If $a^{2}, b^{2}, c^{2}$ are in $A . P .$ show that $b + c, c + a, a + b$ are in $H . P .$

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Solution

$a^{2} + c^{2} = 2 b^{2}$
if $a, b, c$ are in H.P
$\frac{1}{a} + \frac{1}{c} = \frac{2}{b}$
$\frac{1}{b + c} + \frac{1}{a + b} = \frac{2}{a + c}$
$(a + 2 b + c) (a + c) = 2 (b + c) (a + b) ⟹ a^{2} + 2 a b + a c + a c + 2 b c + c^{2} = 2 a b + 2 a c + 2 b^{2} + 2 b c ⟹ a^{2} + c^{2} = 2 b^{2}$

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