TITLE: If a,b,c are in AP and $a^2,b^2,c^2$ are in HP, then prove either $a=b=c$ or $a,b,- \frac c2 $ are in GP
QUESTION [5 upvotes]: As the title says. 

Although first part of the proof is obvious, I'm still able to prove it.
And for the second part, I'm essentially trying to prove $b^2=-c/a$ (which is possible only when c<0 Xor a<0).
The relations found by me are: $a^2+c^2+2ac=4b^2$ and $b^2=\left(\frac {2a^2c^2}{a^2+c^2} \right)$. 
Which provides me with:
$(a^2+c^2+2ac)/4=(2a^2c^2)/(a^2+c^2)$
I don't think this would lead me to the answer. Any help would be appreciated.

REPLY [4 votes]: The AP condition gives $2b = a+c$ and the HP condition gives
$$\frac2{b^2}= \frac1{a^2}+\frac1{c^2} \iff \frac8{(a+c)^2}=\frac{a^2+c^2}{a^2c^2} \iff (a-c)^2(a^2+4ac+c^2)=0$$
Now either $(a-c)^2=0 \implies a=b=c$
or $a^2+4ac+c^2 = 0 \implies (a+c)^2+2ac=0 \implies 2b^2+ac=0 \implies a, b, -\dfrac{c}2$ are in GP.