Known: u≤v

Because of the monotonicity of f, f(u) ≤ f(v)

It is also known that f(v) ≤ g(v)

Then: f(u) ≤ g(v)

That is, f(f(x)) ≤ g(g(x))

The same can be proved: g(g(x)) ≤ h(h(x))