Proving $\def\n#1{\left(\frac12+\sum\limits_{k=1}^n{#1}^{k^2}\right)}\n{a}\n{b}\ge{\n{(ab)}}^2$

Let $n$ be an even postive integer, and $a,b\in (-1,1)$, $a+b\ge0$. Show that $$\left(\frac12+\sum_{k=1}^na^{k^2}\right)\left(\frac12+\sum_{k=1}^nb^{k^2}\right)\ge\left(\frac12+\sum_{k=1}^n(ab)^{k^2}\right)^2\tag{1}$$

It seems promising to use Cauchy-Schwarz inequality to prove it or other inequality: If $a,b>0$, then $$\left(\frac{1}{2}+\sum_{k=1}^{n}a^{k^2}\right)\left(\frac{1}{2}+\sum_{k=1}^{n}b^{k^2}\right)\ge\left(\frac{1}{2}+\sqrt{\sum_{k=1}^{n}(a)^{k^2}\sum_{k=1}^{n}(b)^{k^2}}\right)^2,$$ but this is different from the RHS of $(1)$. Thanks.

This is not an answer, but a possible avenue for proving the result:

$1)$ Consider instead the polynomial

$$f(a,b)=\left(\frac12+\sum_{k=1}^{2n}a^{k^2}\right)\left(\frac12+\sum_{k=1}^{2n}b^{k^2}\right)-\left(\frac12+\sum_{k=1}^{2n}(ab)^{k^2}\right)^2$$

in the region $-b\leq a\leq 1$ and $-1\leq b\leq 1$. We must show that $f(a,b)\geq 0$ in this region. Note that we have switched $n$ even to $n\in\mathbb{N}$.

$2)$ Next, consider the boundary of this region:

$$i)\ f(a,-a)\text{ for }-1\leq a\leq 1$$

$$ii)\ f(a,1)\text{ for }-1\leq a\leq 1$$

$$iii)\ f(1,b)\text{ for }-1\leq b\leq 1$$

We must show that $f(a,b)$ is greater than or equal to $0$ on the boundary. There is obvious symmetry between $ii)$ and $iii)$.That is, if we can show boundary $ii)$ is greater than or equal to $0$, then $iii)$ is also greater than or equal to zero

$3)$ The final step is to show that $f(a,b)$ has no local minimum in the aforementioned region.

Now, what progress have I made on this plan. Consider boundary $ii)$:

$$f(a,1)=\left(\frac12+\sum_{k=1}^{2n}a^{k^2}\right)\left(\frac12+\sum_{k=1}^{2n}1^{k^2}\right)-\left(\frac12+\sum_{k=1}^{2n}(a\cdot 1)^{k^2}\right)^2$$

$$=\left(\frac12+\sum_{k=1}^{2n}a^{k^2}\right)\left(\frac12+2n\right)-\left(\frac12+\sum_{k=1}^{2n}a^{k^2}\right)^2$$

Consider the statement

$$\frac12+\sum_{k=1}^{2n}a^{k^2}\geq 0\text{ for }-1\leq a\leq 1$$

This seems to be true (and in fact the greater than or equal to can probably be changed to a strictly greater than), and graphically checks out for the first few cases. However, I am unable to prove it as

$$\lim_{n\to\infty}\text{min}_{-1\leq a\leq 1 }\left(\frac12+\sum_{k=1}^{2n}a^{k^2}\right)=0$$

(at least, that's what it seems to do graphically). That is, the minimum over $[-1,1]$ of $\sum_{k=1}^{2n}a^{k^2}$ seems to rapidly approach $\frac12$. This issue warrants further study.

Moving on, if that statement could be proved, then we would have

$$0=2n-2n\leq 2n -\sum_{k=1}^{2n}a^{k^2}=2n+\frac{1}{2}-\left(\frac12+\sum_{k=1}^{2n}a^{k^2}\right)$$

$$\Rightarrow 0\leq \left(\frac12+\sum_{k=1}^{2n}a^{k^2}\right)\left(\frac12+2n\right)-\left(\frac12+\sum_{k=1}^{2n}a^{k^2}\right)^2$$

(if $\frac12+\sum_{k=1}^{2n}a^{k^2}=0$, then the result clearly follows) which concludes the case for boundary $ii)$. Now, consider boundary $i)$. For the first few $n$, these become polynomials in $a$:

$$n=1: f(a,-a)=a^{10}(2-a^6)$$

$$n=2: f(a,-a)=a^{26}(2+2a^8-a^{10}-2a^{14}+2a^{24}-a^{38})$$

$$n=3: f(a,-a)$$ $$=a^{50}(2+2a^8-a^{14}-2a^{18}+2a^{24}-2a^{30}+2a^{32}+2a^{40}-a^{50}-2a^{54}+2a^{72}-a^{94})$$

