Basic Algebra Test

We rewrite the expression into

$\dfrac{a}{b+c}+\dfrac{b}{a+c}+\dfrac{c}{a+b}=\dfrac{a+b+c}{a+b+c}$ .

Multiplying both sides by $a+b+c$ , we get

$\dfrac{a^{2}}{b+c}+a+\dfrac{b^{2}}{a+c}+b+\dfrac{c^{2}}{a+b}+c=a+b+c$ .

Subtracting $a+b+c$ from both sides,

$\dfrac{a^{2}}{b+c}+\dfrac{b^{2}}{a+c}+\dfrac{c^{2}}{a+b}=\boxed{0}$ .

Edit: Oops! I didn't mean to copy the comment of Mr. Sahil. My computer was just so slow that I couldn't see the comment that time. So I thought there was no "proper" solution for this. And instead placed mine. Hope this resolves any misconceptions (I guess).

$\frac { { a }^{ 2 } }{ b+c } +\frac { ab }{ a+c } +\frac { ac }{ a+b } =a$

$\frac { ba }{ b+c } +\frac { { b }^{ 2 } }{ a+c } +\frac { bc }{ a+b } =b$

$\frac { ca }{ b+c } +\frac { cb }{ a+c } +\frac { { c }^{ 2 } }{ a+b } =c$

Add the equations up to get

$\frac { { a }^{ 2 } }{ b+c } +\frac { { b }^{ 2 } }{ a+c } +\frac { { c }^{ 2 } }{ a+b } +\frac { a(b+c) }{ b+c } +\frac { b(a+c) }{ a+c } +\frac { c(a+b) }{ a+b } =a+b+c$

Therefore $\frac { { a }^{ 2 } }{ b+c } +\frac { { b }^{ 2 } }{ a+c } +\frac { { c }^{ 2 } }{ a+b } +a+b+c=a+b+c$

So $\frac { { a }^{ 2 } }{ b+c } +\frac { { b }^{ 2 } }{ a+c } +\frac { { c }^{ 2 } }{ a+b } =0$

$(\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}) \times (a+b+c)=(\frac{a^2}{b+c}+\frac{b^2}{a+c}+\frac{c^2}{a+b})+(a+b+c)$ $\Rightarrow \frac{a^2}{b+c}+\frac{b^2}{a+c}+\frac{c^2}{a+b} = (a+b+c) [(\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b})-1]=\boxed{0}$

First of all, my solution uses that they're asking us for a number. If we have some values $a, b, c$ that satisfy the condition, then $-a, -b, -c$ also satisfy (because $\frac { a }{ b+c } =\frac { -a }{ -b-c }$ , and the same to the other fractions. But, in the other fractions, the numbers are sqared, so $\frac { { a }^{ 2 } }{ b+c } =-\frac { { (-a) }^{ 2 } }{ -b-c }$ , and combining for the three fractions, we have $\frac { { a }^{ 2 } }{ b+c } +\frac { { b }^{ 2 } }{ a+c } +\frac { c^{ 2 } }{ a+b } =-\left( \frac { { a }^{ 2 } }{ b+c } +\frac { { b }^{ 2 } }{ a+c } +\frac { c^{ 2 } }{ a+b } \right)$ , which implies $\frac { { a }^{ 2 } }{ b+c } +\frac { { b }^{ 2 } }{ a+c } +\frac { c^{ 2 } }{ a+b } =0$

The denominators are sums which we don't really like. We turn them into single terms by letting $b+c=s$ , $a+c=t$ , $a+b=l$ , and $a+b+c=k$ . Then we have: $\frac{k-s}{s} + \frac{k-t}{t} + \frac{k-l}{l} = 1$ , which implies $\frac{k}{s} + \frac{k}{t} + \frac{k}{l} =4$ . Then we can write $\frac{a^2}{b+c} + \frac{b^2}{a+c} + \frac{c^2}{a+b}$ $=\frac{(k-s)^2}{s} + \frac{(k-t)^2}{t} + \frac{(k-l)^2}{l}$ $=\frac{k^2}{s} + s - 2k + \frac{k^2}{t} + t - 2k + \frac{k^2}{l} + l - 2k$ $=4k-6k+2k=\boxed{0}$ And we are done.

