1. The problem asks to prove the inequality $$\frac{a+9}{2} + \frac{a+16}{2} > 7\sqrt{a}$$ for $a > 0$. 2. First, combine the fractions on the left side: $$\frac{a+9}{2} + \frac{a+16}{2} = \frac{(a+9) + (a+16)}{2} = \frac{2a + 25}{2} = a + \frac{25}{2}$$ 3. So the inequality becomes: $$a + \frac{25}{2} > 7\sqrt{a}$$ 4. To prove this, rearrange it as: $$a - 7\sqrt{a} + \frac{25}{2} > 0$$ 5. Let $x = \sqrt{a}$, then $a = x^2$, and the inequality is: $$x^2 - 7x + \frac{25}{2} > 0$$ 6. Multiply both sides by 2 to clear the fraction: $$2x^2 - 14x + 25 > 0$$ 7. Consider the quadratic function: $$f(x) = 2x^2 - 14x + 25$$ 8. The discriminant is: $$\Delta = (-14)^2 - 4 \cdot 2 \cdot 25 = 196 - 200 = -4 < 0$$ 9. Since $\Delta < 0$ and the leading coefficient $2 > 0$, the quadratic is always positive for all real $x$. 10. Because $x = \sqrt{a} > 0$, the inequality holds for all $a > 0$. **Final answer:** The inequality $$\frac{a+9}{2} + \frac{a+16}{2} > 7\sqrt{a}$$ is true for all $a > 0$.