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$f(y)\le f(x)$ and $x<y$ where $f$ is convex function and $g$ is increasing convex function.
Then $(1-\lambda)g(\lambda f(x)+(1-\lambda)f(y)) \le (1-\lambda) g(f(y))$.

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?

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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.

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Then this is true? : Since $g$ is increasing, $g(f(\lambda x+(1-\lambda)y)) \le g(\lambda f(x)+(1-\lambda)f(y))$. –  cowik Dec 14 '12 at 10:44
    
This is indeed true, since $f$ is convex and $g$ increasing. –  Nick Papadopoulos Dec 14 '12 at 10:51
    
Hmm, this is neither of the two inequalities you mentioned in your question, but it is true anyway. –  user1551 Dec 14 '12 at 10:52
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