I'm going to prove to you that your lecturers actions were indeed correct.
Let $\displaystyle C = (x+1)(x+1)^k $
Let $\displaystyle B = \frac{1}{2}k(k-1)x^3 $.
Let $\displaystyle A = 1+kx+\frac{1}{2}k(k-1)x^2+x+kx^2 $
The equation you have, up until the point you don't understand, is:
$\displaystyle C \geq A + B $
We know that $\displaystyle B \geq 0 $
Therefore we can conclude that A+B is always bigger or equal to A. (If B = 0, then A+B =A, if B>0, then A+B>A): $\displaystyle A +B \geq A$
So we have:
$\displaystyle A+B \geq A $
And the original statement was:
$\displaystyle C \geq A+B $
So C is greater than or equal to A + B, and A + B is greater than or equal to A, we can put this into one single inequality:
$\displaystyle C \geq A+B \geq A $
$\displaystyle \therefore C \geq A $
From this we can conclude that the first term (C) is the biggest in the inequality, and the last term (A) is the smallest, and that that term in the middle is between the two (A+B). What we can conclude from that is:
$\displaystyle \therefore C \geq A $
Or you can go about it more algebraically:
$\displaystyle C \geq A+B \geq A $
$\displaystyle C-B \geq A \geq A-B $
$\displaystyle C-B \geq A-B $
$\displaystyle C \geq A $
Which confirms your lecturers actions.