General Inequality for Convex Functions Suppose we have a convex function $f:[-1,1]\to \mathbb{R}$ with $f(-1)=f(1)=0$ and $\{t_i\}_{i=1}^k$ is a sequence of numbers such that $k\ge 2$ and
$$
\sum_{i=1}^k \frac{t_i}{f(t_i)+f(-t_i)}=0\ \ \ \ \ \ \ \ \ \ \ \ \ \ \       (A)
$$
Is the following necessary?
$$
\sum_{i=1}^k \frac{f(t_i)}{f(t_i)+f(-t_i)}\ge 1\ \ \ \ \ \ \ \ \ \ \ \ \ \ \       (B)
$$
I suspect $1$ can be replaced by $k/2$, in which case it works for $k=1$ and arguably $k=0$, too. I would also like to generalize this to n dimensions eventually.
PROGRESS

f(x) EVEN
(B) is true without (A) when $f$ is even since $k/2\ge 1$. Specifically, notable is when $f$ is a semicircle. I believe this implies (B) is also true when $f$ describes an (oblique) semi-ellipse.
f(x) LINEAR
If $f(x)$ is linear in a region containing all $t_i$, (A) becomes
$$
\sum_{i=1}^k\frac{t_i}{2f(0)}=0
$$
which gives (B) through
$$
\sum_{i=1}^k\frac{f(t_i)}{2f(0)}=\sum_{i=1}^k\frac{(\textrm{slope})t_i+f(0)}{2f(0)}=k/2\ge 1
$$
Maybe next is where all $t_i$ are contained on a few linear pieces, although I've run into muck that way.
INEQUALITY?
If there is a good inequality to prove (B) in the form
$$
\sum_{i=1}^k \frac{t_i}{f(t_i)+f(-t_i)}\cdot \frac{f(t_i)}{t_i}\ge 1
$$
since we know the left factors sum to $0$.
REPARAMETRIZE
Finally, I tried reparametrizing in terms of $u(t)=\frac{t}{f(t)+f(-t)}$. First, we need to show that $u$ is invertible. Write $F(t):=f(t)+f(-t)$.
$$
u^\prime(t)=\frac{F(t)-t(F^\prime(t))}{F^2(t)}
$$

The numerator is strictly negative (always $\le F(0)$) and the denominator is positive (excluding $t=\pm 1$). So, $u$ is strictly decreasing, hence invertible. Equation (B) becomes for any $\{c_i\}_{i=1}^k$ such that $\sum c_i=0$,
$$
\sum_{i=1}^k c_i\frac{f(u^{-1}(c_i))}{u^{-1}(c_i)}\ge 1
$$
Then, we would be quickly finished if we could show $|f(u^{-1}(s))/u^{-1}(s)|$ were convex on $(-\infty,0)$ and on $(0,\infty)$, as we could merge all the $c_i>0$ to one value and all the $c_i<0$ to the opposite value (no harder than $k=2$), and maintain the inequality the whole time. So, we take derivatives. Note $d/ds\  u^{-1}(s)=1/u^\prime(u^{-1}(s))$.
JUNK SCRATCH
$$
u^{\prime}(t)=\frac{F-tF^\prime}{F^2}
$$
$$
u^{\prime\prime}(t)=\frac{-tF^{\prime\prime}F-2F^\prime(F-tF^\prime)}{F^3}
$$
$$
\left.\frac{d}{ds}\frac{f(u^{-1}(s))}{u^{-1}(s)}=\frac{f^\prime(t)t-f(t)}{u^\prime(t)\cdot t^2}\right\vert_{t=u^{-1}(s)}
$$
$$
\left.\frac{d^2}{ds^2}\frac{f(u^{-1}(s))}{u^{-1}(s)}=\frac{f^{\prime\prime}(t)t\cdot u^\prime(t)t^2-(f^\prime(t)t-f(t))(u^{\prime\prime}(t)t^2+2tu^\prime(t))}{(u^\prime(t))^3\cdot t^4}\right\vert_{t=u^{-1}(s)}
$$
Recall that $u^\prime <0$. Thus (B) will follow from (A) if for all $|t|\le 1$,
$$
f^{\prime\prime}\cdot t^3\cdot u^\prime\le (f^\prime\cdot t-f)(u^{\prime\prime}\cdot t^2+2tu^\prime)
$$
or,
$$
f^{\prime\prime}\cdot t^3\cdot (F-tF^\prime)\le (f^\prime\cdot t-f)((-tF^{\prime\prime}-2F^\prime+2t(F^\prime)^2/F)\cdot t^2+2t(F-tF^\prime))
$$
$$
\frac{f^{\prime\prime}\cdot t^3}{f-tf^\prime}\le\frac{F^{\prime\prime}\cdot t^3}{F-tF^\prime}-2t\frac{F-tF^\prime}{F}
$$
If this is going to be true for all convex $f$, it needs to be true when $F=2f$, which says $0\le t(F-tF^\prime)$..... ah, the sign change at $t=0$! This is why I split left and right.
