2018/006) Show that if $a+b$, $c+a$, $b+c$ are in HP $a^2,b^2,c^2$ are in AP

Because $a+b$, $c+a$, $b+c$ are in HP
hence
 $\frac{1}{a+b}$, $\frac{1}{c+a}$, $\frac{1}{b+c}$ are in AP
or
 $\frac{1}{b+c}-\frac{1}{c+a}= \frac{1}{c+a}-\frac{1}{a+b}$
or
$\frac{(c+a)-(b+c)}{(b+c)(c+a)}= \frac{(a+b)-(c+a)}{(c+a)(a+b)}$
or
$\frac{a-b}{(b+c)(c+a)}= \frac{b-c}{(c+a)(a+b)}$
or $(a-b)(a+b)=(b-c)(b+c)$
or $a^2-b^2 = b^2 -c^2$
or $a^2,b^2,c^2$ are in AP