Creative solution :D

By Nesbitt's Inequality, we have $\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}\geq\frac{3}{2}$ for all positive real numbers $a,b,c$ .

If $a,b,c$ are all negative, then $-a,-b,-c$ are all positive, then we will arrive at the same equation, which is a contradiction. Hence at least one of $a,b,c$ must be $0$ . WLOG set $a=0$ . Then we have $\frac{b}{c}+\frac{c}{b}=1$ $\implies b^2+c^2=bc$

We want to find $\frac{b^2}{c}+\frac{c^2}{b}.$

Note that $\frac{b^2}{c}+\frac{c^2}{b}$ $=\frac{b^3+c^3}{bc}$ $=\frac{(b+c)(b^2+c^2-bc)}{bc}$

Since $b^2+c^2=bc$ , then $\frac{(b+c)(b^2+c^2-bc)}{bc}=\boxed{0}$

$Let\quad x=\frac { a }{ b+c } ,\quad y=\frac { b }{ a+c } \quad and\quad z=\frac { c }{ a+b } .\quad \\ \\ By\quad rewriting\quad the\quad constraint,\quad we\quad have\quad x+y+z=1\\ \\ Since\quad a=x(b+c),\quad b=y(a+c),\quad c=z(a+b),\quad we\quad have\\ a+b+c=x(b+c)+y(a+c)+z(a+b)\\ a+b+c=(a+b+c)(x+y+z)-ax-by-cz\\ a+b+c=a+b+c-ax-by-cz\\ ax+by+cz=0$

In the given equation write 1 as a+b+c/a+b+c and then cross multiply to get the answer ;)

$\dfrac{a}{b+c}+\dfrac{b}{a+c}+\dfrac{c}{a+b}=\dfrac{a}{a+b+c}+\dfrac{b}{a+b+c}+\dfrac{c}{a+b+c}$ so . $\sum{a(\dfrac{1}{b+c}-\dfrac{1}{a+b+c})}=0$ .

hence . $\sum{\dfrac{a^{2}}{(b+c)(a+b+c)}}=0$ .

because $a+b+c$ can't be 0

so $\sum{\dfrac{a^{2}}{b+c}}=0$

Q.E.D

Looking at the solutions given, I came to know how stupid I am.
Here is what I did (not a correct method but interesting):
Take $c=0$ . Then the equation reduces to $\frac{a}{b}+\frac{b}{a}=1$ , solving this we get $\frac{a}{b}=-w$ where $w$ is the complex cube root of 1.
Now, if we take $b=w$ , then $a=-w^{2}$
Substituting these values, the given expression is found to be equal to 0.

I think the least elegant solution would be this.

From $\frac{a}{b+c} + \frac{b}{a+c} + \frac{c}{a+b} = 1$

we deduce $a^3 + b^3 + c^3 + abc = 0$ .

Taking the least common multiplier in the expression

$\frac{a^2}{b+c} + \frac{b^2}{a+c} + \frac{c^2}{a+b}$

and computing the numerator we found it is

$(a + b + c) (a^3 + b^3 + c^3 + abc ) = 0$ .

That's it.

Let $x := a/(b+c),\\ y := b/(c+a),\\ z := c/(a+b),\\ w := ax+by+cz,$ then $a+b+c = (x+y+z)(a+b+c) = w + x(b+c) + y(c+a) + z(a+b) = w + a + b + c,$ which implies $w = 0$ .

Let $s=a+b+c$

$\therefore \dfrac{a}{s-a} + \dfrac{b}{s-b}+\dfrac{c}{s-c} = 1$

Now

$\dfrac{a^2}{s-a} + \dfrac{b^2}{s-b}+\dfrac{c^2}{s-c} = \dfrac{a^2 -sa +sa}{s-a} + \dfrac{b^2 -sb +sb}{s-b}+\dfrac{c^2-sc+sc}{s-c}$

$=\dfrac{a(a -s) +sa}{s-a} + \dfrac{b(b -s) +sb}{s-b}+\dfrac{c(c-s)+sc}{s-c}$

$=\dfrac{a(a -s)}{s-a} + \dfrac{b(b -s) }{s-b}+\dfrac{c(c-s)}{s-c} + s( \dfrac{a}{s-a} + \dfrac{b}{s-b}+\dfrac{c}{s-c} )$

$=-s+s=0$

The answer is easily obtained without doing the math. Since any solutions for a,b,c will work equally well with 2a,2b,2c, then assuming the question is valid and has an answer, 2(a^2+b^2+c^2) = a^2+b^2+c^2, so the answer must be zero.

