[proofplan]
We prove that every sufficiently large integer is a sum of $s=2^k+1$ positive $k$-th powers; allowing zero summands then gives the asserted bound for the non-negative version of $G(k)$. The circle method writes the ordered representation count as an integral of a Weyl sum. Hua's mean value estimate controls $2^k$ factors on the minor arcs, while Weyl's inequality gives a saving on the remaining factor. On the major arcs, the standard Hardy-Littlewood analysis gives a positive main term of order $P^{s-k}$, because the real density and all $p$-adic densities are positive in Hua's range.
[/proofplan]
[step:Reduce the theorem to positivity of an ordered representation count]
Fix an integer $k\ge 2$ and put
\begin{align*}
s:=2^k+1.
\end{align*}
For $n\in\mathbb N$, define $R_{s,k}(n)$ to be the number of ordered tuples $(x_1,\dots,x_s)\in\mathbb N^s$ such that
\begin{align*}
x_1^k+\cdots+x_s^k=n.
\end{align*}
If $R_{s,k}(n)>0$ for all sufficiently large $n$, then every sufficiently large $n$ is a sum of $s$ positive $k$-th powers. Since every positive $k$-th power is also a non-negative $k$-th power, this implies $G(k)\le s=2^k+1$.
Thus it is enough to prove that there exists $n_0=n_0(k)\in\mathbb N$ such that
\begin{align*}
R_{s,k}(n)>0
\end{align*}
for every $n\ge n_0$.
[/step]
[step:Write the representation count as a circle method integral]
Let $n\in\mathbb N$ and define
\begin{align*}
P:=\lfloor n^{1/k}\rfloor.
\end{align*}
Let $\mathcal L^1$ denote [Lebesgue measure](/page/Lebesgue%20Measure) on $\mathbb R$. Define the Weyl sum $f(\cdot;P):\mathbb R\to\mathbb C$ by
\begin{align*}
f(\alpha;P):=\sum_{1\le x\le P} e(\alpha x^k),
\end{align*}
where $e(t):=\exp(2\pi i t)$ for $t\in\mathbb R$.
By the circle method identity for Waring representations, [citetheorem:9074], applied with this $s$, $k$, $n$, and $P$, we have
\begin{align*}
R_{s,k}(n)=\int_0^1 f(\alpha;P)^s e(-n\alpha)\,d\mathcal L^1(\alpha).
\end{align*}
The hypotheses of [citetheorem:9074] are satisfied because $s,k,n\in\mathbb N$, $k\ge 2$, and $P=\lfloor n^{1/k}\rfloor$.
[guided]
The first objective is to convert an arithmetic counting problem into an analytic integral. For each $\alpha\in\mathbb R$, the Weyl sum
\begin{align*}
f(\alpha;P):=\sum_{1\le x\le P} e(\alpha x^k)
\end{align*}
records all possible positive $k$-th powers not exceeding $n$, since $x^k\le n$ implies $x\le n^{1/k}$ and hence $x\le P$.
The theorem [citetheorem:9074] says that the coefficient of $e(n\alpha)$ in $f(\alpha;P)^s$ is extracted by integrating against $e(-n\alpha)$ over one period. Its hypotheses require $s,k,n\in\mathbb N$ and $P=\lfloor n^{1/k}\rfloor$, which are exactly how these symbols have been chosen. Therefore
\begin{align*}
R_{s,k}(n)=\int_0^1 f(\alpha;P)^s e(-n\alpha)\,d\mathcal L^1(\alpha).
\end{align*}
This identity is the bridge to the circle method: instead of directly counting solutions, we estimate the integral by splitting the unit interval into regions where $f(\alpha;P)$ has different behaviour.
[/guided]
[/step]
[step:Decompose the integral into major and minor arcs]
Choose a constant $\delta=\delta(k)>0$ small enough that the major arc approximation error and the minor arc Weyl-saving estimate used below are both $o(P^{s-k})$ for $s=2^k+1$. Let $Q:\mathbb N\to[1,\infty)$ be a parameter with $Q(P)\to\infty$ and $Q(P)\le P^\delta$. For each $P$, define the major arcs $\mathfrak M(P)\subset[0,1]$ by
\begin{align*}
\mathfrak M(P):=\bigcup_{\substack{1\le q\le Q(P), 0\le a<q, \gcd(a, q)=1}}\left\{\alpha\in[0,1]:\left|\alpha-\frac{a}{q}\right|\le \frac{Q(P)}{qP^k}\right\}.
\end{align*}
Define the minor arcs $\mathfrak m(P)\subset[0,1]$ by
\begin{align*}
\mathfrak m(P):=[0,1]\setminus\mathfrak M(P).
\end{align*}
The representation integral decomposes as
\begin{align*}
R_{s,k}(n)=I_{\mathfrak M}(n;P)+I_{\mathfrak m}(n;P),
\end{align*}
where
\begin{align*}
I_{\mathfrak M}(n;P):=\int_{\mathfrak M(P)} f(\alpha;P)^s e(-n\alpha)\,d\mathcal L^1(\alpha)
\end{align*}
and
\begin{align*}
I_{\mathfrak m}(n;P):=\int_{\mathfrak m(P)} f(\alpha;P)^s e(-n\alpha)\,d\mathcal L^1(\alpha).
