Minimizing a function by finding its critical points Let $f_n(x)$ equal:
$$(2^n+2) \left(2x+1-\sqrt{2x^2+2x} \right)^n-x^n-(x+1)^n-\left(3x+1-2\sqrt{2x^2+2x}\right)^n-\left(3x+2-2\sqrt{2x^2+2x}\right)^n$$
Mathematica suggests that this function has two critical points on $(0,\infty)$, namely $x_1=1$ and $x_2=1/\sqrt{2}-1/2$. This seems to be independent of $n$ (well, I checked for all $4\leq n\leq 7$; also $f_1(x),f_2(x),f_3(x)$ identically equal to zero, a fact that I don't quite understand yet).
Problem: Prove that the only critical points of $f_n(x)$, $n\geq 4$, are $x_1=1$ and $x_2=1/\sqrt{2}-1/2$.
My goal is to show that $f_n(x)\geq 0$ for all $x\geq 0$. If I can show the only critical points are $1$ and $1/\sqrt{2}-1/2$, then by noting that $f_n(1)=0$ and $f_n(1\sqrt{2}-1/2)>0$ we have proved the claim. Of course, if you can prove $f_n(x)\geq 0$ in any way, please do share your proof.

The fact that $f_1(x),f_2(x),f_3(x)$ are identically equal to zero is simple (even if a bit tedious) to prove; just expand the expression.
Concerning $f_4(x)$ it reduces to
$$f_4(x)=4 x (x+1) \left(17 x^2+17 x+2\right)-24 \sqrt{2}\, x^{3/2} (x+1)^{3/2} (2 x+1)$$
$$f'_4(x)=8 (2 x+1) \left(17 x^2+17 x+1\right)-12 \sqrt{2}\, x^{1/2} (x+1)^{1/2} (4 x+1) (4 x+3)$$ After one squaring  step, the derivative cancels when
$$(x-1)\left(4 x^2+4 x-1\right) (x+2) \left(2 x^2+2 x+1\right) =0$$ So, since $x >0$, the only acceptable solutions are
$$x_1=1 \qquad \text{and} \qquad x_2=\frac{\sqrt{2}-1}{2} $$ as you already found.
Doing the same for $f'_5(x)$, the derivative cancels for
$$(x-1)  \left(4 x^2+4 x-1\right)(x+2) \left(25 x^4+50 x^3+35 x^2+10 x+4\right)=0$$ and the quartic does not show real roots.
Continuing (it starts to be tedious), for $f'_6(x)$, the derivative cancel for
$$(x-1)\left(4 x^2+4 x-1\right)(x+2) \left(2 x^2+2 x+1\right)  \left(9 x^4+18 x^3+9
   x^2+1\right)$$ and again the quartic does not show real roots.
It seems that the expressions which make the derivative equal to zero show a quite clear pattern. This has been tested up to $n=100$; to confirm, the values of  $f'_n(x_1)$ and $f'_n(x_2)$ have been systematically checked and they are always equal to $0$.
It is clear that $f_n(x_1)=0$. So, now
$$f_n(x_2)=2^{n/2}+2^{1-\frac{n}{2}}-2^{1-n}\Big[ \left(\sqrt{2}-1\right)^n+
   \left(\sqrt{2}+1\right)^n\Big]$$ which is a very fast increasing function.
As @Alexey Burdin commented, this looks to be the result of a (difficult) recurrence relation.
