1. **State the problem:** Show that the value of the integral $$\int_0^1 \sqrt{1 + \cos x} \, dx$$ is less than or equal to $$\sqrt{2}$$. 2. **Recall the Max-Min Inequality for definite integrals:** For a continuous function $$f(x)$$ on $$[a,b]$$, the integral satisfies $$\min_{x \in [a,b]} f(x) \cdot (b - a) \leq \int_a^b f(x) \, dx \leq \max_{x \in [a,b]} f(x) \cdot (b - a).$$ 3. **Find the maximum value of the integrand:** The function inside the integral is $$f(x) = \sqrt{1 + \cos x}$$. Since $$\cos x$$ ranges between $$-1$$ and $$1$$, the maximum of $$1 + \cos x$$ on $$[0,1]$$ is at $$x=0$$ where $$\cos 0 = 1$$. Thus, $$\max f(x) = \sqrt{1 + 1} = \sqrt{2}.$$ 4. **Apply the inequality:** Using the Max-Min Inequality, $$\int_0^1 \sqrt{1 + \cos x} \, dx \leq \sqrt{2} \cdot (1 - 0) = \sqrt{2}.$$ 5. **Conclusion:** The value of the integral is less than or equal to $$\sqrt{2}$$ as required.