amparvardi wrote: Let $ABCDEFGH$ be a parallelepiped with $AE \parallel BF \parallel CG \parallel DH$. Prove the inequality \[AF + AH + AC\le AB + AD + AE + AG.\] In what cases does equality hold? Proposed by France. If we denote $AB=a$ , $BC=b$ and $CG=c$ , then the inequality reduces to : $\sqrt{a^2+b^2}+\sqrt{b^2+c^2}+\sqrt{c^2+a^2}\ \le\ a+b+c+\sqrt{a^2+b^2+c^2}$ . The last inequality was posted here and here .

That is for rectangular parallelepiped And the above equality holds when $abc=0$ that means it's rectangular

See here https://artofproblemsolving.com/community/c6h1998796p14494017.