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if a,b,c are the positive uneven numbers are in H.P, then show that a2+c2>2b2
As a,b,c are in HP, we have 2/b = 1/a+1/c. So 4/b2 = 1/a2+1/c2+2/ac -------(1). Now using AM-GM Inequality a2+c2 > 2ac -----{the equality is removed as a,c cannot be equal as given in the question} 1/a2+1/c2 > 2/ac. Now from (1) 4/b2 > 2/ac + 2/ac or 4/b2 > 4/ac, or ac>b2 So a2+c2 > 2b2.
As a,b,c are in HP, we have 2/b = 1/a+1/c.
So 4/b2 = 1/a2+1/c2+2/ac -------(1).
Now using AM-GM Inequality
a2+c2 > 2ac -----{the equality is removed as a,c cannot be equal as given in the question}
1/a2+1/c2 > 2/ac.
Now from (1) 4/b2 > 2/ac + 2/ac
or 4/b2 > 4/ac, or ac>b2
So a2+c2 > 2b2.
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