Question

Let a,b,c be positive real numbers. If the minimum value of

\(\Large \frac{(b+c-a)^2}{(b+c)^2+a^2} + \frac{(c+a-b)^2}{(c+a)^2+b^2} + \frac{(a+b-c)^2}{(a+b)^2+c^2} \)

can be written as \(\Large \frac{m}{n}\), then find \(\Large m+n\)

Answer 1

I think we should use cauchy-schwarz inequality

Answer 2

Clearly this is symmetric in a,b,c
so the min/max occurs only when a=b=c

Answer 3

1/5 + 1/5 + 1/5 = 3/5

http://www.wolframalpha.com/input/?i=minimize+%28x%2By-z%29%5E2%2F%28%28x%2By%29%5E2%2Bz%5E2%29%2B+%28y%2Bz-x%29%5E2%2F%28%28y%2Bz%29%5E2%2Bx%5E2%29%2B+%28z%2Bx-y%29%5E2%2F%28%28z%2Bx%29%5E2%2By%5E2%29%2Cx%3E0%2Cy%3E0%2Cz%3E0

Answer 4

symmetry argument may not look pleasing for everybody, lets see if we can work it using cauchy-schwarz inequality

Answer 5

using cauchy-schwarz inequality on each of these pieces gives
\(\Large \frac{(b+c-a)^2}{(b+c)^2+a^2} + \frac{(c+a-b)^2}{(c+a)^2+b^2} + \frac{(a+b-c)^2}{(a+b)^2+c^2}\\\Large \le2 [\frac{(b+c)^2-a^2}{(b+c)^2+a^2}+\frac{(c+a)^2-b^2}{(c+a)^2+b^2}+\frac{(a+b)^2-c^2}{(a+b)^2+c^2}] \)

Answer 6

so, what is the next step?

Answer 7

@vishweshshrimali5

Answer 8

One minute I would have to go through all of this ^^^

Answer 9

ok :)

Answer 10

Well @ganeshie8 's method is of course correct. This is a very simple method and works almost all times

Answer 11

Are you sure you applied the inequality correctly ?

Answer 12

Just conforming ...

Answer 13

we guess is the minimum?

Answer 14

Appears so

Answer 15

I am on my way home and don't have any pen/copy. So, I would let you know the solution when I reach home :)

Answer 16

okay, I wait

Answer 17

Sure :)

Answer 18

Okay I have not reached home yet but I have an idea

Answer 19

We assume that 3/5 is the minimum value

Answer 20

I can write the given expression in summation form as:

\[\large{\sum {\cfrac{(b+c-a)^2}{(b+c)^2+a^2}}}\]

And I have assumed that:

\[\large{\sum {\cfrac{(b+c-a)^2}{(b+c)^2+a^2}} \ge \cfrac{3}{5}}\]

Answer 21

*I will have to prove this ^^^

Answer 22

Now I am going to subtract 1 from each term, and thus subtracting 3 from 3/5

Answer 23

I will get:

\[\large{\sum ({\cfrac{(b+c-a)^2}{(b+c)^2+a^2}}) - 1 \ge \cfrac{3}{5} -3}\]

\[\large{\implies \sum (\cfrac{(b+c-a)^2 - (b+c)^2 - a^2}{(b+c)^2+a^2}) \ge \cfrac{-12}{5}}\]

\[\large{\implies \sum (\cfrac{-2ab - 2ac}{(b+c)^2+a^2}) \ge \cfrac{-12}{5}}\]

\[\large{\implies \sum \cfrac{2ab+2ac}{(b+c)^2+a^2} \le \cfrac{12}{5}}\]

Answer 24

Okay I can think of only one thing:

Add 3 on both sides.

\[\large{\implies \sum (\cfrac{2ab+2ac}{(b+c)^2+a^2} +1) \le \cfrac{12}{5} + 3}\]

\[\large{\implies \sum (\cfrac{2a(b +c) + (b+c)^2 + a^2}{(b+c)^2+a^2}) \le \cfrac{27}{5}}\]

\[\large{\implies \sum \cfrac{(a+b+c)^2}{a^2+(b+c)^2} \le \cfrac{27}{5}}\]

\[\large{\implies (a+b+c)^2 \sum \cfrac{1}{a^2+(b+c)^2} \le \cfrac{27}{5}}\]

Answer 25

Okay, I would have to think something more for further steps :)

Answer 26

@ganeshie8 any idea ?

Answer 27

@JungHyunRan what do you think ?

Answer 28

Okay since the inequality is homogeneous, I can normalize the inequality. Yup this may work :)

Answer 29

Without loss of generality, I can assume that:

\[\large{a+b+c = 3}\]
This will help me in simplifying the fraction in RHS into 3/5.

Answer 30

So, I get,

\[\large{\sum \cfrac{1}{(3-a)^2 + a^2} \le \cfrac{3}{5}}\]

Answer 31

Any doubt till this step @JungHyunRan ?

