Proof needed?

Question

Given a,b,c,d.... as positive integers. Is it true that
(a/b)+(b/a)>=2, (a/b)+(b/c)+(c/a)>=3, (a/b)+(b/c)+(c/d)+(d/a)>=4
and so on and so forth
And can anybody give the proofs ?

Answer 1

a1/a2 + a2/a3 + a3/a4 + ... + an/a1 >= a1/P + a2/P + ... an/P >= n/P >= n. (qed)
( with P = a1.a2.a3... an. )

Answer 2

Starting with the first equation...

a/b + b/a = a^2/ab + b^/ab = (a^2 + b^2)/ab

Let us say that a >= b for the sake of convenience (it works for b > a as well). Thus a = b + n, where integer n >= 0.

a/b + b/a = [(b + n)^2 + b^2]/[(b + n)b]
= (2b^2 + 2bn + n^2)/(b^2 + bn)
= 2(b^2 + bn)/(b^2 + bn) + n^2/(b^2 + bn)
= 2 + n^2/(b^2 + bn)

Since the fraction must be positive or zero, then the proof holds that a/b + b/a >= 2

The other equations are solved similarly, but I'll let you work on those. It's probably a matter of applying a recursive proof.

Answer 3

Yes. The numbers only need to be positive, not integers. This can be proved by induction on the number of terms n in the sum. Obviously if there's only one term:

S = a/a = 1 >= 1

So assume the inequality holds for such sums involving n-1 terms and consider the sum:

S = a1/a2 + a2/a3 + ... + a_(n-1)/an + an/a1

Without loss of generality we can assume that a1 >= ai for all 1 <= i <= n. (In other words if originally ai = Max{a1,...an} but i != 1, shift the first i-1 terms to the end of the sum and relabel all the a's cyclically, ai-->a1, a_(i+1)--> a2, etc..)

Then a1/a2 >= 1 and an/a1 <= 1 so that:

(a1/a2 - 1)*(1 - an/a1) >= 0
-->
a1/a2 - 1 - an/a2 + an/a1 >= 0
-->
a1/a2 + an/a1 >= 1 + an/a2

The left side of the above inequality is the first and last terms in S, so we can replace them by the right side (1+an/a2):

S >= a2/a3 +...+ a_(n-1)/an + (1 + an/a2)
= (a2/a3 + a3/a4 + ... + an/a2) + 1
>= n-1 + 1 >= n

using the inductive hypothesis.

Answer 4

a,b,c,d,…. all are integers so a >= 1, b >= 1, c >= 1, d >=1, ….. and so on.

so ab ≠ 0; so ab/ab = 1;

equation is ( a/b + b/a ) >= 2

multiply the LHS by (ab/ab) that is one

LHS = ab/ab ( a/b + b/a)

= ab( a/b + b/a) /ab

= (a2 + b2) / ab

= (a2 + b2 - 2ab + 2ab) / ab

= ( (a-b) 2 + 2ab) / ab

= (a-b)2 //ab + 2

LHS= ( a – b )2 / ab + 2


( a – b ) 2 >= 0 and ab > 1

so ( a – b ) 2 / ab >= 0

so ( a – b ) 2 / ab + 2 >= 2

so LHS >= 2

like this u can multiply other equations by abc/abc, abcd/abcd, …. and so on

Answer 5

Why don't you do the simple thing and apply a value to the unknowns?

a=4
b=2
Using " / ", you are indicating division, aren't you?

4/2+2/4...............=2 ??????????????????HUH?
Am I missing something? I have both oars in the water, I think.

Answer 6

For the first one:
(a-b)(a-b)>=0 is true Then

a*a+b*b-2a*b>=0
a*a+b*b>=2*a*b

(a*a+b*b)/(a*b)>=2
By write this using two fractions, you will get
a/b+b/a>=2

Answer 7

see multiply the l.h.s wit the l.c.m n u get the proof...

Answer 8

umm do your own homework!

Answer 9

say what?????????