I found the least elegant method. I noted that the problem is homogeneous, so a=1 is a possible solution. Substituting a =1 gives us a homogeneous equation in b,c so we can set b=1 as well and solve for c, which turns out to be c= (1+- i rt 7)/2. Using c = (1 + i rt 7)/2 and a=b=1 in the second equation yields 0. Not pretty, but it works. Congratulations to the other solvers who found much nicer solutions!

Just as a side note, it is not hard to guess the numerical value, because the given condition on $a,b,c$ is homogeneous.

so i like joseph varghese's solution (simplify the expression a+b+c); Sean Ty's solution isn't bad either, very clever. I personally just solve problems with approaches I don't need to memorize. It's not efficient, but fairly easy to follow.

for this problem it took me a while to find patterns, but eventually after an hour of mulling over it I came across the geometric first step of representing the denominator of the expressions as a box with lengths, height, and widths of (b+c), (a+b), and (a+c); multiplying this out takes a while, but splitting the volume into 8 pieces, and then multiply the original expression out with common denominator, we get (b+c)(a+c)(a+b)=a^2b+ca^2+ab^2+abc+ac^2+abc+bc^2+cb^2=a^3+a^2b+a^2c+abc+ab^2+b^3+abc+cb^2+abc+bc^2+ac^2+c^3; the multiplication is sort of tedious so this isn't really a clever solution, but canceling terms in this expression yields a nice expression 0=a^3+b^3+c^3+abc;

multiplying the question expression out to a common denominator yields a numerator a^4+b^4+c^4+ba^3+ca^3+ab^3+cb^3+bc^3+ac^3+bca^2+acb^2+abc^2=a(a^3+a^2b+a^2c+b^3+c^3+abc+cb^2+bc^2)+b^4+c^4+cb^3+bc^3=a^3b+a^3c+acb^3+abc^2+b^4+c^4+cb^3+bc^3=b(a^3+acb+ac^2+b^3+cb^2+c^3)=b(ac^2+cb^2)+a^3c+c^4=c(abc+b^3+a^3+c^3)=0

brilliant encourages users to find the most efficient solution possible, so this solution probably won't win you the math olympics, but I find value in solving problems with your own wits first and only look at solutions if you'v done all you can; though this again is not efficient because I'm basically reinventing the "wheel" everytime instead of utilizing what my predecessors have to offer.

I'd recommend everybody to not try to reinvent the "wheel", but understanding how the wheel came to be can be important if I want to understand how the past differs from the present, and history...and so on so forth.

enough ranting anyways, I'd just add that I've finally realized what I've done wrong in the past. If everyone went about life like me, the world might still be in the dark ages lol

Rearranging the original equation:

$\frac{a}{b+c}=1-\frac{b}{a+c}-\frac{c}{a+b}$

$\frac{b}{a+c}=1-\frac{a}{b+c}-\frac{c}{a+b}$

$\frac{c}{a+b}=1-\frac{a}{b+c}-\frac{b}{a+c}$

Substituting into the second equation:

$a(1-\frac{b}{a+c}-\frac{c}{a+b})+b(1-\frac{a}{b+c}-\frac{c}{a+b})+c(1-\frac{a}{b+c}-\frac{b}{a+c})$

Distributing:

$=a-\frac{ab}{a+c}-\frac{ac}{a+b}+b-\frac{ba}{b+c}-\frac{bc}{a+b}+c-\frac{ac}{b+c}-\frac{bc}{a+c}$

Collecting terms:

$=a+b+c-\frac{b}{a+c}(a+c)-\frac{c}{a+b}(a+b)-\frac{a}{b+c}(b+c)$

Cancelling:

$=a+b+c-a-b-c$

$=\boxed{0}$

Again, Mathematica does the trick:

Simplify[ a^2/(b+c)+b^2/(a+c)+c^2/(a+b) /. Solve[a/(b+c)+b/(a+c)+c/(a+b)==1,{a}] ]

Gives

{0,0,0}

As the three possible solutions.

Where r you multiplying with a+b+c on both the sides.