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October 21, 2014

October 21, 2014

Posted by **black_widow** on Thursday, March 21, 2013 at 2:15am.

- maths -
**Steve**, Thursday, March 21, 2013 at 12:19pma^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.

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