What is the minimum value of $\frac{2a}{a+2b+2c} + \frac{2b}{b+2c+2a} +\frac{2c}{c+2a+2b}$ given $ a+b+c=2$? What is the minimum value of $\frac{2a}{a+2b+2c} + \frac{2b}{b+2c+2a} +\frac{2c}{c+2a+2b}$ given $ a+b+c=2$?
I know the answer is going to be $\frac{6}{5}$ because it will occur when there is equality between the three pronumerals.
The issue is this question was given to a group of students who have only learned the AM/GM inequality for two pronumerals.
I can do the question by using the AM/GM inequality with three pronumerals (along with a bit of not immediately obvious algebraic manipulation) however I'm struggling to see how I can solve this using the AM/GM inequality with only two values.
I've tried making 3 separate equations and adding them together, however so far I haven't been able to find how to do it.

This is the most basic solution I can think of. This method is useful even if you only know AM-GM. Also, when I starts to learn the inequalities, I do them like this.
To prove
$$\frac{2a}{a+2b+2c} + \frac{2b}{b+2c+2a} +\frac{2c}{c+2a+2b}\ge 6/5$$
Cancelling out the denominators
$$5(2a(b+2c+2a)(c+2a+2b)+2b(a+2b+2c)(c+2a+2b)+2c(a+2b+2c)(b+2c+2a))\ge 6(a+2b+2c)(b+2c+2a)(c+2a+2b)$$
Expand them all, we have
$$40(a^3+b^3+c^3)+80(a^2b+b^2c+c^2a+ab^2+bc^2+ca^2)+150abc\ge 24(a^3+b^3+c^3)+84(a^2b+b^2c+c^2a+ab^2+bc^2+ca^2)+174abc$$
Or,
$$16(a^3+b^3+c^3)\ge 4(a^2b+b^2c+c^2a+ab^2+bc^2+ca^2)+24abc$$
By AM-GM, we have
$$a^3+ab^2\ge 2\sqrt{a^3\times ab^2}=2a^2b$$
And similarly $b^3+ba^2\ge 2\sqrt{b^3\times ba^2}=2b^2a$. So add them together, we have
$$a^3+ab^2+b^3+ba^2\ge 2a^2b+2b^2a$$
And therefore,
$$a^3+b^3\ge a^2b+b^2a$$
So we have
$$8(a^3+b^3+c^3)=4(a^3+b^3)+4(c^3+b^3)+4(a^3+c^3)\ge 4(a^2b+b^2a)+4(c^2b+b^2c)+4(a^2c+c^2a)=4(a^2b+b^2c+c^2a+ab^2+bc^2+ca^2)$$
So what remains is $$8(a^3+b^3+c^3)\ge 24abc$$
However, by above, we know that
$$8(a^3+b^3+c^3)\ge 4(a^2b+b^2c+c^2a+ab^2+bc^2+ca^2)$$
So we need to prove that
$$4(a^2b+b^2c+c^2a+ab^2+bc^2+ca^2)\ge 24abc$$
Which is also AM-GM:
$$4(a^2b+b^2c+c^2a+ab^2+bc^2+ca^2)=4((a^2b+c^2b)+(b^2c+a^2c)+(c^2a+b^2a))\ge 4(2\sqrt{a^2b\times c^2b}+2\sqrt{b^2c\times a^2c}+2\sqrt{c^2a\times b^2a})=4(2abc+2abc+2abc)=4\times6abc=24abc$$
