If $a$, $b$, $c$ are three positive numbers in harmonic sequence, show that $a^2+c^2>2b^2$.

Question

If $a$, $b$, $c$ are three positive numbers in harmonic sequence, show that $a^2+c^2>2b^2$.

My attempt : $(a-c)^2>0$

$a^2+c^2>2ac$

This is true for any two real positive values of $a$ and $c$.

Again,

If $a$, $b'$, $c$ are in geometric progression then

$b'^{2} = ac$. So

$a^2+b^2>2b'^{2}$

Answer 1

$\implies b=\dfrac{2ca}{c+a}$

$$\left(\dfrac ab\right)^2+\left(\dfrac cb\right)^2=(c+a)^2\left(\dfrac1{4c^2}+\dfrac1{4a^2}\right)$$

Now $(c+a)^2-4ca\ge0\iff(c+a)^2\ge4ca$

and $\dfrac{\dfrac1{4c^2}+\dfrac1{4a^2}}2\ge\dfrac1{4ca}$

Both equality occur if $c=a$

Answer 2

A.M.-G.M.-H.M.

$a^2+c^2\ge 2ac=2(\sqrt{ac})^2\ge 2\left(\frac{2}{\frac{1}{a}+\frac{1}{c}}\right)^2=2b^2$ with equality holds if and only if $a=b=c$.