Question
Let the positive real numbers $a, b, c, x$ be such that the numbers $ax + b , b*x + c$ and $cx + a$ are directly proportional to the numbers $c, a$ and$ b$. Show that $a = b = c$.
My idea
So we know that $\frac{ax+b}{c}=\frac{bx+c}{a}=\frac{cx+a}{b}=\frac{ax+bx+cx+a+b+c}{a+b+c}=\frac{(x+1)(a+b+c)}{a+b+c}=x+1$
$\frac{ax+b}{c}=x+1$
$ax+b=cx+c$
Observing that the equality is cyclice we can suppose WLOG that $a \leq b \leq c$
We let the case of equality and we focus on $a<b<c$
Using this we get that $ax<cx$ and $b<c$ so we get that $ax+b<cx+c$
Which means that we cant have the case $a<b<c$, which means that $a=b=c$
Hope one of you can tell me if my rationament ans proof are ok. Thank you!!!
Since they are in proportion let k be a real number such that $$ ax+b-ck=0\\ bx+c-ak=0\\ cx+a-bk=0\\ $$
Adding these three and cancelling (a+b+c) we get k=x+1 (which in in your solution)
Now we get by assuming no two are equal $$x=\frac{c-b}{a-c}=\frac{a-c}{b-a}=\frac{b-a}{c-b}\\ \ \\x^3=\frac{c-b}{a-c}\frac{a-c}{b-a}\frac{b-a}{c-b}=1\\ \ \\x=1$$
Now we get a,b,c and a,c,b are in A.P $$ a+c=2b\\ a+b=2c\\ c-b=2b-2c\\ b=c$$
Similarly we get a=b=c which contradicts our initial assumption that none of them are pairwise equal which means they are infact all equal
Ps.Its a bit weird using contradiction but its better keeping $\frac{0}{0}$ as far away as possible