Answer 32

I tried using CS inequality but it didn't prove much beneficial

Answer 33

do we have general solution for this type ?

Answer 34

Well for different inequalities there are different methods.

Answer 35

I tried to use Jensen's inequality at this step but since the function is convex I cannot get the maximum value using this inequality.

So, lets move on

Answer 36

Okay I can write the last inequality as:

\[\large{\sum \cfrac{1}{(3-a)^2 + a^2} \le \cfrac{3}{5}}\]

\[\large{\implies \sum \cfrac{5}{(3-a)^2+a^2} \le 3}\]

\[\large{\implies \sum (\cfrac{5}{(3-a)^2 + a^2} -1) \le 0}\]

\[\large{\implies \sum (1-\cfrac{5}{(3-a)^2+a^2}) \ge 0}\]

Okay stuck again :P

Answer 37

I have to prove my assumption.

Answer 38

So, I am basically moving backwards. I assumed that the minimum value is 3/5. Now, to prove my assumption, I try to use this minimum value to get a step which is true because of given information in problem. If I can arrive at one such step, then I would be able to prove that my assumption was correct and that 3/5 is the minimum value.

Answer 39

i see

Answer 40

Okay lets see if this works out:

\[\large{\sum 1 - \cfrac{5}{(3-a)^2+a^2}}\]

\[\large{= \sum 1 - \cfrac{5}{9+2a^2-6a}}\]

Now, I got 3/5 in the very first step by assuming that:

\[\large{a=b=c}\]

Then, in a later step during normalization I assumed that:

\[\large{a+b+c = 3}\]

Thus, I can write that I have assume:

\[\large{a = b= c =1}\]

Answer 41

No that's not correct

Answer 42

\(\color{blue}{\text{Originally Posted by}}\) @vishweshshrimali5
Okay lets see if this works out:

\[\large{\sum 1 - \cfrac{5}{(3-a)^2+a^2}}\]

\[\large{= \sum 1 - \cfrac{5}{9+2a^2-6a}}\]

Now, I got 3/5 in the very first step by assuming that:

\[\large{a=b=c}\]

Then, in a later step during normalization I assumed that:

\[\large{a+b+c = 3}\]

Thus, I can write that I have assume:

\[\large{a = b= c =1}\]

\[\color{red}{\text{FORGET IT!!}}\]
\(\color{blue}{\text{End of Quote}}\)

Answer 43

This is what that happens whenever I try to apply more knowledge than necessary :P

Answer 44

^^

Answer 45

Yeah...

Okay may be something else may help ... Will have to think :)

Answer 46

you are so awesome, so smart :D

Answer 47

Okay:


\[\large{\sum(1-\cfrac{5}{2a^2-6a+9})}\]
\[\large{=2(a^2-3a+2)\sum\cfrac{1}{2a^2-6a+9}}\]
\[\large{=2(1-a)(2-a)\sum\cfrac{1}{2a^2-6a+9}}\]
\[\large{=2(1-a)\sum \cfrac{-5a+10}{5(2a^2-6a+9)}}\]
\[\large{=2(1-a)\sum \cfrac{-5a+10}{5(2a^2-6a+9)}}\]
\[\large{=2(1-a)\sum \cfrac{-5a+10}{5(2a^2-6a+9)}}\]
\[\large{=2(1-a)\sum \cfrac{2a^2-6a+9 + a + 1 -2a^2}{5(2a^2-6a+9)}}\]
\[\large{=2(1-a)\sum \cfrac{2a^2-6a+9 + (a+1-2a^2)}{5(2a^2-6a+9)}}\]
\[\large{=2(1-a)\sum \cfrac{2a^2-6a+9 + (1-a)(2a+1)}{5(2a^2-6a+9)}}\]
\[\large{=\sum \cfrac{2(1-a)(2a^2-6a+9) + 2(a-1)^2(2a+1)}{5(2a^2-6a+9)}}\]
\[\large{=\sum \cfrac{2(1-a)}{5} + \cfrac{2(a-1)^2(2a+1)}{5(2a^2-6a+9)}}\]

Answer 48

\[\large{\ge \sum \cfrac{2(1-a)}{5}}\]

\[\large{=0}\]

Thus,

\[\large{\sum(1-\cfrac{5}{2a^2-6a+9}) \ge 0}\] is true.

Hence,

\[\large{\sum \cfrac{1}{(3-a)^2 + a^2} \le \cfrac{3}{5}}\] is true

Hence, 3/5 is the minimum value of the given expression.

Thus, m+n = 8

Answer 49

And thanks @JungHyunRan

Answer 50

@ganeshie8 @JungHyunRan I would be offline for 30 - 45 minutes till then can you please check out the solution

Answer 51

ok :)

Answer 52

I am back! I am back ! :D

Answer 53

Any doubt @JungHyunRan in the solution ?

Answer 54

no, :)
I see the solution