The problem is: Prove $L_1,L_2:\mathbb{R}^n\rightarrow \mathbb{R}$ are convex, smooth and superlinear (that is $L(x)/|x|\rightarrow\infty$ as $|x|\rightarrow\infty$). Show that $$ \min\limits_{v\in\mathbb{R}^n}(L_1(v)+L_2(v))=\max\limits_{p\in\mathbb{R}^n}(-H_1(p)-H_2(-p)), $$ where $H_1=L_1^*,H_2=L_2^*$.
Here, $$ L^*(v)=\sup_{p\in\mathbb{R}^n}v\cdot p-L(v). $$ My try: From the superlinearity of $L_1$ and $L_2$, we know that the min on left side and max on right side are attainable.
By definition of the Legendre transform, for any $v,p\in\mathbb{R}^n$, \begin{equation} L_1(v)+L_2(v)\geq p\cdot v-H_1(p)+(-p)\cdot v-H_2(-p)=-H_1(p)-H_2(-p) \label{12-1} \end{equation} which implies that $$ L_1(v)+L_2(v)\geq \max\limits_{p\in\mathbb{R}^n}(-H_1(p)-H_2(-p)). $$ Take min over $v$, $$ \min\limits_{v\in\mathbb{R}^n}(L_1(v)+L_2(v))\geq \max\limits_{p\in\mathbb{R}^n}(-H_1(p)-H_2(-p)). $$
But how to prove the other direction?