1. **State the problem:** Solve the inequality $$\sqrt{2x - 3} + \sqrt{4x - 2} \leq \sqrt{6x - 5}$$. 2. **Identify domain restrictions:** - For $$\sqrt{2x - 3}$$, require $$2x - 3 \geq 0 \Rightarrow x \geq \frac{3}{2}$$. - For $$\sqrt{4x - 2}$$, require $$4x - 2 \geq 0 \Rightarrow x \geq \frac{1}{2}$$. - For $$\sqrt{6x - 5}$$, require $$6x - 5 \geq 0 \Rightarrow x \geq \frac{5}{6}$$. The overall domain is $$x \geq \frac{3}{2}$$ since it is the most restrictive. 3. **Square both sides to eliminate square roots:** $$\left(\sqrt{2x - 3} + \sqrt{4x - 2}\right)^2 \leq (\sqrt{6x - 5})^2$$ Expanding the left side: $$2x - 3 + 4x - 2 + 2\sqrt{(2x - 3)(4x - 2)} \leq 6x - 5$$ Simplify: $$6x - 5 + 2\sqrt{(2x - 3)(4x - 2)} \leq 6x - 5$$ 4. **Subtract $$6x - 5$$ from both sides:** $$2\sqrt{(2x - 3)(4x - 2)} \leq 0$$ Since square roots are non-negative, the only way this inequality holds is if: $$\sqrt{(2x - 3)(4x - 2)} = 0$$ 5. **Solve for when the product inside the root is zero:** $$(2x - 3)(4x - 2) = 0$$ Set each factor to zero: - $$2x - 3 = 0 \Rightarrow x = \frac{3}{2}$$ - $$4x - 2 = 0 \Rightarrow x = \frac{1}{2}$$ 6. **Check which solutions are in the domain $$x \geq \frac{3}{2}$$:** Only $$x = \frac{3}{2}$$ is valid. 7. **Verify the original inequality at $$x = \frac{3}{2}$$:** $$\sqrt{2(\frac{3}{2}) - 3} + \sqrt{4(\frac{3}{2}) - 2} = \sqrt{3 - 3} + \sqrt{6 - 2} = 0 + 2 = 2$$ $$\sqrt{6(\frac{3}{2}) - 5} = \sqrt{9 - 5} = \sqrt{4} = 2$$ Equality holds. 8. **Conclusion:** The inequality holds only at $$x = \frac{3}{2}$$. **Final answer:** $$\boxed{x = \frac{3}{2}}$$