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

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(C) Given that $a, b, c$ are in $H.P.$ (Harmonic Progression).By definition,$b = \frac{2ac}{a+c}$.We know that for any two positive numbers $a$ and $c

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

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(C) Given that $a, b, c$ are in $H.P.$ (Harmonic Progression).By definition,$b = \frac{2ac}{a+c}$.We know that for any two positive numbers $a$ and $c$,the Arithmetic Mean $(A.M.)$ is greater than the Harmonic Mean $(H.M.)$.$A.M. = \frac{a+c}{2}$ and $H.M. = b$.So,$\frac{a+c}{2} > b$.Using the Power Mean Inequality,for $n=2$,the quadratic mean is greater than the harmonic mean.Alternatively,consider the expression $a^2 + c^2 - 2b^2$.Substituting $b = \frac{2ac}{a+c}$,we get $a^2 + c^2 - 2(\frac{2ac}{a+c})^2 = a^2 + c^2 - \frac{8a^2c^2}{(a+c)^2}$.$= \frac{(a^2+c^2)(a+c)^2 - 8a^2c^2}{(a+c)^2} = \frac{(a^2+c^2)(a^2+c^2+2ac) - 8a^2c^2}{(a+c)^2}$.$= \frac{(a^2+c^2)^2 + 2ac(a^2+c^2) - 8a^2c^2}{(a+c)^2} = \frac{(a^2-c^2)^2 + 2ac(a^2+c^2-2ac)}{(a+c)^2} = \frac{(a^2-c^2)^2 + 2ac(a-c)^2}{(a+c)^2}$.Since this expression is always positive for $a 
eq c$,we have $a^2 + c^2 > 2b^2$.

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

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