%$$
%(f^{\prime\prime}\cdot t^3-2t^2f^\prime+2tf)\cdot (F-tF^\prime)\le (f^\prime\cdot t-f)((-tF^{\prime\prime}-2F^\prime+2t(F^\prime)^2/F)\cdot t^2)
%$$
%$$
%f^{\prime\prime}\cdot t\cdot u^\prime\le (f/t)^\prime(u^\prime t^2)^\prime
%$$
 A: Let's work where $-f(t)$ is convex instead of $f(t)$.
Note that $t_i=0$ can be omitted, as it adds nothing to (A) but increases (B).
First, we will only work with $t_i>0$ and the other half will work symmetrically. Call $$c_i:=\frac{1}{f(t_i)+f(-t_i)}$$ and let $T=\sum_{t_i>0}c_it_i$ be fixed. We can show $f(t)/t$ is strictly decreasing by $$\left(\frac{f}{t}\right)^\prime=\frac{-(f-tf^\prime)}{t^2}$$ and thinking of $f-tf^\prime\ge f(0)$ as the "Legendre transform" as in the OP graph. By the same argument, $t/(f(t)+f(-t))$ is strictly increasing.
Now, we would like to bound the quantity
$$
\frac{\sum_{t_i>0} c_it_i\cdot f(t_i)/t_i}{\sum_{t_i>0} c_it_i} \ \ \ \ \ \ \ (B')
$$
which is a weighted average of values of $f(t)/t$. Since $t/(f(t)+f(-t))$ is increasing, the largest possible $t_i$ under fixed $T$ occurs when $\#\{t_i>0\}=1$, i.e. $T=c_mt_m$ for some $m$. So, (B') finds its minimum under fixed $T$ at $c_m f(t_m)$.
Similarly, the minimum over $\{t_i<0\}$ occurs when $\#\{t_i<0\}=1$. The only configurations under $k=2$ are when $t_1=-t_2$, due to invertibility of $u(t):=t/(f(t)+f(-t))$. It follows that the minimum value of (B) when there are both positive and negative $t_i$ values is $1$.
I suspect the $1$ bound cannot be replaced by $t/2$ since that would appear to be true iff $|sf(u^{-1}(s))/u^{-1}(s)|$ were convex on each side $(-1,0)$ and $(0,1)$ separately.
It's interesting to note that $$\frac{d^2}{dt^2}\left|\frac{f(t)}{t}\right|=-\text{sgn}(t)\frac{t^2f^{\prime\prime}(t)-2f^\ast(t)}{t^3}$$ is positive when $1/2 f^{\prime\prime}t^2<f^\ast(t)$, which looks like it could be formed into a physics problem. If $f(t)$ is position over time, $1/2 f^{\prime\prime}t^2$ is a displacement under constant acceleration equal to $f^{\prime\prime}$. Also, $f^\ast$ represents a displacement under no accelaration. Or, rather, $1/2 f^{\prime\prime} t^2 \le f^\ast=tf^\prime-f$ when the best quadratic approximation $y=1/2 f^{\prime\prime}(x-t)^2+f^\prime(x-t)+f$ has its $y$-intercept on the same side of the x-axis as the graph of $f$.
