Help with arguing against symmetry proofs. I am asking for the help of mathematicians here to show my colleague that her arguments are wrong. I tried my best to explain that inequalities and equations can be solved like that, but maybe my arguments were not good enough.
Some of the examples of her solutions:
Example 1. Over the real numbers, if $x+y=2$, what is the maximum value of $xy$?
Solution: By symmetry, there must be a maximum or minimum when $x=y$ so $x=y=1$ and $xy=1$. This is clearly a maximum, since if say $x=0$ and $y=2$, you get a lower value.
Example 2. Let $x, y$ and $z$ be positive real numbers such that $x + y + z = 1$. Prove that $\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2} \geq 27$.
Solution: By symmetry, there is a minimum at x=y=z hance all = 1/3 and the expression equals 27.
Example 3. For positive numbers $a,b,c$ prove that $\sqrt{\frac{a}{a+b}}+\sqrt{\frac{b}{b+c}}+\sqrt{\frac{c}{c+a}} \leq \frac{3\sqrt{2}}{2}$?
Solution: Another of those symmetry arguments.  The LHS is a maximum for $a=b=c$ and then equals the RHS.

It is very clear, that these are not even close to be called solutions. When I (or my other collegaues) try to explain that it not that simple to show that extreme points will be at $x=y=z$ we get answers 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.


have repeatedly said that symmetry arguments do not always work.  However, it must be obvious to any bright 18 year old that the argument does work in this particular case.


Much to my amazement, many people in this group do not understand how to solve problems using symmetry. In England, this is taught in schools.

I am not sure what will help us to explain that it does not work like that. If we will get many reponces here, I believe it will help to convince my colleagues. Also, if you are from UK or have degree from prestigous college it will be helpfull to mention that.
 A: 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  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.
A: Example 1 does not seem so bad at first sight. It is true that by symmetry (and smoothness of the functions) $x=y$ is a stationary point, hence it probably forms a local maximum or minimum.
Anyway, exhibiting a lower point is not enough to prove that we have a global maximum.
If we absorb the constraint with $x=1-t,y=1+t$, we only have to maximize $1-t^2$ and it turns out that the unique stationary point is a maximum, which favors your colleague.
Now, on the opposite, the function $t^2-t^4$, corresponding to the maximization of
$$(4xy+1)(x^2+y^2)-x^4-y^4-6x^2y^2-2xy$$ under $x+y=2$, induces an asymmetric solution.
