$ f(x^n)\le f(x)f(x^2)^{\frac{1}{2}}\dots f(x)^{\frac{1}{n}} $ or $ f(x^n)\le f(x)f(x^2)^{\frac{1}{2}}\dots f(x^n)^{\frac{1}{n}} $ ?

nice problem!

I've checked the CMO1993 problem list. Ysymyth is right! Set $F_n(x)=f(x)f(x^2)^{\frac{1}{2}}\cdots f(x^n)^{\frac{1}{n}}$ Use Mathematical Induction to prove \[\forall n\in \mathbb{N}^+,f(x^n)\le F_n(x) \]

ysymyth wrote: $ f(x^n)\le f(x)f(x^2)^{\frac{1}{2}}\dots f(x)^{\frac{1}{n}} $ or $ f(x^n)\le f(x)f(x^2)^{\frac{1}{2}}\dots f(x^n)^{\frac{1}{n}} $ ? Obviously it must be a typo and it has been corrected now.

smy2012 wrote: I've checked the CMO1993 problem list. Ysymyth is right! Set $F_n(x)=f(x)f(x^2)^{\frac{1}{2}}\cdots f(x^n)^{\frac{1}{n}}$ Use Mathematical Induction to prove \[\forall n\in \mathbb{N}^+,f(x^n)\le F_n(x) \] How do you do the induction step?

Nobody???

Maybe it will help you. I think about proof too. For $n=3$ the procedure is $f(x^3)^\frac{1}{3}f(x^2)^\frac{1}{2}f(x)\ge f(x^3)\iff f(x^2)^3f(x)^6\ge f(x^3)^4$ and that's true because $LHS\ge f(x^2)^4f(x)^4\ge f(x^3)^4$

Inductive step is basically $f(x^{n+1})^n\le f(x^n)^{n+1}$ which is trivial for $n=1$ but for $n\ge 2$ it's hard

Nobody??

i dunno how to use induction, but the below ineq might be helpful? $f(x)f(x^{n-1})\geq f(x^n)$

$f(x^{n+1})^n\le f(x^n)^{n+1}$ is too strong and your inequalities won't help here.

ThE-dArK-lOrD wrote:

Thank you so much.

Let $F_n(x) = f(x)f(x^2)^{\frac{1}{2}} \cdots f(x^n)^{\frac{1}{n}}$. We make the following claim: Claim: For $n \geq 3$, $F_n(x) \geq f(x^n)$. Proof: The base case for $n = 3$ is as follows: \begin{align*} & f(x^2) \leq f(x)^2 \\ \implies& f(x^2)^4 \leq f(x)^2 f(x^2)^3 \\ \implies& f(x^2)^4 f(x)^4 \leq f(x)^6 f(x^2)^3 \\ \implies& f(x^3)^4 \leq f(x^2)^4 f(x)^4 \leq f(x)^6 f(x^2)^3 \\ \implies& f(x^3)^6 \leq f(x)^6 f(x^2)^3 \\ \implies& f(x^3) \leq F_3(x). \phantom{c} \square \end{align*} Observe that $F_{n + 1}(x)^{n + 1} = F_n(x)^{n + 1}f(x^{n + 1})$. Now for all $k \leq n$, assume $F_k(x) \geq f(x^k)$. We show that $F_{n + 1} (x) \geq f(x^{n + 1})$: \begin{align*} F_{n + 1}(x)^{n + 1} &= F_n(x)^{n + 1}f(x^{n + 1}) \\ &= F_n(x)^n F_n(x) f(x^{n + 1}) \\ &= F_{n - 1}(x)^{n - 1} {[F_{n - 1}(x)f(x^n)]}{[F_n(x)f(x^{n + 1})]} \\ &\vdots \\ &= F_1(x) {[F_1(x)f(x^2)]} \cdots {[F_{n - 1}(x)f(x^n)]} {[F_n(x)f(x^{n + 1})]} \\ &= {[f(x)f(x^2) \cdots f(x^{n + 1})]} \cdot {[F_1(x)F_2(x) \cdots F_n(x)]} \\ &\geq {[f(x)f(x^2) \cdots f(x^n)]} \cdot f(x^{n + 1}) \cdot {[f(x)f(x^2) \cdots f(x^{n + 1})]}, \text{ by our induction hypothesis.} \\ &= \underbrace{{[f(x)f(x^n)]}{[f(x^2)f(x^{n - 1})]} \cdots {[f(x^n)f(x)]}}_{\text{n - pairs}} f(x^{n + 1}) \\ &\geq f(x^{n + 1})^{n + 1}, \end{align*}so that $F_{n + 1}(x) \geq f(x^{n + 1})$, as desired. $\blacksquare$