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.
 A: Here's a "Only 2-variable AM-GM" approach.
(Per the comments: Since the expression is homogenous, I'm working with $a+b+c = 3 $ to simplify the working. There isn't a huge difference with working with $a+b+c = 2$, but the constants get uglier.
Also I'm restricting to positive values, since otherwise there is no minimum like when $a \rightarrow 6^+$.)
Some of the takeaways of using AM-GM (I have a much longer list from my class.)

*

*We can apply it termwise to a sum (likewise product).

*We can cancel a denominator with a numerator, so we might want to force that term in.

*Equality holds when the terms of the AM-GM are equal. If not, try to tweak with constants.

Given those specific ideas, what ways can we bound $ \frac{ 2a} {6-a}$ from below via AM-GM, with equality at $ a = 1$, in the domain $ a \in [ 0, 3 ] $?
Notice that $ \frac{2a}{6-a} = \frac{12}{6-a} - 2. $
Let's apply AM-GM to $ \frac{12}{6-a} +  ( 6-a) \times k \geq 2\sqrt{12\times k } $. We want equality at $ a = 1$, which means $ k = \frac{12}{25}$. Thus:
$$ \frac{12 }{6-a} + (6-a) \times \frac{12}{25} \geq \frac{24}{5}$$
As such, this gives us
$$ \frac{2a}{6-a} \geq \frac{24}{5} - (6-a)\times \frac{12}{25} - 2 = \frac{12a}{25} - \frac{2}{25}. $$
Finally, sum it up to get
$$ \sum \frac{2a}{6-a} \geq \sum \left( \frac{12a}{25}  - \frac{2}{25} \right) = \frac{ 12\times 3 }{ 25} - 3 \times \frac{2}{25} = \frac{6}{5}.$$
Notes:

*

*A more direct approach is to realize that the inequality $ \frac{ 2a}{6-a} \geq \frac{ 12a}{25} - \frac{2}{25} $ is known as the "Tangent Line Method". It can easily be verified if we knew the inequality beforehand (EG polynomial inequality), but deriving it takes some work (EG Calculus, or the above).

*It's hard to AM-GM $\frac{2a}{6-a}$ directly, since we likewise can't easily manipulate $\frac{6-a}{2a}$, and using $ k\times (6-a)$ leaves us with $ \sqrt{a}$ that we can't manipulate. Hence, the switch to a term where the numerator is constant, which allows us to do$ \frac{12}{6-a} + k \times ( 6 -a ) \geq 2 \sqrt{12 \times k } $.

*For a further challenge, find the minimum subject to $ a^2 + b^2 + c^2 = 3$ using the above ideas.

A: 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:
$$4(a^2b+b^2c+c^2a+ab^2+bc^2+ca^2)=4((a^2b+c^2b)+(b^2c+a^2c)+(c^2a+b^2a))\ge 4(2\sqrt{a^2b\times c^2b}+2\sqrt{b^2c\times a^2c}+2\sqrt{c^2a\times b^2a})=4(2abc+2abc+2abc)=4\times6abc=24abc$$
A: $a+b+c=2$ Minimize: $\frac{2a}{a+2b+2c}+\frac{2b}{2a+b+2c}+\frac{2c}{2a+2b+c}$
$a+2b+2c=4-a.$
So goal is to minimize $\frac{2a}{4-a}+\frac{2b}{4-b}+\frac{2c}{4-c}$
Note the constraint function nor the target condition are effected by swapping any two of the variables.
Lagrange Multipliers can be used to derive 3 separate equations.
$1=\lambda(\frac{8}{(4-a)^2})$
Because of the symmetry of the problem, something identical can be said about $b$ and $c$. So they are equal to each other, and, by the constraint, 2/3.
