Part b)

Solution.

Consider the second inequality first. Note that since \(x,y > 0\), \(\sqrt{x}, \sqrt{y} > 0\), and so \[x + y \leq x + 2\sqrt{x}\sqrt{y} + y = (\sqrt{x} + \sqrt{y})^2.\] Square rooting this result gives us that \[\sqrt{x + y} \leq \sqrt{x} + \sqrt{y}.\] Next, we have that \[\begin{align*} \left(\sqrt{\frac{x}{2}} + \sqrt{\frac{y}{2}}\right)^2 &= \frac{x}{2} + 2\sqrt{\frac{x}{2}}\sqrt{\frac{y}{2}} + \frac{y}{2},\\ &\leq \frac{x}{2} + 2\left(\frac{x}{2} + \frac{y}{2}\right) + \frac{y}{2} \;\;\; \text{(by part a)},\\ &= x + y. \end{align*}\] Again, square rooting gives us that \[\sqrt{\frac{x}{2}} + \sqrt{\frac{y}{2}} \leq \sqrt{x + y}.\]