The eta function is defined as follows: $\eta(s) = \sum_{n=1}^\infty \frac{(-1)^{n+1}}{n^s}$, where $s$ is a complex number $a+bi$. $\eta(s)$ can be split into the sum of a real series and an imaginary series, as so: $\eta(s) = \sum_{n=1}^\infty \frac{(-1)^{n+1}}{n^a}\cos(b\ln(n)) + i\sum_{n=1}^\infty \frac{(-1)^{n+1}}{n^a}\cos(b\ln(n)+\frac{\pi}{2})$
In the critical strip, $\eta(s)$ has all of the same zeros as the Riemann zeta function. This means that its zeros in the critical strip are symmetric about the line $a = 1/2$. Therefore, if $a+bi$ is a zero of $\eta(s)$ in the critical strip, $1-a+bi$ is also a zero of $\eta(s)$. If $a+bi$ and $1-a+bi$ are zeros of $\eta(s)$, then the distance in the complex plane between $\eta(a+bi)$ and $\eta(1-a+bi)$ must be zero, because $\eta(a+bi)$ and $\eta(1-a+bi)$ are both at the origin.
The distance $M$ between $\eta(a+bi)$ and $\eta(1-a+bi)$, computed simply by using the Pythagorean theorem, can be represented by the following function: $M(s) = \sqrt{[\sum_{n=1}^\infty (-1)^{n+1}(\frac{1}{n^a}-\frac{1}{n^{1-a}})\cos(b\ln(n))]^2 + [\sum_{n=1}^\infty (-1)^{n+1}(\frac{1}{n^a}-\frac{1}{n^{1-a}})\cos(b\ln(n)+\frac{\pi}{2})]^2}$
If the derivative with respect to $a$ of $M(s)$, $M'(s)$, is negative for all combinations of $a$ and $b$ with $0\lt$$a$$\lt1/2$ and $b$$\gt$$k$, for some positive finite constant $k$, then all zeros, with $b$$\gt$$k$ and in the critical strip, of $\eta(s)$ are on the critical line. Here is why: If $\eta(s)$ is zero at $a+bi$ for some $a$ strictly between $0$ and $1/2$ (not on the critical line), $M'(s)$ cannot be negative at that point. This is because $M'(s)$ represents the change in the distance between $\eta(a+bi)$ and $\eta(1-a+bi)$ with respect to $a$. $M'(s)$ cannot be negative because the distance between $\eta(a+bi)$ and $\eta(1-a+bi)$ is already zero, because $\eta(a+bi)$ and $\eta(1-a+bi)$ are both at the origin. The distance between the two points $\eta(a+bi)$ and $\eta(1-a+bi)$ cannot decrease any further. Therefore, if $M'(s)$ is negative for all combinations of $a$ and $b$ with $0\lt$$a$$\lt1/2$ and $b$$\gt$$k$, there can be no zeros of $\eta(s)$ in the critical strip (with $b$$\gt$$k$) that are not on the critical line.
Calculating the derivative with respect to $a$ of $M(s)$ requires the use of the following functions: $A = \sum_{n=1}^\infty (-1)^{n+1}(\frac{1}{n^a}-\frac{1}{n^{1-a}})\cos(b\ln(n))$; $B = \sum_{n=1}^\infty (-1)^{n+1}(-\frac{\ln(n)}{n^a}-\frac{\ln(n)}{n^{1-a}})\cos(b\ln(n))$; $C = \sum_{n=1}^\infty (-1)^{n+1}(\frac{1}{n^a}-\frac{1}{n^{1-a}})\cos(b\ln(n)+\frac{\pi}{2})$; $D = \sum_{n=1}^\infty (-1)^{n+1}(-\frac{\ln(n)}{n^a}-\frac{\ln(n)}{n^{1-a}})\cos(b\ln(n)+\frac{\pi}{2})$
Using these functions, $M'(s)$ can be defined in the following way: $M'(s) = \frac{AB+CD}{M(s)}$ Because $M(s)$ represents distance, it is non-negative. Therefore, it has two cases in terms of its sign: positive or zero. In the case that $M(s)$ is positive, it does not affect the sign of $M'(s)$. In the case that $M(s)$ is zero, $M'(s)$ is undefined. However, if the numerator $AB+CD$ is negative for all combinations of $a$ and $b$ with $0\lt$$a$$\lt1/2$ and $b$$\gt$$k$, $M(s)$ cannot be zero in this region. The reasoning is as follows: If $AB+CD$ is negative for all combinations of $a$ and $b$ with $0\lt$$a$$\lt1/2$ and $b$$\gt$$k$ in which $M(s)$ is not zero, any zero of $M(s)$ in this region would be followed by a further decrease in $M(s)$ as $a$ increases. This is impossible, since $M(s)$ is already zero. Therefore, if $AB+CD$ is negative for all combinations of $a$ and $b$ with $0\lt$$a$$\lt1/2$ and $b$$\gt$$k$, then $M'(s)$ is negative in this region as well.
Here is a graph in the $b$,$y$ plane of $AB$ in red, $CD$ in blue, and $AB+CD$ in purple with $a$ set to $\frac{1}{8}$ (with the series accurate to $120$ terms):
$a$ set to Here is the same thing as the previous graph but with $a$ set to $\frac{1}{4}$:
$a$ set to Finally, here is the same thing as the previous graph but with $a$ set to $\frac{3}{8}$:
$a$ set to There are a few things to note here. Firstly, $AB$ and $CD$ seem to be almost entirely negative. Secondly, $AB$ almost always goes down when $CD$ goes up, and vice versa. This seems to make $AB+CD$ very unlikely to go above the $b$ axis into positive territory (at least when $b$ is greater than $10$ or so, which is why there is the stipulation that $b$$\gt$$k$). These imprecise, anecdotal observations are certainly not proof that $AB+CD$ is always negative in the specified region, but it does give hope that perhaps it can be proven.
So, is it possible to prove that AB+CD is always negative in the region mentioned above? Also, is it correct to say that proving this would imply that all zeros in the critical strip of the eta function, and therefore all zeros in the critical strip of the zeta function, are on the critical line?