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

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(C) Given $a, b, c, d$ are in $H.P.$,their reciprocals $\frac{1}{a}, \frac{1}{b}, \frac{1}{c}, \frac{1}{d}$ are in $A.P.$For any three terms in $H.P.$

Vedclass · Accepted Answer

$ac + bd > b^2 + c^2$

Vedclass · Answer

(C) Given $a, b, c, d$ are in $H.P.$,their reciprocals $\frac{1}{a}, \frac{1}{b}, \frac{1}{c}, \frac{1}{d}$ are in $A.P.$For any three terms in $H.P.$,say $x, y, z$,we know that $y = \frac{2xz}{x+z}$. Since $A.M. > H.M.$,we have $\sqrt{xz} > y$,which implies $xz > y^2$.Applying this to the consecutive terms of the $H.P.$:$1$. For $a, b, c$ in $H.P.$,we have $ac > b^2$.$2$. For $b, c, d$ in $H.P.$,we have $bd > c^2$.Adding these two inequalities,we get:$ac + bd > b^2 + c^2$.

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

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