10.3. Let the expression $E=\frac{3(x+y)(z+t)}{8} \cdot\left(\frac{1}{x y}+\frac{1}{z t}\right)$, where $x, y, z, t \in[1 ; 3]$. Determine the maximum and minimum possible values of the expression $E$.
Solution. The expression $\frac{8 E}{3}=(x+y)(z+t)\left(\frac{1}{x y}+\frac{1}{z t}\right)=\frac{(x z+y t+y z+x t)(z t+x y)}{x y z t}=$ $=\frac{x z^{2} t+x^{2} y z+y z t^{2}+x y^{2} t+y z^{2} t+x y^{2} z+x z t^{2}+x^{2} y t}{x y z t}=\frac{z}{y}+\frac{x}{t}+\frac{t}{x}+\frac{y}{z}+\frac{z}{x}+\frac{y}{t}+\frac{t}{y}+\frac{x}{z}=$ $=\left(\frac{x}{z}+\frac{z}{x}\right)+\left(\frac{z}{y}+\frac{y}{z}\right)+\left(\frac{y}{t}+\frac{t}{y}\right)+\left(\frac{t}{x}+\frac{x}{t}\right)$.

We know that $\frac{a}{b}+\frac{b}{a} \geq 2, \forall a, b>0$, with equality when $a=b$. Since $x, y, z, t \in[1 ; 3]$, i.e., they are positive, it follows that $\frac{8 E}{3} \geq 4 \cdot 2=8$, i.e., $E \geq 3$. Equality holds when $x=y=z=t$.

Let $a, b \in[1 ; 3]$. Since $a \leq 3$ (1) and $b \geq 1 \Leftrightarrow-3 b \leq-3$ (2), adding (1) and (2), we get the relation $a-3 b \leq 0$ (3). Furthermore, we have $a \geq 1 \Leftrightarrow 3 a \geq 3$ (4) and $b \leq 3 \Leftrightarrow-b \geq-3$ (5), adding (4) and (5), we obtain $3 a-b \geq 0$ (6). From relations (3) and (6), it follows that $(a-3 b)(3 a-b) \leq 0$. But this is equivalent to $3 a^{2}-a b-9 b a+3 b^{2} \leq 0 \Leftrightarrow 3\left(a^{2}+b^{2}\right) \leq 10 a b \Leftrightarrow \frac{a b>0}{a b} \leq \frac{a^{2}}{a b} \leq \frac{10}{3} \Leftrightarrow \frac{a}{b}+\frac{b}{a} \leq \frac{10}{3}$, with equality when $a=1, b=3$ or $a=3, b=1$. Therefore, $\frac{8 E}{3} \leq 4 \cdot \frac{10}{3}=\frac{40}{3}$, i.e., $E \leq 5$. Equality holds, for example, for $x=y=1, z=t=3$.

Thus, we have $\max _{[1 ; 3]} E=5$ and $\min _{[1 ; 3]} E=3$.