We often encounter inequalities of symmetric expressions, i.e. the expression doesn't change if the variables in it are interchanged, with the prior knowledge of a certain relation between those variables. In all such cases that I have encountered thus far, we can find the extremum of the expression by letting the variables equal. Here are a few examples:
$(1)$ Given $a + b + c = 3$ ($a,b,c\in\Bbb{R}$), the extremum of $ab + bc + ca$ is achieved when $a = b= c$. Hence the extremum is $3$, which is expected.
$(2)$ In a triangle $ABC$, the extremum of $\cos A + \cos B + \cos C$ is achieved when $A = B = C$. With the knowledge that $A + B + C = \pi$, the extremum is $\frac32$, which is, again, expected.
$(3)$ Similar to the previous one, in a triangle $ABC$, the extremum of $\sin \frac{A}2\sin\frac{B}2\sin\frac{C}2$ is achieved when $A = B = C = \frac{π}{3}$. The extremum is $\frac{1}{8}$. Expected, once again.
What I have noticed is this technique doesn't work when expressions remain invariant under a cyclic shift of the variables. So, with the groundwork laid, here are my questions. If this technique IS valid, then
$1.$ How can one go about proving that the technique works?
$2.$ How do I know if the extremum is a maximum or a minimum. Moreover, are these local extrema or global extrema?
$3.$ Does this technique have an "official" name?
If the technique is NOT valid, please provide explanations and counterexamples.
EDIT: The values encountered as extrema above were expected since we have proofs for the individual inequalities. So please don't provide proofs for them, as answers. Instead, what I am specifically looking for is a general proof that expressions completely symmetric in their variables indeed achieve their extrema when the variables equal each other.
For example.
$\max(xy+xz+yz)=3$ because $xy+xz+yz\leq\frac{1}{3}(x+y+z)^2\Leftrightarrow$ $(x-y)^2+(x-z)^2+(y-z)^2\geq0$ with equality for $x=y=z$.
The extremum indeed achieved when $x=y=z=1$, but without proof like my it's nothing!
$\max(\cos\alpha+\cos\beta+\cos\gamma=\frac{3}{2}$ because $$\cos\alpha+\cos\beta+\cos\gamma\leq\frac{3}{2}\Leftrightarrow$$ $$\Leftrightarrow1-2\sin^2\frac{\alpha}{2}+2\sin\frac{\alpha}{2}\cos\frac{\beta-\gamma}{2}\leq\frac{3}{2}\Leftrightarrow$$ $$\Leftrightarrow4\sin^2\frac{\alpha}{2}-4\sin\frac{\alpha}{2}\cos\frac{\beta-\gamma}{2}+1\geq0,$$ which is obvious. The extremum indeed occurs for $\alpha=\beta=\gamma$, but without proof it's nothing!
The following inequality is symmetric, but the equality does not occur for equality case of variables.
Let $a$, $b$ and $c$ be non-negative numbers such that $a+b+c=3$. Prove that: $$(a^2-ab+b^2)(a^2-ac+c^2)(b^2-bc+c^2)\leq12$$ The maximum value is $12$, but for $a=b=c=1$ we get $3$.