For $t>0$, <positive homogeneity> and <concavity> give
$$
\begin{aligned}
f(X+tY)
&=(1+t)f\left(\frac{X}{1+t}+\frac{tY}{1+t}\right)\\
&\geq(1+t)\left(\frac1{1+t}f(X)+\frac t{1+t}f(Y)\right)\\
&=f(X)+tf(Y).
\end{aligned}
$$
After subtracting $f(X)$, dividing by $t$, and taking the one-sided <directional derivative> at zero,
$$
\boxed{\left.\frac d{dt}\right|_{t=0^+}f(X+tY)\geq f(Y)}.
$$

Solution

Past exam of the mathematics course of the University of Cambridge / 2025 / iii / Paper 323 / 3 / ii / c / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2025/iii/paper-323/3/ii/c/solution