For any three positive real numbers a,b,c: 9(25a^2+b^2)+25(c^2-3ac)=15b(3a+c).

Question

For any three positive real numbers $a,b,c$: $9(25a^2+b^2)+25(c^2-3ac)=15b(3a+c)$. Then:

Accepted Answer

Expanding: $225a^2+9b^2+25c^2-75ac=45ab+15bc$, i.e.
$(15a)^2+(3b)^2+(5c)^2-45ab-15bc-75ac=0.$
This is $\tfrac12\big[(15a-3b)^2+(3b-5c)^2+(5c-15a)^2\big]=0$, which forces
$$15a=3b,\quad 3b=5c,\quad 5c=15a.$$
So $b=5a$ and $c=3a$ (from $3b=5c$: $b=\tfrac{5c}{3}$, and $a=\tfrac{c}{3}$), giving $a+b = \tfrac{c}{3}+\tfrac{5c}{3}=2c$. Hence $b,c,a$ are in A.P.

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