Where is my solution wrong for this AM-GM Maxima question?

Question

Question

For $x> 0$, maximize $f(x)=(1+x)(1+x)(1-x)$.

My solution Using AM-GM for two variables, $a$ and $b$, $a = (1+x)^4$ and $b = (1-x)^2$, we get f(x) is maximum when $a=b$.

But when I solve for $x$ by equating $a=b$, it has a negative root and 2 complex roots.

The correct answer is $1.185$.

Where is my solution wrong? How can we solve this using two variables by AM-GM inequality?

Thank you.

Answer 1

$$2f(x) = (1+x)(1+x)(2-2x)\leq \Big({1+x+1+x+2-2x\over 3}\Big)^3 = \Big({4\over 3}\Big)^3$$

Equality is iff $1+x=2-2x$, i.e. $x=1/3$ and $y_{\max} = {32\over 27}$

Answer 2

Perhaps given this is a cubic, calculus is simpler. You are maximizing $$ \begin{split} f(x) &= -x^3-x^2 + x + 1\\ f'(x) &= -3x^2 -2x + 1\\ f''(x) &= -6x-2 = -2(x+3) \end{split} $$ and $f'(x) = 0$ yields via the quadratic formula $$ x_\pm = \frac{2 \pm 4}{-6} = \{-1,1/3\}. $$ Note that $f''(x) < 0$ for $x>-3$. Can you finish?

Answer 3

You did it correctly, but your AM is not a constant, but a variable. $$(1+x)^4+(1-x)^2\ge 2(1+x)^2|1-x|,$$ the equality occurs for $(1+x)^4=(1-x)^2 \Rightarrow x=-3,0$. Indeed, the equality occurs, but it does not mean it is the maximum value.

Similarly, you can consider $a=\frac38(1+x)^4,b=\frac83(1-x)^2$: $$\frac38(1+x)^4+\frac83(1-x)^2\ge 2(1+x)^2|1-x|,$$ the equality occurs for $a=b \Rightarrow x=-5,\frac13$ and this time the maximum occurs at $x=\frac13$. How do I know? I found the answer other way and chose such $a$ and $b$ deliberately. What is the other way? You can see other answers. Note that in Aqua's answer the AM (the right hand side) is a constant number, so it is true for all values of $x$. Here I just wanted to answer your question: "Where is my solution wrong?"