Bound the distance between two vectors on the probability simplex

Question

Let $a,b$ be two vectors with strictly positive elements and $\delta = 1 - \frac{\langle a,b \rangle}{\|a\|\|b\|}$. Bound the following optimization problem as a function of $\delta$

$$\sup_{x>0} \frac{\langle a^2,x^2 \rangle}{\langle a,x \rangle^2} + \frac{\langle b^2,x^2 \rangle}{\langle b,x \rangle^2} - 2 \frac{\langle a \circ b,x^2 \rangle}{\langle a,x \rangle \langle b,x \rangle }$$

Here $a^2$ is the vector $a$ with all of its elements squared and $a\circ b$ is the element-wise product of $a$ and $b$.

If $\delta=0$, then the optimal value is 0. If $\delta \to 1$, then the optimal value is at most 2. Another way to express for the optimization problem is:

$$\sup_{x>0} \bigg\|\frac{a \circ x}{\langle a,x \rangle} - \frac{b \circ x}{\langle b,x \rangle} \bigg\|_2^2$$

Notice that $\frac{a \circ x}{\langle a,x \rangle}$ and $\frac{b \circ x}{\langle b,x \rangle}$ can be interpreted as two points on the probability simplex.

Answer 1

It turns out that $\delta = 0$ is the only case with a bound better than 2. Let $a = (1,t), b = (1,\epsilon^2)$ and $x = (\epsilon,1)$ for some $t,\epsilon > 0$. Then $$\bigg\|\frac{a \circ x}{\langle a,x \rangle} - \frac{b \circ x}{\langle b,x \rangle} \bigg\|_2^2 = \left(\frac{1}{1+\epsilon}-\frac{\epsilon}{t+\epsilon}\right)^2+\left(\frac{t}{t+\epsilon}-\frac{\epsilon}{1+\epsilon}\right)^2 = 2-O(\epsilon).$$ Hence, by letting $\epsilon \to 0$ we get arbitrarily close to the upper bound of 2. By selecting appropriate $t$, we get any correlation $\frac{\langle a, b\rangle}{\|a\|\|b\|} \in (0,1)$. So there is no better bound in terms of the correlation $1-\delta$.