1. **Problem Statement:** We are given four expressions: $z^2$, $\sqrt{5}z$, $\frac{z}{4}$, and $\frac{1}{z}$. We need to determine which inequality comparing two of these expressions is false for $\frac{1}{5} \leq z \leq \frac{2}{3}$.\n\n2. **Inequalities to test:**\n- $\sqrt{5}z \geq z^2$\n- $z^2 \geq \frac{1}{z}$\n- $\frac{1}{z} > \frac{z}{4}$\n- $\frac{z}{4} \leq \sqrt{5}z$\n\n3. **Step-by-step analysis:**\n\n**Inequality 1: $\sqrt{5}z \geq z^2$**\nRewrite as $\sqrt{5}z - z^2 \geq 0$.\nFactor out $z$: $z(\sqrt{5} - z) \geq 0$.\nSince $z \geq \frac{1}{5} > 0$, $z$ is positive.\nCheck if $\sqrt{5} - z \geq 0$ for $z \leq \frac{2}{3}$.\n$\sqrt{5} \approx 2.236$, and $\frac{2}{3} \approx 0.666$, so $\sqrt{5} - z > 0$.\nTherefore, inequality 1 is true for the interval.\n\n**Inequality 2: $z^2 \geq \frac{1}{z}$**\nRewrite as $z^2 - \frac{1}{z} \geq 0$. Multiply both sides by $z$ (positive in interval): $z^3 - 1 \geq 0$.\nCheck $z^3 - 1$ for $z \in [\frac{1}{5}, \frac{2}{3}]$.\nAt $z=\frac{1}{5}$, $\left(\frac{1}{5}\right)^3 - 1 = \frac{1}{125} - 1 = -\frac{124}{125} < 0$.\nAt $z=\frac{2}{3}$, $\left(\frac{2}{3}\right)^3 - 1 = \frac{8}{27} - 1 = -\frac{19}{27} < 0$.\nSo $z^3 - 1 < 0$ in the interval, meaning $z^2 < \frac{1}{z}$.\nInequality 2 is false in the interval.\n\n**Inequality 3: $\frac{1}{z} > \frac{z}{4}$**\nRewrite as $\frac{1}{z} - \frac{z}{4} > 0$. Multiply both sides by $4z$ (positive): $4 - z^2 > 0$.\nSince $z^2 \leq \left(\frac{2}{3}\right)^2 = \frac{4}{9} < 4$, this is true.\nInequality 3 is true.\n\n**Inequality 4: $\frac{z}{4} \leq \sqrt{5}z$**\nRewrite as $\frac{z}{4} - \sqrt{5}z \leq 0$. Factor $z$: $z\left(\frac{1}{4} - \sqrt{5}\right) \leq 0$.\nSince $z > 0$ and $\frac{1}{4} - \sqrt{5} < 0$, the product is negative, so inequality 4 is true.\n\n4. **Conclusion:** The only false inequality in the interval $\frac{1}{5} \leq z \leq \frac{2}{3}$ is $z^2 \geq \frac{1}{z}$.