For real nums a,b,c prove that
(i) (a+b)(1/a +1/b)>=4
(ii)(a+b+c)(1/a +1/b +1/c) >=9
(iii) well, if this seems trivial prove in general for reals nums a_1,a_2,...a_n prove that
(a_1 +a_2 +...+a_n) (1/ a_1 +1/ a_2 +...+1/ a_n)>=n^2
(iv) well you may still go further proving (1+1/2 +1/3 +....) is not convergent.[S]
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2 comments:
(i)=> (a +b)^2 >=4ab (*)
=> (a - b )^2 >=0
which is true.
(ii) this would need slightly different technique.
(iii) => (a/b + b/a ) + (b/c + c/b ) + ( a/c + c/a ) <=6
now let me digress a little proving you the AM-GM inequailty.
inequality says
(a+b)/2 >=(ab)^1/2
proof : squaring lands us in (*)
by AM-GM inequailty
(a/b + b/a)>=2 and similarly to all three parenthesis
we obtain the inequality.
(iv) general case may be solved using the technique of n=3 case.
Does anybody know to solve the inequality using induction or any other method ?
You said real numbers. :-)
if a = -1 and b = 1
then a+b = 0 and so is 1/a + 1/b
Works only for positive reals.
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