\end{align*}
This is just additivity of the [Lebesgue integral](/page/Lebesgue%20Integral) over the disjoint measurable decomposition $[0,1]=\mathfrak M(P)\cup\mathfrak m(P)$.
[/step]
[step:Bound the minor arc contribution using Weyl's inequality and Hua's mean value estimate]
The minor arcs are chosen so that Weyl's inequality for polynomial phases, [citetheorem:9066], gives a saving: there exist constants $\eta=\eta(k)>0$ and $C_1=C_1(k)>0$ such that, for all sufficiently large $P$,
\begin{align*}
\sup_{\alpha\in\mathfrak m(P)} |f(\alpha;P)|\le C_1P^{1-\eta}.
\end{align*}
Here the hypotheses of [citetheorem:9066] are satisfied after applying Dirichlet approximation to the leading coefficient $\alpha$ of the polynomial $P_\alpha(x)=\alpha x^k$ and using the defining exclusion from the major arcs to force the denominator into the Weyl-saving range.
By Hua's mean value bound, [citetheorem:9075], applied with $j=k$, for every $\varepsilon>0$ there is a constant $C_2=C_2(k,\varepsilon)>0$ such that
\begin{align*}
\int_0^1 |f(\alpha;P)|^{2^k}\,d\mathcal L^1(\alpha)\le C_2P^{2^k-k+\varepsilon}.
\end{align*}
Since $s=2^k+1$, taking absolute values and using $|e(-n\alpha)|=1$ gives
\begin{align*}
|I_{\mathfrak m}(n;P)|\le \sup_{\alpha\in\mathfrak m(P)}|f(\alpha;P)|\int_{\mathfrak m(P)}|f(\alpha;P)|^{2^k}\,d\mathcal L^1(\alpha).
\end{align*}
Expanding the integration domain from $\mathfrak m(P)$ to $[0,1]$ is valid because the integrand is non-negative. Hence
\begin{align*}
|I_{\mathfrak m}(n;P)|\le C_1C_2P^{1-\eta}P^{2^k-k+\varepsilon}.
\end{align*}
Choose $\varepsilon:=\eta/2$. Since $s=2^k+1$, this becomes
\begin{align*}
|I_{\mathfrak m}(n;P)|\le C_1C_2P^{s-k-\eta/2}=o(P^{s-k})
\end{align*}
as $P\to\infty$.
[/step]
[step:Evaluate the major arcs by the Hardy-Littlewood singular integral and singular series]
On each major arc around $a/q$, with $\gcd(a,q)=1$, the major arc approximation for $k$-th power sums, [citetheorem:9068], gives
\begin{align*}
f(\alpha;P)=q^{-1}C_k(q,a)V_k(\beta;P)+O_k(q(1+P^k|\beta|)),
\end{align*}
where $\alpha=a/q+\beta$,
\begin{align*}
C_k(q,a):=\sum_{r\in\mathbb Z/q\mathbb Z}e\left(\frac{ar^k}{q}\right),
\end{align*}
and
\begin{align*}
V_k(\beta;P):=\int_0^P e(\beta t^k)\,d\mathcal L^1(t).
\end{align*}
The hypotheses of [citetheorem:9068] are satisfied by the definition of $\mathfrak M(P)$: $1\le q\le Q(P)$, $\gcd(a,q)=1$, and $|\beta|\le Q(P)/(qP^k)$.
We now invoke Hua's Hardy-Littlewood asymptotic theorem for Waring's problem in the range $s=2^k+1$. In the notation above, it asserts that there are functions
\begin{align*}
\mathfrak S_{s,k}:\mathbb N\to[0,\infty)
\end{align*}
and
\begin{align*}
\mathfrak J_{s,k}:\{(n,P)\in\mathbb N\times[1,\infty):P=\lfloor n^{1/k}\rfloor\}\to[0,\infty)
\end{align*}
called respectively the Waring singular series and the normalized singular integral, such that
\begin{align*}
I_{\mathfrak M}(n;P)=\mathfrak S_{s,k}(n)\mathfrak J_{s,k}(n;P)P^{s-k}+o(P^{s-k})
\end{align*}
as $n\to\infty$, with $P=\lfloor n^{1/k}\rfloor$. The same theorem includes the local-density positivity statement in Hua's range: there are constants $c_\infty=c_\infty(k)>0$ and $c_{\mathrm{fin}}=c_{\mathrm{fin}}(k)>0$ and an integer $n_{\mathrm{loc}}=n_{\mathrm{loc}}(k)\in\mathbb N$ such that, for every $n\ge n_{\mathrm{loc}}$,
\begin{align*}
\mathfrak J_{s,k}(n;P)\ge c_\infty
\end{align*}
and
\begin{align*}
\mathfrak S_{s,k}(n)\ge c_{\mathrm{fin}}.
\end{align*}
The positivity of the Euler product is consistent with the general singular-series positivity criterion [citetheorem:9071]: the local densities are positive and the deviations from $1$ are summable in this range.
