<s>[INST] If $a+b+c=3,$ show $a+b+c\geq ab+bc+ac$ by using AM-GM A am having issues with a problem that states: let $a, b, c$ be positive real numbers such that $a + b + c = 3$. Show that $$a^bb^cc^a\le1$$
As this was an example problem, a solution path was provided but I am confused as to why the solution works. It writes:
$$1=\frac{a+b+c}{3}$$
$$\ge\frac{ab+bc+ca}{a+b+c}$$
$$\ge (a^bb^cc^a)^{\frac{1}{a+b+c}}$$
I understand how weighted AM-GM could show the second inequality but I can't seem to understand how one can show $a+b+c \ge ab+bc+ca$ which is what the first inequality implies.
Btw: at this point in the text, they have only proved AM-GM and weighted AM-GM. I can think of a way to show this using Chebyshev's Inequality but I was hoping to try to understand what the text meant by this specific solution path.
 [/INST] Hint
$$(a+b+c)^2=a^2+b^2+c^2+2(ab+ac+bc)$$
but, by $MA\ge MG$:  $a^2+b^2\ge 2ab, a^2+c^2\ge 2ac, b^2+c^2\ge 2bc$, what gives you
$$a^2+b^2+c^2\ge ab+ac+bc.$$
Can you finish?
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