At which values ​of the parameter $k$, there is no solution to the inequality $(k+1)x^2-2kx+2k+2<0$? 
Problem:
At which values ​​of the parameter  $k$, there is no solution to the inequality $$(k+1)x^2-2kx+2k+2<0.$$

The solution in my textbook is as follows:
$a=k+1, D=4\left[k^2-2(k+1)^2\right]$
$\begin{cases} k+1>0 \\k^2-2(k+1)^2<0 \end{cases} \Longrightarrow \begin{cases} k>-1 \\ -k^2-2k-2 <0 \end{cases} \Longrightarrow -1<k<+\infty.$

Answer: That is, there is no solution to the inequality of $ (k + 1) x ^ 2-2kx + 2k + 2 <0$ for the $ k $ `s that satisfy the $ -1 <k <+ \infty $ condition.

Firstly, I understand the question as follows:

At which values ​​of the parameter  $k$, there is no solution to the inequality $(k+1)x^2-2kx+2k+2<0$ , for all $x\in\mathbb{R}.$
The last sentence is logically equivalent to :
At which values ​​of the parameter  $k$, the inequality  $(k+1)x^2-2kx+2k+2\geq 0$ holds on for all $x\in\mathbb{R}.$

If I understand the question correctly, here is my solution:

It is obvious that, for $k=-1$ is not a solution.
$\color{black}{\large\text{Case} \thinspace  1:}$ $k+1>0$
We have,
$$x^2-\dfrac{2k}{k+1}x+\dfrac{2k+2}{k+1}\geq 0 \\ \left(x-\dfrac{k}{k+1} \right)^2-\dfrac{k^2}{(k+1)^2}+\dfrac{2k+2}{k+1}\geq 0 \\ \left(x-\dfrac{k}{k+1} \right)^2 + \dfrac{2k+2}{k+1}-\dfrac{k^2}{(k+1)^2}\geq 0 \\ \left(x-\dfrac{k}{k+1} \right)^2+\dfrac {2(k+1)^2-k^2}{(k+1)^2} \geq 0 \\ \left(x-\dfrac{k}{k+1} \right)^2+\dfrac{k^2+4k+2}{(k+1)^2}\geq 0, ∀ x\in\mathbb {R}$$
Then, applying $x=\dfrac{k}{k+1}$ we get, $\dfrac{k^2+4k+2}{(k+1)^2}\geq 0$. We have,
$$\begin{cases} k+1>0 \\ \dfrac{k^2+4k+2}{(k+1)^2}\geq 0 \end{cases} \Longrightarrow k\geq -2+\sqrt2.$$
$\color{black}{\large\text{Case} \thinspace  2:}$ $k+1<0$
We have,
$$x^2-\dfrac{2k}{k+1}x+\dfrac{2k+2}{k+1}\leq 0 \\ \left(x-\dfrac{k}{k+1} \right)^2-\dfrac{k^2}{(k+1)^2}+\dfrac{2k+2}{k+1}\leq 0 \\ \left(x-\dfrac{k}{k+1} \right)^2 + \dfrac{2k+2}{k+1}-\dfrac{k^2}{(k+1)^2}\leq 0 \\ \left(x-\dfrac{k}{k+1} \right)^2+\dfrac {2(k+1)^2-k^2}{(k+1)^2} \leq 0 \\ \left(x-\dfrac{k}{k+1} \right)^2+\dfrac{k^2+4k+2}{(k+1)^2}\leq 0, ∀ x\in\mathbb {R}$$
For sufficiently large $ x $ `s, we have $\left(x-\dfrac{k}{k+1} \right)^2+\dfrac{k^2+4k+2}{(k+1)^2}\geq 0$, which gives a contradiction.
Finally we deduce that, for all  $x\in\mathbb{R}$ satisfying the condition $k\in\mathbb[\sqrt 2-2; +\infty)$, there is no solution to the inequality $(k+1)x^2-2kx+2k+2<0.$

It does not match the solution in my book. Probably, maybe I misunderstand the question or my  solution is wrong. Or the book says it wrong.
Did I get the question right? If so, is my solution correct?
Thank you very much.
 A: There seems some wrong points in your 'textbook solution' you posted, but I don't know if it is your typo or the textbook. Your solution is correct.
Using the textbook solution, we can achieve same result. 
$a=k+1, D=4\left[k^2-2(k+1)^2\right]$
If $k + 1 < 0$, we can be sure that for sufficiently large $x$, we always will have solution. we hence assume that $k+1 \geq 0$.
$$\begin{cases} k+1>0 \\k^2-2(k+1)^2 \leq 0 \end{cases} \Longrightarrow \begin{cases} k>-1 \\ -k^2-4k-2 \leq0 \end{cases} \Longrightarrow k \geq \sqrt{2}-2$$
Hence solution is $k \in [\sqrt{2}-2, \infty)$.
A: Here is an alternative, perhaps simpler, approach. Rewrite the given inequality $(k+1)x^2-2kx+2k+2<0$ as 
$$k< -\frac{x^2+2}{x^2-2x+2}$$
In order for the inequality not to hold, the values of $k$ has to be great than or equal to the maximum value of the RHS, which can be obtained by evaluating its derivative and setting it zero, i.e.
$$\frac{2(x^2-2)}{(x^2-2x+2)^2}=0\implies x=\pm\sqrt2$$
It is straightforward to verify that the maximum $-2+\sqrt2$ occurs at $x = -\sqrt2$. Thus, 
$$k\ge-2+\sqrt2$$
A: Well, in your textbook as can be seen in your post there are two inconsistencies. 
k+1>0 is not k>-2 and
more importantly
 k^2−2(k+1)^2<0 isn't equal to −k^2−2k−2<0 but it's similar to condition which you got in your solution i.e. k^2+4k+2. 
So, you have done it correctly. However, your textbook method if corrected is shorter.