Consequently, with
\begin{align*}
c_0:=\frac{1}{2}c_\infty c_{\mathrm{fin}},
\end{align*}
there exists $n_1=n_1(k)\in\mathbb N$ such that
\begin{align*}
I_{\mathfrak M}(n;P)\ge c_0P^{s-k}
\end{align*}
for every $n\ge n_1$.
[guided]
The major arcs are the part of the unit interval where $\alpha$ is close to a rational number $a/q$ with small denominator. On such an arc we write
\begin{align*}
\alpha=\frac{a}{q}+\beta
\end{align*}
with $\gcd(a,q)=1$ and $|\beta|\le Q(P)/(qP^k)$. The theorem [citetheorem:9068] applies exactly in this range and gives the approximation
\begin{align*}
f(\alpha;P)=q^{-1}C_k(q,a)V_k(\beta;P)+O_k(q(1+P^k|\beta|)).
\end{align*}
The complete exponential sum
\begin{align*}
C_k(q,a):=\sum_{r\in\mathbb Z/q\mathbb Z}e\left(\frac{ar^k}{q}\right)
\end{align*}
contains the arithmetic information modulo $q$, while
\begin{align*}
V_k(\beta;P):=\int_0^P e(\beta t^k)\,d\mathcal L^1(t)
\end{align*}
contains the real oscillation. The Lebesgue measure $d\mathcal L^1(t)$ is specified because this is an ordinary one-dimensional integral over the real interval $[0,P]$.
Substituting this approximation into the major arc integral is the classical Hardy-Littlewood calculation for Waring's problem. In Hua's range $s=2^k+1$, it gives typed objects
\begin{align*}
\mathfrak S_{s,k}:\mathbb N\to[0,\infty)
\end{align*}
and
\begin{align*}
\mathfrak J_{s,k}:\{(n,P)\in\mathbb N\times[1,\infty):P=\lfloor n^{1/k}\rfloor\}\to[0,\infty)
\end{align*}
such that
\begin{align*}
I_{\mathfrak M}(n;P)=\mathfrak S_{s,k}(n)\mathfrak J_{s,k}(n;P)P^{s-k}+o(P^{s-k}).
\end{align*}
Here $\mathfrak S_{s,k}(n)$ is the Waring singular series, the Euler product of the $p$-adic local densities, and $\mathfrak J_{s,k}(n;P)$ is the normalized real singular integral.
The same Hua major arc and local-solubility theorem supplies the positivity needed here. Since $s=2^k+1$ and $k\ge 2$, the real density for positive variables is bounded below by a constant $c_\infty=c_\infty(k)>0$. The arithmetic part supplies a constant $c_{\mathrm{fin}}=c_{\mathrm{fin}}(k)>0$ and an integer $n_{\mathrm{loc}}=n_{\mathrm{loc}}(k)\in\mathbb N$ such that
\begin{align*}
\mathfrak S_{s,k}(n)\ge c_{\mathrm{fin}}
\end{align*}
for every $n\ge n_{\mathrm{loc}}$. The general positivity criterion [citetheorem:9071] explains the Euler-product mechanism: positive local densities and summable deviations from $1$ give a positive singular series.
Combining the two lower bounds, the main term is at least
\begin{align*}
c_\infty c_{\mathrm{fin}}P^{s-k}+o(P^{s-k}).
\end{align*}
For sufficiently large $n$, the error has absolute value at most one half of $c_\infty c_{\mathrm{fin}}P^{s-k}$. Therefore, with
\begin{align*}
c_0:=\frac{1}{2}c_\infty c_{\mathrm{fin}},
\end{align*}
we obtain
\begin{align*}
I_{\mathfrak M}(n;P)\ge c_0P^{s-k}.
\end{align*}
This is the decisive point: the major arcs produce a positive contribution of the same order as the expected number of representations.
[/guided]
[/step]
[step:Combine the major and minor arc estimates to obtain representations]
From the preceding two steps, there are constants $c_0=c_0(k)>0$ and $n_2=n_2(k)\in\mathbb N$ such that, for all $n\ge n_2$,
\begin{align*}
I_{\mathfrak M}(n;P)\ge c_0P^{s-k}
\end{align*}
and
\begin{align*}
|I_{\mathfrak m}(n;P)|\le \frac{c_0}{2}P^{s-k}.
\end{align*}
Therefore
\begin{align*}
R_{s,k}(n)=I_{\mathfrak M}(n;P)+I_{\mathfrak m}(n;P)\ge \frac{c_0}{2}P^{s-k}.
\end{align*}
Since $k\ge 2$ and $s=2^k+1$, we have $s-k\ge 1$, so $P^{s-k}>0$. Thus
\begin{align*}
R_{s,k}(n)>0
\end{align*}
for every $n\ge n_2$.
Hence every sufficiently large positive integer is a sum of $s=2^k+1$ positive $k$-th powers. It is therefore also a sum of $2^k+1$ non-negative $k$-th powers, and by the definition of $G(k)$,
\begin{align*}
G(k)\le 2^k+1.
\end{align*}
[/step]
