= Solution
Put $q=\lambda a+(1-\lambda)b$. Since $0<\lambda<1$ and $a<b$, $a<q<b$. Apply the <mean value theorem> separately to $[a,q]$ and $[q,b]$. There exist $x\in(a,q)$ and $y\in(q,b)$ such that
$$
\boxed{f(q)=f(a)+(1-\lambda)(b-a)f'(x)
=f(b)-\lambda(b-a)f'(y).}
$$
These are the requested points, and $x<y$. Part (i) makes $f'(x)\leq f'(y)$, so the corresponding secant slopes satisfy
$$
\frac{f(q)-f(a)}{(1-\lambda)(b-a)}
\leq\frac{f(b)-f(q)}{\lambda(b-a)}.
$$
Multiply by the positive denominators and collect $f(q)$ to obtain
$$
\boxed{f(\lambda a+(1-\lambda)b)\leq\lambda f(a)+(1-\lambda)f(b).}
$$
This is precisely the defining chord inequality for a <convex function>. The proof uses only existence and nonnegativity of $f''$, with no additional regularity hypothesis on that derivative.
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