A beautiful Inequality

[jay_shark]

Here is an inequality that lends itself to a multitude of solutions and is truly one of my favorites . In fact , this problem went unsolved in the 1994 USA IMO training camp . Even though this problem was well known to some , it was in fact asked again in a Balkan Math Competition and surprising
not many figured it out . It's like asking to prove the Pythagorean theorem all over again :P

If a,b,c >0 Show that

1/(a(b+1)) + 1/(b(c+1)) + 1/(c(a+1)) >= 3/(1+abc)

See what you come up with , as there are many different approaches and then I'll share a beautiful solution that is not one of my own .

[jay_shark]

bump

[thylacine]

[ QUOTE ]
Here is an inequality that lends itself to a multitude of solutions and is truly one of my favorites . In fact , this problem went unsolved in the 1994 USA IMO training camp . Even though this problem was well known to some , it was in fact asked again in a Balkan Math Competition and surprising
not many figured it out . It's like asking to prove the Pythagorean theorem all over again :P

If a,b,c >0 Show that

1/(a(b+1)) + 1/(b(c+1)) + 1/(c(a+1)) >= 3/(1+abc)

See what you come up with , as there are many different approaches and then I'll share a beautiful solution that is not one of my own .

[/ QUOTE ]

Okay got it. I won't write out the solution but instead will post hints in white for anyone that wants to try it.

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Hint 1: <font color="white"> Put everything over a common denominator, collect terms on one side and discard denominator (which is positive) giving a polynomial in a,b,c. </font>

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Hint 2: <font color="white"> Rearrange as a sum of terms, each of which is obviously positive, that is, a product of terms that are a, b, c, or squares. </font>


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[jay_shark]

It takes a brave person to work it out that way :P

[m_the0ry]

He wouldn't say it's "beautiful" if there wasn't a way to solve it that is more simplistic than doing partial fractions. I think working it out graphically may be a better option with a triangle representing the three nonzero quantities.

[thylacine]

[ QUOTE ]
It takes a brave person to work it out that way :P

[/ QUOTE ]

In a test with a time limit you have to quickly find a strategy that will work, and this can sometimes take some instinct and intuition. I was pretty sure it would work, so I bit the bullet and did the algabra, and everything fell into place quickly (about 10 minutes, though it would take longer to fully write out the solution). I think it is pathetic that `this problem went unsolved in the 1994 USA IMO training camp.' It must have been a weak year.

I'd be curious to see other solutions.

Also I'd be curious to see if there is a wrong solution that assumes there is complete symmetry between the variables, whereas there is just a cyclic symmetry.

[djames]

Bump until we see the "beautiful" solution. The common denominator solution isn't beautiful.

[jay_shark]

Ok here is the solution that I'm very fond of .

re-write the inequality as

(1+abc)[1/a(b+1) + 1/b(c+1) +1/c(a+1)]&gt;=3

Now notice that
(1+abc)/a(b+1) = (1+a+ab+abc)/a(b+1) -1 = (1+a)/a(b+1) +b(1+c)/(1+b) -1
Likewise do the same for the other two .

Therefore we have to show that

(1+a)/a(1+b) + b(1+c)/(1+b) +(1+b)/b(1+c) +c(1+a)/(1+c)+ (1+c)/c(1+a) +a(1+b)/(1+a) &gt;= 6

But this follows immediately from the Am/Gm inequality .

What a beauty !!

[djames]

Hmm...

As the "now notice that..." equalities are not necessarily immediately seen by the reader, I have to honestly say that I'm more fond of the traditional proof which is only a few lines longer and doesn't require knowledge of another inequality.

But, nice proof, nonetheless.

[thylacine]

[ QUOTE ]
Bump until we see the "beautiful" solution. The common denominator solution isn't beautiful.

[/ QUOTE ]

What is beautiful is the underlying theory of how to prove polynomial inequalities in general (not just this particular one), which is what I was using.

[djames]

Absolutely. I'm sorry if I slanted the method you used (which is also what I used) wasn't beautiful. I shouldn't have been quick to do so.

I was hoping for some geometric representation or a graph theory combinatorial proof though. So can you accept this is "relatively" ugly? /images/graemlins/smile.gif