How do you find $ max(f)$ by AM-GM?

Well, I think the point of AM-GM is, that $ \left(\frac {a}{\lambda} - 1\right)(a\lambda - 1) \leq (a - 1)(a - 1)$ for every $ \lambda > 1$, which can be easily proved by the obvious inequality $ \frac {1}{\lambda} + \lambda\geq 2$. Therefore maximum of $ f$ is, when all $ x_{i}$ are equal.

Sorry what's the theory ?

britishprobe17 wrote: Sorry what's the theory ? wdym

I think L-b is wrong, as the variable ( the line ) is not bigger then 1, but is 1.

Umm. I did not say that

peelybonehead wrote: britishprobe17 wrote: Sorry what's the theory ? wdym i got that, my bad

Rephrase the problem as follows: Given $b_1, b_2, \cdots, b_n \in \mathbb{R}^+$ with $\prod_i (b_i + 1) = 3^n$, find the possible values of $\prod_i b_i$. Now note that by taking $b_2$ till $b_n$ tending to 0, and $b_1$ tending to $3^n - 1$, $b_1 \cdots b_n > 0$, where we can get arbitrarily close to 0.

the answer is $(0, 2^n]$.

Solved with KISAH_SANGJUARA. All the possible values that $p(1)$ can take is in the interval $[0,2^n]$. First we prove that $p(1) \in [0,2^n]$. Since all the roots of $p(x)$ is $\le 1$, we can write $$p(x)=(x+x_1-1)(x+x_2-1)\cdots(x+x_n-1)$$for some nonnegative values $x_1,x_2,\cdots x_n$. We have $$p(2)=(1+x_1)(1+x_2)\cdots(1+x_n) \Rightarrow 3^n=(1+x_1)(1+x_2)\cdots(1+x_n)$$By AM-GM, we have $$1+x_i = 1+\frac{x_i}{2}+\frac{x_i}{2}\ge 3\sqrt[3]{\frac{x^2_i}{4}}$$. Multiplying this for $i=1,2,\cdots,n$ yields $$3^n=(1+x_1)(1+x_2)\cdots(1+x_n) \ge 3^n\sqrt[3]{\frac{(x_1x_2\cdots x_n)^2}{4^n}}$$$$\Rightarrow 2^n \ge x_1x_2\cdots x_n$$. Clearly $p(1)=x_1x_2\cdots x_n \ge 0$, hence $p(1) \in [0,2^n]$. Now we prove that every value in $[0,2^n]$ is achievable. Lemma. There exist nonnegative reals $x_1,x_2$ satisfying $$(1+x_1)(1+x_2)=9$$and $x_1x_2=p$ for all $p \in [0,4]$. Proof. Notice that the equation $$p+x_1+\frac{p}{x_1}=8 \Rightarrow x^2_1+(p-8)x_1+p=0$$has a positive root. Indeed, just choose $$x_1=\frac{8-p+\sqrt{(p-16)(p-4)}}{2}$$. It follows that $x_2=\frac{p}{x_1}$ is nonnegative, also $x_1,x_2$ satisfies the required condition. For $n=2$, it follows from the lemma that every value in $[0,2^n]$ is achievable For $n>2$, choose $x_i=2$ for every $i=3,\cdots,n$. This reduces the case to $n=2$, then it also follows from the lemma that every value in $[0,2^n]$ is achievable. As desired.