Correct solutions were submitted by Jens Voß, Lisa Schechner, Dan Dima. 

It's fun to use a little calculus.  Since f  is twice differentiable,

So it is enough to show that , or equivalently,  f ' ( t + x ) ≤  f ' ( t ) for 0 ≤ x , t ≤ y.  By the Mean Value Theorem applied to the differentiable function f ' we have 

f ' ( t + x ) - f ' ( t ) = f '' (c) ∙((t+x) - t) = f '' (c) ∙ x ≤ 0,

from which the desired property follows.

Note that the result is false without some restriction on the values of x and y as the function  f (x) = -x², with x = -1 = y demonstrates.  The result is also false without the assumption that f (0) = 0 as the function  f (x) = -x² - 1 with x = 0 = y demonstrates.  Finally, the function is only assumed to be twice differentiable, so power series arguments, though suggestive, are not wholly valid.