The random variables $W_1,\dots,W_n$ are all equal in distribution since $X_1,\dots,X_n$ are identically distributed. Thus $E(W_1) = \dots = E(W_n)$. This is the "by symmetry" part of the argument.
Let $\mu$ denote the law of $X_1$ (which is the same as the law of any of $X_2,\dots,X_n$ by assumption). The intuition for why $W_1,\dots,W_n$ are equal in expectation is that we form $W_1,\dots,W_n$ by taking $n$ random samples from $\mu$, and then picking any one of the samples as the numerator, and using the sum of the samples as the denominator. Since the numerator samples are all coming from the same distribution, we expect $E(W_i)$ to be a quantity that does not depend on the particular index $i$.
You can prove for instance that $E(W_1) = E(W_2)$ as follows. By independence of $X_1,\dots,X_n$, the pushforward $(X_1,\dots,X_n)_*P = \mu\otimes\dots\otimes \mu$ ($n$ times), so by Fubini's theorem and the change of variable
$$
(x_1,x_2,\dots,x_n)\mapsto (x_2,x_1,\dots,x_n),
$$
we have:
\begin{align*}
E(W_1) &= E\bigg(\frac{X_1}{X_1+X_2+\dots+X_n}\bigg) \\
&= \int_{\mathbf R^{n}}\frac{x_1}{x_1+x_2+\dots+x_n}\,\mu(dx_1)\mu(dx_2)\dots\mu(dx_n)\\
&= \int_{\mathbf R^n}\frac{x_2}{x_1+x_2\dots+x_n}\,\mu(dx_2)\mu(dx_1)\dots\mu(dx_n)\\
&= E(W_2).
\end{align*}
The issue is that your colleague seems to be confusing a useful heuristic with a proof. If the principle of why something 'must be true' by symmetry cannot be articulated and only applied ex post, then that's a red flag that it's not a proof. One approach is to suppose your colleague is right in 2 simple separate cases, then combine them (via multiplication in this case) and show that the 'principle' breaks.
A better way to view 'symmetry arguments' when looking at inequalities involving symmetric functions like example 1, is through majorization. The deeper reason example 1 holds is that $x_1\cdot x_2$ is symmetric and in fact a Schur concave function and the constraint $x_1+x_2=2$ ensures that $\mathbf x =\mathbf 1$ is majorized by any other candidate. By Schur concavity this is a maximum.
Alternatively consider $\mathbf x \in \mathbb R^2_{\geq 0}$ and $\mathbf x\mapsto x_1^4 + x_2^4$ with the constraint $x_1+x_2=2$. This is symmetric and in fact a Schur convex function. The candidate $\mathbf x =\mathbf 1$ is a minimum since it is majorized by all other allowed choices of $\mathbf x$ and Schur convexity ensures that is a minimum.
Finally consider consider $\mathbf x \in \mathbb R^2_{\geq 0}$ and $\mathbf x\mapsto (x_1\cdot x_2)\cdot(x_1^4 + x_2^4)$ with the constraint $x_1+x_2=2$. This is a symmetric function but neither (Schur) concave nor convex. Being majorized by all choices of $\mathbf x$ (what your colleague calls the symmetric solution) doesn't tell you anything definitive. Hence applying a heuristic like "It's a symmetry thing. If you imagine that say 𝑎 tends to 0 or infinity you see that the LHS is going to decrease" misleads you despite the fact that this is a symmetric function. And $\mathbf x =\mathbf 1$ is neither a maximum or a minimum, as $0$ is a clear minimum e.g. as $x_1\to 0$ but e.g. $x_1=1.5, x_2=0.5$ is much larger than the constant solution.
conclusion: your colleague's heuristic is true when dealing with symmetric functions that are Schur convex (concave). But not all functions are Shur convex (concave); understanding this is vital. The theory of majorization is a well developed mathematical theory quite distinct from declaring things are obvious.
Best Answer
A cubic has a point of symmetry at the inflection.
Taking the second derivative,
$$12x-2b=0,$$
so $\left(\dfrac b6,-\dfrac{b^3}{54}+\dfrac{bc}6\right)$.