Consider a function $ F : [a; b] \times [a; b] \longrightarrow [0, \infty)$ such that $F(x; y) = F(y; x)$ and $F(x; y) \leq F(x; z) + F(z; y)$ for all $x, y, z \in [a; b].$ Prove that $$ \sup_{x \in [a,b]}(\sup_{y \in [a,b]}F(x,y)) \leq 2\inf_{x \in [a,b]}(\sup_{y \in [a,b]} F(x,y)). $$
I did something like $$\sup_{y \in [a,b]}F(x,y) \leq \sup_{z \in [a,b]}F(x,z) + \sup_{z \in [a,b]}F(y,z)$$ $$ \sup_{x \in [a,b]}(\sup_{y \in [a,b]}F(x,y)) \leq \sup_{z \in [a,b]}F(x,z) + \sup_{z \in [a,b]}F(y,z)$$ $$ \sup_{x \in [a,b]}(\sup_{y \in [a,b]}F(x,y)) \leq \inf_{x, y \in [a,b]}(\sup_{z \in [a,b]}F(x,z) + \sup_{z \in [a,b]}F(y,z))\leq \inf_{x \in [a,b]}\sup_{z \in [a,b]}F(x,z)\\ + \inf_{y\in [a,b]}\sup_{z \in [a,b]}F(y,z)$$
You need to be more careful with your variables, but you essentially got the idea. It is just a matter of compatibility of supremum and infimum with inequalities. Starting from the 'triangular inequality' you are provided: $$\forall x,y,z,\ F(x,y)\leq F(z,x) + F(z,y)$$ Taking the supremum over $y$: $$\forall x,z,\ \sup_y F(x,y) \leq \sup_y (F(z,x) + F(z,y))=F(z,x) + \sup_y F(z,y)$$ Taking the supremum over $x$: $$\forall z,\ \sup_x\sup_y F(x,y) \leq \sup_x\left(F(z,x) + \sup_y F(z,y)\right) = \sup_x F(z,x) + \sup_y F(z,y)$$ The last term is equal to $2\sup_x F(z,x)$ since $x$ and $y$ are both bound variables. That is: $$\forall z,\ \sup_x\sup_y F(x,y) \leq 2\sup_x F(z,x)$$ Taking the infimum over $z$ gives you your inequality (with $z$ and $x$ replacing $x$ and $y$, but this is the same quantity since they are bound variables).