If a, b, c are in H.P.,then | vedclass

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(C) Given that $a, b, c$ are in $H.P.$,we have $b = \frac{2ac}{a+c}$.Since $a, b, c$ are positive real numbers,by the property of $A.M. > H.M.$,we hav

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$a^2 + c^2 > 2b^2$

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(C) Given that $a, b, c$ are in $H.P.$,we have $b = \frac{2ac}{a+c}$.Since $a, b, c$ are positive real numbers,by the property of $A.M. > H.M.$,we have $\frac{a+c}{2} > b$.Squaring both sides,we get $\left(\frac{a+c}{2}\right)^2 > b^2$,which implies $\frac{a^2 + 2ac + c^2}{4} > b^2$.Since $b = \frac{2ac}{a+c}$,we have $ac = \frac{b(a+c)}{2}$.Alternatively,using the power mean inequality,for $n=2$,$\frac{a^2+c^2}{2} > \left(\frac{a+c}{2}\right)^2$.Since $\frac{a+c}{2} > b$,it follows that $\left(\frac{a+c}{2}\right)^2 > b^2$.Therefore,$\frac{a^2+c^2}{2} > b^2$,which simplifies to $a^2 + c^2 > 2b^2$.

If $a, b, c$ are in $H.P.$,then

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