# h.p.

if a,b,c, are in h.p. prove that a^2+c^2> 2b^2 ?
answer= 1/a,1/b, and 1/c are in h.p.
=1/b-1/a=1/c-1/b
=1/b+1/b=1/c+1/a
=b+b/b^2=a+c/ac
=2/b^2=a+c/ac
=b^2(a+c)=2ac
=(a+c)ac=2b^2
=a^2+b^2 >2b^2

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