FE ^_^

For discussing Olympiad Level Algebra (and Inequality) problems
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Atonu Roy Chowdhury
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Location: Chittagong, Bangladesh

FE ^_^

Unread post by Atonu Roy Chowdhury » Thu May 04, 2017 7:59 pm

Find all $f:\mathbb{R}^+ \rightarrow \mathbb{R}^+$ such that for all distinct $x,y,z$
$f(x)^2-f(y)f(z)=f(x^y)f(y)f(z)[f(y^z)-f(z^x)]$
This was freedom. Losing all hope was freedom.

User avatar
Atonu Roy Chowdhury
Posts: 63
Joined: Fri Aug 05, 2016 7:57 pm
Location: Chittagong, Bangladesh

Re: FE ^_^

Unread post by Atonu Roy Chowdhury » Thu May 04, 2017 8:05 pm

Multiplying $f(x)$ we get $f(x)^3 = f(x)f(y)f(z)[1+f(x^y)-f(y^z)]$ and two similar expressions for $f(y)^3$ and $f(z)^3$ . Adding them, we get $ f(x)^3+ f(y)^3+ f(z)^3 = 3f(x)f(y)f(z)$ . Equality case of AM-GM! So, $f(x) = c$ where $c \in \mathbb{R}^+$
This was freedom. Losing all hope was freedom.

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