt{\frac{7}{6}}+\sqrt{\frac{96}{7}}, \sqrt{\frac{8}{6}}+\sqrt{\frac{96}{8}}, \sqrt{\frac{9}{6}}+\sqrt{\frac{96}{9}}, \ldots, \sqrt{\frac{95}{6}}+\sqrt{\frac{96}{95}}
$$
Solution

Remember that $(x-y)^{2} \geq 0$ for all real numbers $x$ and $y$. Thus, we can rewrite the inequality as $\frac{x^{2}+y^{2}}{2} \geq x y$, and by substituting $x=\sqrt{a}$ and $y=\sqrt{b}$, with $a$ and $b$ being non-negative real numbers, we have $\frac{a+b}{2} \geq \sqrt{a b}$. Now, observe that all terms are of the form $\sqrt{\frac{n}{6}}+\sqrt{\frac{16 \cdot 6}{n}}=$ $\sqrt{\frac{n}{6}}+4 \sqrt{\frac{6}{n}}$, with $n$ being an integer and greater than or equal to 7. Taking $a=\sqrt{\frac{n}{6}}$ and $b=4 \sqrt{\frac{6}{n}}$, we can do

$$
\begin{aligned}
\frac{\sqrt{\frac{n}{6}}+4 \sqrt{\frac{6}{n}}}{2} & \geq \sqrt{\sqrt{\frac{n}{6}} \cdot 4 \sqrt{\frac{6}{n}}} \\
\sqrt{\frac{n}{6}}+4 \sqrt{\frac{6}{n}} & \geq 2 \cdot 2 \cdot \sqrt{\sqrt{\frac{n}{6} \cdot \frac{6}{n}}} \\
& =4
\end{aligned}
$$

The minimum value of each term is 4 and occurs when $\sqrt{\frac{n}{6}}=4 \sqrt{\frac{6}{n}}$, that is, $n=24$. The term we are looking for is $\sqrt{\frac{24}{6}}+\sqrt{\frac{96}{24}}$.