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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
8 years ago

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