<s>[INST] 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.
 [/INST] Some thoughts
It suffices to prove that $f_n(x) \ge 0$ for $x > 0$.
With the substitution $x = \frac{2}{u^2+2u-1}$ for $u > \sqrt{2} - 1$ (correspondingly, $u = \sqrt{2 + \frac{2}{x}} - 1$), we have
$$f_n(x) = \frac{1}{(u^2+2u-1)^n}
[(2^n+2)(u^2+1)^n - 2^n - (u+1)^{2n} - (u-1)^{2n} - 2^n u^{2n}].$$
It suffices to prove that, for $u > 0$,
$$(2^n+2)(u^2+1)^n - 2^n - (u+1)^{2n} - (u-1)^{2n} - 2^n u^{2n} \ge 0.$$
Denote the LHS by $g_n(u)$. Note that $u^{2n} g_n(\frac{1}{u}) = g_n(u)$ for $u > 0$.
Thus, it suffices to prove that, for $u \ge 1$,
$$(2^n+2)(u^2+1)^n - 2^n - (u+1)^{2n} - (u-1)^{2n} - 2^n u^{2n} \ge 0.$$
Let $u = 1 + v$. It suffices to prove that, for $v \ge 0$,
$$(2^n+2)((1+v)^2+1)^n - 2^n - (v+2)^{2n} - v^{2n} - 2^n (1+v)^{2n} \ge 0.$$
Denote the LHS by $h_n(v)$. It suffices to prove that
the polynomial $h_n(v)$ has non-negative coefficients.
I did some numerical experiments which support this claim.
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