# Challenge Math Challenge by QuantumQuest #4

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1. Jul 2, 2017

### Greg Bernhardt

Submitted by: @QuantumQuest
Challenge Level: High School

RULES:

1) In order for a solution to count, a full derivation or proof must be given. Answers with no proof will be ignored.
2) It is fine to use nontrivial results without proof as long as you cite them and as long as it is "common knowledge to all mathematicians". Whether the latter is satisfied will be decided on a case-by-case basis.
3) If you have seen the problem before and remember the solution, you cannot participate in the solution to that problem.
4) You are allowed to use google, wolframalpha or any other resource. However, you are not allowed to search the question directly. So if the question was to solve an integral, you are allowed to obtain numerical answers from software, you are allowed to search for useful integration techniques, but you cannot type in the integral in wolframalpha to see its solution.

CHALLENGE:

If a, b and c are positive real numbers show that $a^{a}b^{b}c^{c} \geq (abc)^{\frac{a + b + c}{3}}$

2. Jul 3, 2017

### .Scott

I just noticed, this is an HS challenge. So I am deleting my post.

3. Jul 3, 2017

### Greg Bernhardt

Just use the spoiler tag

4. Jul 3, 2017

### .Scott

consider this:
K = (2a-b-c)log(a)+ (2b-a-c)log(b)+ (2c-a-b)log(c)
If a, b, and c are all equal, then K=0.
Now make one large - for example let's increase 'a' by a positive amount 'd'.
So the first factor of each term will increase by 2d, -d, and -d. But the second factors will increase by (log(a+d)-log(a)), 0, and 0.
So the increase in the 1st term will more than offset the decreases in the 2nd and 3rd term. Thus K will go positive.
Similar logic can be use to show that decreasing one of them will also force K positive - so long as a, b, and c remain positive.

Any combination of a,b, and c can be created by starting with a,a,a and applying 'd's to get a,b,c. K will be zero when they are all equal and positive otherwise.

So K>=0 and
(2a-b-c)log(a)+ (2b-a-c)log(b)+ (2c-a-b)log(c) >= 0

adding (a+b+c)log(a) + (a+b+c)log(b) + (a+b+c)log(c) to both sides:

(3a)log(a)+ (3b)log(b)+ (3c)log(c) >= (a+b+c)log(a) + (a+b+c)log(b) + (a+b+c)log(c)

dividing both sides by 3:

(a)log(a)+ (b)log(b)+ (c)log(c) >= ((a+b+c)/3) (log(a) + log(b) + log(c))

exponentiation:

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

QED

5. Jul 4, 2017

### QuantumQuest

Interesting solution and a good one. The most interesting point is how to figure out / reach the starting expression.

6. Jul 6, 2017

It's a good problem, even for people above the high school level to try to solve. (@QuantumQuest showed it to a couple of us before he posted it , and it is a fun problem.) :) :)

7. Sep 7, 2017

### ddddd28

Since a legitimate solution was handed in a long time ago, but definitely not one that a highschool student can come up with, I think that it's ok to ask what is the solution of @QuantumQuest, or at least a hint.
Personally, I put quite a lot of efforts trying to solve it, but didn't succeed. I tried to play with substituting c=a*d, which seemed promising because I managed to simplify it to b^b*d^d=(bd)^(b+d+1)/3, but then got stuck. After that I thought in the direction of induction, but again couldn't carry it out.

8. Sep 7, 2017

### QuantumQuest

The problem is at the high school level but it needs to think what to use (i.e. be somewhat creative), otherwise it would not be a challenge. If I give you what I came up with when I solved it (i.e the idea to start with) I'll effectively give you the solution which won't help learning. As a hint, try to find an inequality that involves some relevant to the problem powers and go from there.