Given non-negative real numbers, $a, b, c$, show that $2a + 2ab + abc \leq 18$ when $a + b + c = 5$

Question

Given non-negative real numbers $a, b, c$, show that $$2a + 2ab + abc \leq 18$$ when $a + b + c = 5$

I have already started with AM-GM inequality, but now I am stuck.

From AM-GM inequality, I found that

$$abc \leq \frac{125}{27}$$

but now I am not sure what I have other two terms to finish the proof.

Maybe I am going in the wrong direction. Please help and thank you. It will be great if you can show a step-by-step solution.

Answer 1

The idea is to reduce the number of variables, adapting from this other solution.

1. Suppose that $b+c = k$, how can we maximize $2b + bc $? What is the maximum value in terms of $k$?

$ 2b + bc = b (2+c) = b ( 2 + k - b ) \leq ( \frac{2+k}{2} ) ^2$.
The maximum occurs when $ b = 2 + c = \frac{k+2}{2}$, with maximum value $ \frac{1}{4} ( k+2)^2$.

2. Hence, for fixed $a$, what is the maximum of $ a(2+2b+bc)$ subject to $b+c = 5 - a$?

It is $ a ( 2 + \frac{1}{4} ( 7-a)^2)$.

3. For $ a \in [ 0, 5 ]$, what is the maximum of the above cubic?

Using calculus, the maximum occurs when $a = 3$ (with $ b = 2, c = 0$) and has value 18.
If you don't want a calculus approach, see the link for how to maximize a cubic.