Problem 1468

1. Consider the given inequality:
$$a^{\sin x} \cdot (a+1)^{\cos x} \ge a \quad \forall x \in \left[0, \frac{\pi}{2}\right].$$

2. Take the natural logarithm on both sides of the inequality:
$$\ln \left( a^{\sin x} \cdot (a+1)^{\cos x} \right) \ge \ln a.$$

3. Using the properties of logarithms, we can separate the terms:
$$\sin x \cdot \ln a + \cos x \cdot \ln (a+1) \ge \ln a.$$

4. Let $A = \ln a$ and $B = \ln (a+1)$. The inequality becomes:
$$A \sin x + B \cos x \ge A.$$

5. To simplify, we introduce the angle $\alpha$ such that:
$$\cos \alpha = \frac{A}{\sqrt{A^2 + B^2}} \quad \text{and} \quad \sin \alpha = \frac{B}{\sqrt{A^2 + B^2}}.$$

6. Substitute these into the inequality:
$$A \sin x + B \cos x = \sqrt{A^2 + B^2} \left( \frac{A}{\sqrt{A^2 + B^2}} \sin x + \frac{B}{\sqrt{A^2 + B^2}} \cos x \right).$$

7. This can be rewritten using the angle sum identity for sine:
$$A \sin x + B \cos x = \sqrt{A^2 + B^2} \sin (x + \alpha).$$

8. Therefore, the inequality becomes:
$$\sqrt{A^2 + B^2} \sin (x + \alpha) \ge A.$$

9. Divide both sides by $\sqrt{A^2 + B^2}$:
$$\sin (x + \alpha) \ge \frac{A}{\sqrt{A^2 + B^2}}.$$

10. The minimum value of $\sin (x + \alpha)$ over the interval $x \in \left[0, \frac{\pi}{2}\right]$ is $\sin \alpha$, since $\alpha$ is a fixed angle and $x$ varies.

11. We need to show that:
$$\sin \alpha = \frac{B}{\sqrt{A^2 + B^2}} \ge \frac{A}{\sqrt{A^2 + B^2}}.$$

12. Since $B = \ln (a+1)$ and $A = \ln a$, and given that $a$ is a positive real number, it follows that $B > A$ because $\ln (a+1) > \ln a$.

13. Therefore:
$$\frac{B}{\sqrt{A^2 + B^2}} > \frac{A}{\sqrt{A^2 + B^2}}.$$

14. Hence, the inequality $\sin (x + \alpha) \ge \frac{A}{\sqrt{A^2 + B^2}}$ holds true for all $x \in \left[0, \frac{\pi}{2}\right]$.

$\blacksquare$