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Prove : (a^2+b^2+c^2+d^2)(1/a^2+1/b^2+1/c^2+1/d^2)greater equal to 16Asked by black_widow
Prove : (a^2+b^2+c^2+d^2)(1/a^2+1/b^2+1/c^2+1/d^2)greater equal to 16
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Answered by
Steve
a^2 * 1/a^2 >= 1
assume
(a1^2+...+ak^2)(1/a1^2+...+1/ak^2) >= k^2
(a1^2+...+ak+1^2)(1/a1^2+...+1/ak+1^2)
= (a1^2+...+ak^2)(1/a1^2+...+1/ak+1^2)
+ ak+1^2((1/a1^2+...+1/ak+1^2)
= (a1^2+...+ak^2)(1/a1^2+...+1/ak^2)
+ (a1^2+...+ak^2)*1/ak+1^2
+ ak+1^2((1/a1^2+...+1/ak^2)
+ ak+1^2 * 1/ak+1^2
now, by hypothesis,
>= k^2
+ (a1^2+...+ak^2)*1/ak+1^2
+ ak+1^2((1/a1^2+...+1/ak^2)
+ 1
Hmmm. We have to prove that the two middle terms >= 2k, then we have proved the induction step. Gotta go now, but this may help some.
If you can show it, then the given problem is just a special case where k=4.
assume
(a1^2+...+ak^2)(1/a1^2+...+1/ak^2) >= k^2
(a1^2+...+ak+1^2)(1/a1^2+...+1/ak+1^2)
= (a1^2+...+ak^2)(1/a1^2+...+1/ak+1^2)
+ ak+1^2((1/a1^2+...+1/ak+1^2)
= (a1^2+...+ak^2)(1/a1^2+...+1/ak^2)
+ (a1^2+...+ak^2)*1/ak+1^2
+ ak+1^2((1/a1^2+...+1/ak^2)
+ ak+1^2 * 1/ak+1^2
now, by hypothesis,
>= k^2
+ (a1^2+...+ak^2)*1/ak+1^2
+ ak+1^2((1/a1^2+...+1/ak^2)
+ 1
Hmmm. We have to prove that the two middle terms >= 2k, then we have proved the induction step. Gotta go now, but this may help some.
If you can show it, then the given problem is just a special case where k=4.
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