Inequalities of expressions completely symmetric in their variables 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.
 A: For example.


*

*$\max(xy+xz+yz)=3$ not because it happens for $x=y=z=1$.


$\max(xy+xz+yz)=3$ 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!


*$\max(\cos\alpha+\cos\beta+\cos\gamma=\frac{3}{2}$ not because it happens for $\alpha=\beta=\gamma=60^{\circ}$. 


$\max(\cos\alpha+\cos\beta+\cos\gamma=\frac{3}{2}$ 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 $3$.
