Maximum value of $x+2 x y+4 x y z$ Given that $x,y,z \in R^+$ such that $x+y+z=1.5$. Without using calculus find the maximum value of $x+2 x y+4 x y z $
My try:
We can write $S=x+2 x y+4 x y z =x(1+2y+4yz)$
$\implies$
$$S=x((1+2y)(1+2z)-2z)$$
$\implies$
$$S=\frac{1}{2}(1+2 x-1)(1+2 y)(1+2 z)-2 x z$$
$\implies$
$$S=\frac{1}{2}(1+2 x)(1+2 y)(1+2 z)-\frac{1}{2}(1+2 y)(1+2z)-2xz$$
By $AM-G.M$, we have
$$(1+2 x)(1+2 y)(1+2 z) \leq\left(\frac{3+2(x+y+z)}{3}\right)^{3}$$
$\implies$
$$(1+2 x)(1+2 y)(1+2 z) \leq 8$$
Any hint from here?
 A: The idea is to reduce the number of variables.
EG We could use the substitution $ z = 3/2 - y - x $ to make this a 2 variable inequality.
How can we make this a 1 variable inequality?
Work through the following. If you're stuck, explain what you've tried.


*

*(In hindsight, because) The maximum occurs at $ (1, 0.5, 0)$, so we will allow for non-negative reals (or deal with finding the supremum instead, your choice).


*We use a substitution $ 2x = a, 2y = b, 2c = z$ to simplify the problem

$a+b+c = 3$ are non-negative reals, what is the maximum of $a+ab+abc$?



*Suppose that $  b + c = k$, how can we maximize $ b+ bc$? What is it's value in terms of $k$?


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



*Hence, what is the maximum of $ a ( 1 + b+bc) $, subject to $ b+c = 3-a$?


 We want to find the maximum of $ a ( 1 + \frac{1}{4} ( 4 - a ) ^2 ) $, subject to $ 0 \leq a \leq 3$.

Use your favorite approach to maximize the above cubic in a restricted range.
The goal is to find the turning points of the cubic. Once we have those, we can check against the end points to find the max/min.
I personally would use calculus here, but that's not absolutely necessary:

 Consider $Y(a) = a^3 - 8a^2 +20a $, which is our expanded cubic after multiplying by 4.
 We want to find a $D$ such that $Y - D = a^3 - 8a^2 + 20a - D$ can be written in the form $ (a - p)(a-q)^2 $, then $ a = q$ will be a turning point.
 Expanding and equating terms, we have $ p+2q = 8, q(2p+q) = 20, pq^2 = D$.
 Solving the first 2 equations and substituting it into the 3rd, we have 2 solutions: $ ( p,q, D ) = (4, 2, 16), ( 4/3, 10/3, 400/27)$.
 So the local maximum is at $a = 2$, the local minimum is at $ a = 10/3$.
 Finally, the maximum of $ Y(a)$ on $[0,3]$ is the max of $ Y(0), Y(2), Y(3)$, which is 16.

Hence, the maximum occurs at $ a = 2, b = 1, c = 0$.
Substituting this back, we get   $ x + 2xy + 4xyz \leq 2$ with equality at $ (1, 0.5, 0)$.
