If a, b, c, d be in H.P., then

a^{2} + c^{2} > b^{2} + d^{2}

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a^{2} + d^{2} > b^{2} + c^{2}

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a c + b d > b^{2} + c^{2}

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a c + b d > b^{2} + d^{2}

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Solution

The correct option is C $a c + b d > b^{2} + c^{2}$
a,b,c,d be in H.P., then $\frac{1}{a}, \frac{1}{b}, \frac{1}{c}, \frac{1}{d}$ will be in A.P.
Therefore $\frac{1}{b} - \frac{1}{a} = \frac{1}{c} - \frac{1}{b} = \frac{1}{d} - \frac{1}{c} \Rightarrow b = \frac{2 a c}{a + c}$
G.M. between a and b = $\sqrt{a c} .$
Now as G.M > H.M., so here $\sqrt{a c} > b$ or $a c > b^{2} .$
Similarly $\sqrt{b d} > c$ or $b d > c^{2}$
Adding, $a c + b d > b^{2} + c^{2}$

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Similar questions

If a, b, c, d are in G.p., prove that :

(i) $(a^{2} + b^{2}), (b^{2} + c^{2}), (c^{2} + d)^{2}$ are in G.P.

(ii) $(a^{2} - b^{2}), (b^{2} - c^{2}), (c^{2} - d)^{2}$ are in G.P.

(iii) $\frac{1}{a^{2} + b^{2}}, \frac{1}{b^{2} + c^{2}}, \frac{1}{c^{2} + d^{2}}$ are in G.P.

(iv) $(a^{2} + b^{2} + c^{2}), (a b + b c + c d), (b^{2} + c^{2} + d^{2})$

a, b, c, d

\sqrt{a^{2} + c^{2}} : \sqrt{b^{2} + d^{2}} = \sqrt{a c + \frac{c^{3}}{a}} : \sqrt{b d + \frac{d^{3}}{b}}

\frac{{(a + b)}^{2} - {(c + d)}^{2}}{{(a + b)}^{2} - {(c - d)}^{2}} \times \frac{{(a + b + c)}^{2} - d^{2}}{{(a + b - c)}^{2} - d^{2}}

(a^{2} + b^{2}), (b^{2} + c^{2}), (c^{2} + d^{2})

\frac{a}{b} = \frac{c}{d} = \frac{e}{f}

(\frac{a^{2}}{b^{2}} + \frac{c^{2}}{d^{2}} + \frac{e^{2}}{f^{2}}) = (\frac{a c}{b d} + \frac{c e}{d f} + \frac{a e}{b f})

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