$$\vdots$$

It is easy to check that these individual cases satisfy $f(a,-a)\geq 0$. For example, for $n=2$ note that

$$2>a^{10}$$

$$2a^8\geq 2a^{14}$$

$$2a^{24}>a^{38}$$

which implies

$$2+2a^8-a^{10}-2a^{14}+2a^{24}-a^{38}>0$$

and therefore

$$n=2:\ f(a,-a)=a^{26}(2+2a^8-a^{10}-2a^{14}+2a^{24}-a^{38})\geq 0$$

Unfortunately, I have not been able to prove this fact in general. Finally, one would need to prove that $f(a,b)$ has no local minimum inside the region. It would seem plausible that there might be some tools available to help with this problem, as this is fundamentally a question about a $2$D polynomial, but I do not have the expertise needed to find a solution.

In conclusion, this is a possible path forward towards proving this theorem, but many difficulties remain in order to make a proof reality.

For the last part of this post, I will prove that the conjecture is true for the first case. First, note that

$$\sum_{k=1}^2 a^{k^2}=a^4+a$$

which has a minimum at $a=-2^{-2/3}$. Then

$$\text{min}_{-1\leq a\leq 1}\left(\frac12+\sum_{k=1}^2 a^{k^2}\right)$$ $$=\left.\left(\frac12+\sum_{k=1}^2 a^{k^2}\right)\right|_{a=-2^{-2/3}}=\frac{1}{8} \left(4-3 \sqrt[3]{2}\right)=0.0275296>0$$

Thus, we are assured that boundary $ii)$ and $iii)$ are greater than or equal to $0$. For boundary $i)$, it is obvious that

$$f(a,-a)=a^{10}(2-a^6)$$

is greater than or equal to $0$. Finally, to show that $f(a,b)$ has no local minimum in the region, solve

$$f_a(a,b)=f_b(a,b)=0$$

This gives us the critical points

$$\left( \begin{array}{cc} -0.654048 & 1.28016 \\ 1.28016 & -0.654048 \\ 0.853753 & 0.853753 \\ \end{array} \right)$$

The only point in the region is $(a,b)=(0.853753 ,0.853753)$. Since

$$\left.f_{aa}(a,b)f_{bb}(a,b)-f_{ab}(a,b)^2\right|_{(a,b)=(0.853753 ,0.853753)}=-313.356<0$$

this is a saddle point. Thus, $f(a,b)$ has no local minimum in the region and the conjecture is true for $n=1$ ($n=2$ as originally written). In fact, this process is easily repeated for as many $n$ as one wishes (as long as one isn't to squeamish about numerical results). I have verified this up to $n=3$ ($n=6$ in the original post) before it became to difficult for my computer to verify.

Partial answer

WLOG, assume that $a \ge b$. The inequality is written as $$\frac{1}{2}(\sum a^{k^2} + \sum b^{k^2}) + \sum a^{k^2} \sum b^{k^2} \ge \sum (ab)^{k^2} + (\sum (ab)^{k^2})^2. \tag{1}$$

If $b > 0$, the proof is easy. Indeed, since $0\le ab \le 1$, we have (by AM-GM) $$\sum a^{k^2} + \sum b^{k^2} \ge \sum 2(\sqrt{ab})^{k^2} \ge \sum 2(ab)^{k^2}$$ and (by CBS) $$\sum a^{k^2} \sum b^{k^2} \ge (\sum (\sqrt{ab})^{k^2})^2 \ge (\sum (ab)^{k^2})^2. $$ So, the desired inequality in (1) is true.

It remains to prove the inequality in (1) under the condition $0\le -b \le a \le 1$.

1. If $-b = 0$, clearly the inequality in (1) is true.

2. If $a = 1$, it suffices to prove that $$\frac{1}{2}(n + \sum b^{k^2}) + n \sum b^{k^2} \ge \sum b^{k^2} + (\sum b^{k^2})^2.$$ It suffices to prove that $\sum b^{k^2} \ge -\frac{1}{2}$.
   If $-b = 1$, clearly $\sum b^{k^2} = 0 \ge -\frac{1}{2}$.
   If $0 < -b < 1$, clearly $f(2m) \triangleq \sum_{k=1}^{2m} b^{k^2}$ is non-increasing.
   Also, $\sum_{k=1}^\infty b^{k^2} = \frac{1}{2}\vartheta_3(0, b) - \frac{1}{2} \ge -\frac{1}{2}$ where $\vartheta_3(z, q) = 1 + 2 \sum_{k=1}^\infty q^{k^2}\cos (2k z)$ is the Jacobi theta function (by using the property $\vartheta_3(0, q)= \prod_{n=1}^\infty (1-q^{2n})(1+q^{2n-1})^2$).
   Thus, we have $\sum b^{k^2} \ge -\frac{1}{2}$.

3. If $-b = a$, to be continued.

4. If $0 < -b < a < 1$, to be continued.