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.
 A: 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 $0\le -b \le a \le 1$.

*

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


*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 $\sum b^{k^2} \ge -\frac{1}{2}$.


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


*If $0 < -b < a < 1$, to be continued.
A: Too long for a comment for $a,b>0$ and the constraint of the OP :
Put $a=\exp(y)$ and $b=\exp(z)$
We have for $u<0$:
$$f(u)=\ln(0.5+\sum_{k=1}^{n}\exp(u{k^2}))$$
The second derivative of $f(u)$ is positive because after simplification the numerator have only positives terms (using a computer).
The conclusion is straightforward using Jensen's inequality .We get also a generalisation.


Conjecture for the case where $n=2p$ and $-b=a>0$:
Define the functions :
$$f(x)=(0.5-e^{x}+e^{4x}+\cdots -e^{(n-1)^2x}+e^{n^2x})(0.5+e^{x}+e^{4x}+\cdots e^{(n-1)^2x}+e^{n^2x})$$
$$g(x)=(0.5+e^{x}+e^{4x}+\cdots e^{(n-1)^2x}+e^{n^2x})$$
$$h(x)=(0.5-e^{x}+e^{4x}+\cdots -e^{(n-1)^2x}+e^{n^2x})$$
Claim :
Denotes by $a$ the solution for the equation $f'(x)=0$ and by $a'$ the solution for the equation $f''(x)=0$ then :
$$a'<a$$
Second claim :
if $f(x)$ is convex  by Jensen's inequality on$(a',0-\varepsilon_n)$:
$$f(\ln(a))+f(\ln(a^2))\geq 2f(1.5\ln(a))\geq h(1.5\ln(a))\geq 2(h(2\ln(a)))^2$$
