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Let $f$ be a real function defined on $[1, +\infty)$ and convex from a number on.

Is it true that $f$ is monotone from a number on?

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For $x_1 < x_2 < x_3$ from the definition of convex function

$$\dfrac {f \left({x_2}\right) - f \left({x_1}\right)} {x_2 - x_1} \le \dfrac {f \left({x_3}\right) - f \left({x_2}\right)} {x_3 - x_2} $$ so that $$f \left({x_3}\right) - f \left({x_2}\right) \ge \dfrac {(x_3 - x_2)(f \left({x_2}\right) - f \left({x_1}\right))} {x_2 - x_1} $$

So if $f \left({x_2}\right) \ge f \left({x_1}\right)$ then $f \left({x_3}\right) \ge f \left({x_2}\right)$. If there is no such $x_1, x_2$, then your function is monotonically decreasing.

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    @ Andrei Kulunchakov. Do you mean if $f$ is convex on $[M,\infty)$ then $f$ is monotone on it? – 2017-02-25
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    @M.H.Hooshmand, it is monotone starting from some point on – 2017-02-25
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    @ Andrei Kulunchakov. What is the start point (regarding $M$)? – 2017-02-25
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    @M.H.Hooshmand, it seems to be the minimum of $f$ (if such exists) on the ray. After this minimum it monotonically increasing, If this minimum does not exist, then the function is monotonically decreasing on the whole ray. $M$ has nothing to do with it. – 2017-02-25
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    @ Andrei Kulunchakov. How do you conclude it form your answer above (let $x_1=M$)? – 2017-02-25
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    @M.H.Hooshmand, if there are two points $x_1,x_2$ s.t. $f(x_2)\ge f(x_1)$, then $\forall x_3>x_2$ we have $f(x_3)\ge f(x_2)$ (from my answer above, as the right side of the centered inequality will be positive). Maximum possible $x_2$ (s.t. $\forall x_1$ we have $f(x_2)< f(x_1)$) is the minimum of $f$. Therefore, after this $x_2$ the function will increase. – 2017-02-25
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    @ Andrei Kulunchakov. Good, that's right. – 2017-02-25