Proving a sum of three squared terms, cyclic in #a,b,c#, is equal to 1

[brotherbobby]

Homework Statement

If $a+b+c=0$, show that $\boxed{\frac{a^2}{2a^2+bc}+\frac{b^2}{2b^2+ca}+\frac{c^2}{2c^2+ab}=1}$

Relevant Equations

1. If $a+b+c=0$, then $a^2+2ab+b^2=c^2$
2. If $a+b+c=0$, then $a^2+b^2+c^2=-2(ab+bc+ca)$ (not sure how useful this would be)

Problem statement : I copy and paste the statement of the problem from the text.
(Given $\boldsymbol{a+b+c=0}$)

Attempt : I am afraid I couldn't make any meaningful progress. With $a = -(b+c)$, I substituted for $a$ in the whole of the L.H.S, both numerators and denominators. I multiplied the denominators and multiplied the numerators by the "other" denominators, much as we would do for the case $\dfrac{A}{X}+\dfrac{B}{Y}+\dfrac{C}{Z} = \dfrac{AYZ+BZX+CXY}{XYZ}$. That gave me three terms with sizable numbers of terms in the numerator and denominator, ranging from $b^6\rightarrow c^6$ and terms of decreasing and increasing powers of $b,c$ respectively in them. It was clumsy and tedious, not to mention that I didn't obtain the answer 1 due to what must be an error somewhere. I am certain there must be a better way to do the problem.

Request : Not having done my fair share, I would be happy for the barest of hints in the right direction to get me going.

 

[fresh_42]

Looks like $(a+b+c)^2=a^2+b^2+c^2+2ab+2ac+2bc$ could be of help. Those problems usually need a tricky algebraic rephrasing, e.g. by considering $(a+1)(b+1)(c+1)$ instead.

 

[BvU]

If I see that 1+1-2=0 and then the first term (your A/X) does not exist, it seems to me you haven' t given us the complete problem statement.

$\$

 

[BvU]

And I don' see 6th powers emerging ?

 

[brotherbobby]

If I see that 1+1-2=0 and then the first term ( your X) does not exist, it seems to me you haven' t given us the complete problem statement.

$\$

Please see Problem 16 below taken from the text.

And I don' see 6th powers emerging ?

I don't know how clear it is. You will see terms like $-4b^6$ and $-2c^6$ in the last line.

 

[valenumr]

Homework Statement:: If $a+b+c=0$, show that $\boxed{\frac{a^2}{2a^2+bc}+\frac{b^2}{2b^2+ca}+\frac{c^2}{2c^2+ab}=1}$
Relevant Equations:: 1. If $a+b+c=0$, then $a^2+2ab+b^2=c^2$
2. If $a+b+c=0$, then $a^2+b^2+c^2=-2(ab+bc+ca)$ (not sure how useful this would be)

View attachment 296002Problem statement : I copy and paste the statement of the problem from the text.
(Given $\boldsymbol{a+b+c=0}$)

Attempt : I am afraid I couldn't make any meaningful progress. With $a = -(b+c)$, I substituted for $a$ in the whole of the L.H.S, both numerators and denominators. I multiplied the denominators and multiplied the numerators by the "other" denominators, much as we would do for the case $\dfrac{A}{X}+\dfrac{B}{Y}+\dfrac{C}{Z} = \dfrac{AYZ+BZX+CXY}{XYZ}$. That gave me three terms with sizable numbers of terms in the numerator and denominator, ranging from $b^6\rightarrow c^6$ and terms of decreasing and increasing powers of $b,c$ respectively in them. It was clumsy and tedious, not to mention that I didn't obtain the answer 1 due to what must be an error somewhere. I am certain there must be a better way to do the problem.

Request : Not having done my fair share, I would be happy for the barest of hints in the right direction to get me going.

Well, that's just plain nasty. I'm only commenting to follow.

 

[PeroK]

Please see Problem 16 below taken from the text.

View attachment 296041
View attachment 296042

I don't know how clear it is. You will see terms like $-4b^6$ and $-2c^6$ in the last line.

I think you have to be a bit smarter. I took $c = -(a + b)$. And took the third term over to the RHS. I.e. show that $$A + B = 1 - C$$It then looked easier to invert the two fractions and show that $$\frac{1}{A + B} = \frac 1 {1-C}$$That got rid of some factors and the rest was not too bad.

I can't see a quick and easy way to do it, although there must be one.

 

[valenumr]

I think you have to be a bit smarter. I took $c = -(a + b)$. And took the third term over to the RHS. I.e. show that $$A + B = 1 - C$$It then looked easier to invert the two fractions and show that $$\frac{1}{A + B} = \frac 1 {1-C}$$That got rid of some factors and the rest was not too bad.

