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

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

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

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

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

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Solution

The correct option is C $a^{2} + c^{2} > 2 b^{2}$
From the section of inequality, we know that A.M. of $n^{t h}$ powers > $n^{t h}$ power of A.M.
i.e. $\frac{1}{2} (a^{n} + c^{n}) > {(\frac{1}{2} (a + c))}^{n}$
Considering two quantities a and c or $\frac{1}{2} (a^{n} + c^{n}) > (A)^{n} o r \frac{1}{2} (a^{n} + c^{n}) > (H)^{n}$
Since A.M. >H.M.
$\frac{1}{2} (a^{n} + c^{n}) > (b)^{n} \Rightarrow a^{n} + c^{n} > 2 b^{n}$
Putting, n = 2, we have $a^{2} + c^{2} > 2 b^{2} .$

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