An academic and a bit difficult problem.?

Question

'a' and 'b' are two real numbers such that a+b=1.

Show that:

(a+1/a)^2+(b+1/b)^2 >= 25/2

Thanks for your time and consideration.

Answer 1

You should exclude the cases a=0 or b=0.

1) Proof for a > 0 and b > 0, a + b = 1: the easiest way is by applying the inequality between the geometric mean, arithmetic mean and quadratic mean - for every x > 0 and y > 0 is well-known that
(G-A-Q-M) √(xy) ≤ (x + y)/2 ≤ √((x² + y²)/2)
equalities above we have if and only if x = y.
We'll have it proven if we show that
((a + 1/a)² + (b + 1/b)²)/2 ≥ 25/4 = (5/2)²
Let x = a + 1/a, y = b + 1/b, then according (Q-A-M)
and using a + b =1 we obtain:
√(((a+1/a)² + (b+1/b)²)/2) ≥ ((a+1/a) + (b+1/b))/2 ≥
≥ (a + b + 1/a + 1/b)/2 = (1 + (a+b)/ab)/2 =
= (1 + 1/ab)/2
Now according (G-A-M) ab ≤ ((a + b)/2)² = 1/4,
so 1/(ab) ≥ 4 (equality iff a = b= 1/2) and
√(((a+1/a)² + (b+1/b)²)/2) ≥ (1 + 1/ab)/2 ≥ (1 + 4)/2 = 5/2, or
((a + 1/a)² + (b + 1/b)²)/2 ≥ 25/4, or
(a + 1/a)² + (b + 1/b)² ≥ 25/2 as required.

2) If a + b = 1, one of the numbers a, b negative, other positive then
1 = |a + b| ≤ |a| + |b|, the left side
(a + 1/a)² + (b + 1/b)² = |a + 1/a|² + |b +1/b|²
can become even greater than in case 1), so the inequality holds also.

Answer 2

Let a = -1 and b = 2 then
a + b = 1 and
(a+1/a)^2+(b+1/b)^2 = 9/4 which is less than 25/2 = 50/4

So it is not true.

Edit: Thanks for showing the error of my ways Raistlin.
I've been reading too many questions where people forget parentheses and write x+1/a when they mean (x+1)/a. Now I'm beginning to think like them. Scary. But then again maybe that's what the original asker DID mean? :-D

Answer 3

I have the answer in the last 2 paragraphs you may skip to there if you don't want to see my trial and error...

How about a=3 and b= -2

So that makes:
((3)+1/(3))^2 + ((-2)+1/(-2))^2 > 12.5
(4/3)(4/3) + (-1/-2)(-1/-2) >12.5
(4/3)(4/3) + (1/2)(1/2) >12.5
16/9 + 1/4 <12.5
64/36 + 9/36 = 73/36
2+1/36 is NOT greater than 25/2

Ok.... what about a=100 and b= -99
So we have:
(101/100)(101/100) + (98/99)(98/99)
10201/10000 + 9604/9801
1060200001/980100000
1 + 80100001/980100000 is NOT Greater than 25/2

Ok... what about a=1 and b=0
So we have:
(2/1)(2/1) + (1/0)(1/0)
Doesn't work...

Ok... what about a = 2 and b= -1
So we have:
(3/2)(3/2) + (0/-1)(0/-1)
Doesn't work either...

What about a = .5 and b = .5
So we have:
(1.5/.5)(1.5/.5) + (1.5/.5)(1.5/.5)
2.25/.25 + 2.25/.25
4.5/.25 = 18
CLOSER!

a= .05 and b= .95
(1.05/.05)(1.05/.05) + (1.95/.95)(1.95/.05)
1.1025/.0025 + 3.8025/.9025
Not sure about fractions but using a calculator it equals 445.2132 which IS Greater than 25/2

Answer 4

"Let a = -1 and b = 2 then
a + b = 1 and
(a+1/a)^2+(b+1/b)^2 = 9/4 which is less than 25/2 = 50/4

So it is not true."

huh??
the first term works out to be
-.5 * -.5 which is .25
the second term is
2.5 squared which is 6.25
so it comes out to 6.5 which is STILL less hat 12.5
so you're wrong and right at the same time :)

Answer 5

first we have to give equal chance to a, b.
so a=1/2, b=1/2 .
then,
(1/2 + 2)^2 + (1/2 +2)^2
= 25/4 + 25/4
=25/2

Answer 6

I think so you have to write that a and b are two positive real numbers and if so the equation will be true.

Answer 7

let a=0

1/a is not defined ...

and (a+1/a)^2+(b+1/b)^2 is "not defined either"


when a or b = 0, we have singularities