I can't see a quick and easy way to do it, although there must be one.

It almost seems like the pattern of the equation suggests a "trick".

 

[BvU]

I don't know how clear it is. You will see terms like $-4b^6$ and $-2c^6$ in the last line.

Ah, it's because of the substitution of $a$, is suppose

What about $(a,b,c)=(1,1,-2)$ that has $a+b+c=0$ but no $a^2 \over 2a^2+bc$ as far as I can tell ?

I must be wrong, because here it gets a result of $1$, too !

Anyway,

Not having done my fair share

I'd say you definitely have done your fair share !

$\$

My path of least resistance was to work out your $AYZ+BZX+CXY=XYZ$ starting with the LHS :$$\begin{align*}
&a^2(2b^2+ca)(2c^2+ab)+b^2(2a^2+bc)(2c^2+ab)+c^2(2a^2+bc)(2b^2+ca)\quad=\\
&a^2(4b^2c^2+2b^3a+2c^3a+a^bc) \quad+\\
&b^2(4a^2c^2+2a^3b+2c^3b+ab^2c)+\quad\\
&c^2(4a^2b^2+2a^3c+2b^3c+abc^2)\qquad=\\
&12a^2b^2c^2+4b^3a^3+4c^3a^3+4c^3b^3+a^4bc+b^4ac+c^4ab\end{align*}
$$and for the RHS: XYZ=$$\begin{align*}
&(2a^2+bc)(2b^2+ca)(2c^2+ab)\quad=\\
&(2a^2+bc)(4b^2c^2+2c^3a+2b^3a+a^2bc)\quad=\\
&8a^2b^2c^2 + 4c^3a^3 + 4b^3a^3 + 2a^4bc\quad + \\
&\qquad 4b^3c^3 + 2abc^4 + 2b^4ac + a^2b^2c^2\end{align*}
$$Remove equal terms left and right and collect to get LHS=RHS$\Leftrightarrow$ $$
\begin{align*}
3a^2b^2c^2 &=a^4bc+b^4ac+c^4ab\\
3abc&=a^3+b^3+c^3
\end{align*}
$$now it seems a good time to substitute $a=-(b+c)\Rightarrow a^3 = -(b^3+3b^2c+3bc^2+c^3)$ :$$
\begin{align*}
-3(b+c)bc &\stackrel{?}{=} -(b+c)^3 +b^3+c^3 \\

-3b^2c-3bc^2 &\stackrel{?}{=} -3b^2c-3bc^2
\end{align*}$$which concludes things.

But it's brute force and dumb work, so I sure look forward to seeing the quickest way !

Nevertheless, nice exercise (apart from the ##(1,1,-2)#3 and such ) !

$\$

 

[PeroK]

Here's an idea. Still messy, but the best I can do. All of these steps are backwards implications (or "enough to show"):
$$\frac{a^2}{2a^2 +bc} + \frac{b^2}{2b^2 +ac} = \frac{c^2 + ab}{2c^2 +ab}$$Now invert both sides:
$$\Leftarrow \ \frac{(2a^2 +bc)(2b^2 + ac)}{a^2(2b^2 +ac) + b^2(2b^2 +bc)} = \frac{2c^2 + ab}{c^2 +ab}$$$$\Leftarrow \ \frac{4a^2b^2 + 2c(a^3 + b^3) + abc^2}{4a^2b^2 + c(a^3 + b^3)} = 1 + \frac{c^2}{c^2 +ab}$$$$\Leftarrow \ 1 + \frac{c(a^3 + b^3) + abc^2}{4a^2b^2 + c(a^3 + b^3)} = 1 + \frac{c^2}{c^2 +ab}$$Cancel the $1$ and invert again!
$$\Leftarrow \ \frac{4a^2b^2 + c(a^3 + b^3)}{c(a^3 + b^3) + abc^2} = \frac{c^2 +ab}{c^2}$$$$\Leftarrow \ 1 + \frac{ab(4ab - c^2)}{c(a^3 + b^3) + abc^2} = 1 + \frac{ab}{c^2}$$$$\Leftarrow \ \frac{(4ab - c^2)}{c(a^3 + b^3) + abc^2} = \frac{1}{c^2}$$$$\Leftarrow \ c(4ab - c^2)= a^3 + b^3 + abc$$And, finally, we set $c = -(a + b)$ to confirm that last equality

 

[valenumr]

I think you have to be a bit smarter. I took $c = -(a + b)$. And took the third term over to the RHS. I.e. show that $$A + B = 1 - C$$It then looked easier to invert the two fractions and show that $$\frac{1}{A + B} = \frac 1 {1-C}$$That got rid of some factors and the rest was not too bad.

I can't see a quick and easy way to do it, although there must be one.

I see now. It looks much nicer substituting for c = (-a -b) in all the denominators and factoring them.

