I will provide a solution only for b), as a) is trivial. We want to prove that: $$ \int_{0}^{1} f(\phi(x)) \text{dx} + \int_{0}^{1} \phi(x)f'(\phi(x)) \text{dx} \leq \int_{0}^{1} f'(\phi(x)) \text{dx}$$ It s easier to work with $f'$ than with $f$, as $f'$ is a function that is just nonnegative, so we will write the first integral in the L.H.S as: $$\int_{0}^{1} f(\phi(x)) \text{dx} = \int_{0}^{1} x' f(\phi(x)) \text{dx} = f(1) - \int_{0}^{1} xf'(\phi(x)) \phi'(x) \text{dx}=\int_{0}^{1} f'(x) \text{dx} - \int_{0}^{1} xf'(\phi(x)) \phi'(x) \text{dx}=\int_{0}^{1} f'(\phi(x))\phi'(x) \text{dx} - \int_{0}^{1} xf'(\phi(x)) \phi'(x) \text{dx}= \int_{0}^{1} f'(\phi(x))\phi'(x)(1-x) \text{dx}$$ where the second equality is from integrating by parts, and the fourth one is from making the change of variables $x=\phi(t)$ in $\int_{0}^{1} f'(x) \text{dx}$ Now it remains to prove that: $$\int_{0}^{1} f'(\phi(x)) \phi'(x)(1-x) \text{dx} + \int_{0}^{1} \phi(x)f'(\phi(x)) \text{dx} \leq \int_{0}^{1} f'(\phi(x)) \text{dx}$$, which is equivalent with $$\int_{0}^{1} f'(\phi(x))(\phi'(x)(1-x)+\phi(x)-1) \text{dx} \leq 0$$ But, $\phi$ is convex, so we have $$\frac{1-\phi(x)}{1-x}=\frac{\phi(1)-\phi(x)}{1-x}\geq \phi'(x) \ \forall x \in [0,1)$$, so $\phi'(x)(1-x)+\phi(x)\leq 1$, and as $f'(\phi(x))\geq 0 \ \forall x \in [0,1]$, we have that, $\int_{0}^{1} f'(\phi(x))(\phi'(x)(1-x)+\phi(x)-1) \text{dx} \leq 0$. $\square$

Here is a solution for a) Note that $g(x) =\frac{d}{dx} (x-1)f(x)$. Thus, $\int_0^1 g(x) = [(x-1)f(x)]_0^1 =0$