The problem states:

If $a > 0$ and $b > 0$, which is greater: $\left(\sqrt{a} + \sqrt{b}\right)^2 \quad \text{or} \quad \sqrt{a^2} + \sqrt{b^2}?$

Step 1: Simplify both expressions

1. Expand $\left(\sqrt{a} + \sqrt{b}\right)^2$: $\left(\sqrt{a} + \sqrt{b}\right)^2 = a + b + 2\sqrt{ab}.$

2. Simplify $\sqrt{a^2} + \sqrt{b^2}$: Since $a > 0$ and $b > 0$, $\sqrt{a^2} = a$ and $\sqrt{b^2} = b$, so: $\sqrt{a^2} + \sqrt{b^2} = a + b.$

Step 2: Compare the two expressions

• From Step 1, the two expressions are: $\left(\sqrt{a} + \sqrt{b}\right)^2 = a + b + 2\sqrt{ab},$ and $\sqrt{a^2} + \sqrt{b^2} = a + b.$

• Clearly, $\left(\sqrt{a} + \sqrt{b}\right)^2$ has an extra term $2\sqrt{ab}$, which is always positive since $a > 0$ and $b > 0$.

Conclusion:

$\left(\sqrt{a} + \sqrt{b}\right)^2 > \sqrt{a^2} + \sqrt{b^2}.$

Would you like further clarification or additional examples? 😊

Follow-up Questions:

1. What happens if $a = b$? How does that affect the comparison?
2. Can you think of a geometric interpretation for $\sqrt{a}$ and $\sqrt{b}$?
3. How does the inequality change if $a$ or $b$ equals zero?
4. What role does the term $2\sqrt{ab}$ play in determining which is larger?
5. How would the result change if $a, b < 0$?

Tip: When comparing expressions, simplify and group terms systematically to identify differences easily.