$f(y)\le f(x)$ and $x
I can't understand well.
I guess, since $g$ is increasing function,
the inequality should be $(1-\lambda)g(\lambda f(x)+(1-\lambda)f(y)) \le (1-\lambda) g(f(x))$ because $f(y)\le f(x)$.
Why does that inequality hold?
$f(y)\le f(x)$ and $x
I can't understand well.
I guess, since $g$ is increasing function,
the inequality should be $(1-\lambda)g(\lambda f(x)+(1-\lambda)f(y)) \le (1-\lambda) g(f(x))$ because $f(y)\le f(x)$.
Why does that inequality hold?
The inequality, as you have found out, is not true in general. One can easily construct a counterexample (e.g. $\lambda=\frac12,\, x=0,\, y=1,\, f(t)=1-t,\, g(t)=t$). There must be a typo.