 

[brotherbobby]

I have bad news. I have been trying to splve using @PeroK 's solution earlier above (post #7). I used $c=-(a+b)$ on both sides and reduced both L.H.S and R.H.S exactly as asked but am stuck.

Attempt : I present my attempt below. I hope it's readable.

A help at the last step would be welcome.

 

[PeroK]

See post #10, where I did some fancy footwork to simplify the two sides!

 

[brotherbobby]

See post #10, where I did some fancy footwork to simplify the two sides!

Yes @PeroK - I was trying not to see it and do it myself. I am afraid that I am now forced to take more of your help.

 

[valenumr]

Yes @PeroK - I was trying not to see it and do it myself. I am afraid that I am now forced to take more of your help.

I only substituted for c in the denominators, and then factored those denominators. It resulted in three common factors, so it became really basic algebra after that, something like (2a² + b), (2b² + a), and (a - b) as factors, IIRC. I did later substitute c² as a late step before expanding everything else to show both sides were equal. It was only like five or six steps, and the expansion had cubic terms.

Quick edit: I'm just going from memory, but I don't think now that any factors had a square term, actually I'm pretty certain, so just ignore the squares.

 

[brotherbobby]

Ok. I would like to "close" this thread by doing the problem in an elegant manner. While @PeroK 's method in post #10 above did the job, a look through websites has told me that there is an elegant method to solve the problem. Additionally, I will post a second problem which can be solved by the same technique.

Original problem statement : If $a+b+c=0$, show that $$\boxed{\frac{a^2}{2a^2+bc}+\frac{b^2}{2b^2+ca}+\frac{c^2}{2c^2+ab}=1}$$

Attempt :

Since $a+b+c=0$, anyone of the variables is equal to "negative" of the sum of the other two.

$\begin{equation*} \begin{split} \text{L.H.S}& = \frac{a^2}{2a^2+bc}+\frac{b^2}{2b^2+ca}+\frac{c^2}{2c^2+ab}\\ & = \frac{a^2}{a^2-a(b+c)+bc}+\frac{b^2}{b^2-b(c+a)+ca}+\frac{c^2}{c^2-c(a+b)+ab}\\ & = \frac{a^2}{a^2-ab-ac+bc}+\frac{b^2}{b^2-bc-ab+ca)}+\frac{c^2}{c^2-ca-bc+ab}\\ & = -\left[\frac{a^2}{(a-b)(c-a)}+-\frac{b^2}{(a-b)(b-c)}+-\frac{c^2}{(b-c)(c-a)}\right]\\ & = - \left[\frac{a^2(b-c)+b^2(c-a)+c^2(a-b)}{(a-b)(c-a)(c-a)}\right]^{\mathbf{\Large *}}\\ & = \boxed{1} \end{split} \end{equation*}$

$\mathbf{\Large{{}^*}}$ One of a series of results we have been familiar from school, which I copy and paste below. Of course they can be derived. Our equation of interest is the second in the list.

Problem Statement (New) : If $a+b+c=0$, show that $\dfrac{1}{2a^2+bc}+\dfrac{1}{2b^2+ca}+\dfrac{1}{2c^2+ab}=0$

Solution : It can be solved via the same method as above. The numerator can be shown to vanish.

 

[epenguin]

Apologies if this repeats what others have said, but in terms I hope makes the calculation look less heavy. And apologies for being so late - but you are asked (#5) to shew this result, not to show it, which shows the question was first asked more than a century ago.

The ∑ notation for symmetric products economises somewhat the calculation and writing out the denominator:

$D = 8a^2b^2c^2 + 4∑a^3b^3 +2∑a^4bc + a^2b^2c^2$

$= 9(abc)^2 + 4∑(ab)^3 +2∑a^4bc$. .

For the numerator we have more economy:

$N = ∑a^2(2b^2 + ac)(2c^2 + ab)$

$= ∑4abc + 2∑a^3(b^3 + c^3) + ∑a^4bc$

$= 12(abc)^2 + 4∑(ab)^3 + ∑a^4bc$

(You have to be careful about the numerical coefficients this type of calculation.)

N/D = 1 is (N - D) = 0

We have from the above

N - D = $3(abc)^2 - Σa^4bc$

This has factor $abc$ and so we are left to show that a factor of the rest
$3abc - Σa^3$
is $Σa = 0$ which does not look very difficult though hardly obvious.

Write it out

$3abc - (a^3 + b^3 + c^3)$

$= 3abc - [ a^3 +b^3 - (a + b)^3]$

$= 3abc - [a^3 + b^3 - (a^3 + 3a^2b +3ab^2 +b^3)]$

$= 3( abc + a^2b + ab^2)$

$= 3ab(c + a + b)$

as required.