8
$\begingroup$

Suppose that $a,b,c$ are three real numbers such that $\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}=1$.

What are the possible values for $\dfrac{a^2}{b+c}+\dfrac{b^2}{c+a}+\dfrac{c^2}{a+b}$?

We have $a(c+a)(a+b)+b(b+c)(a+b)+c(b+c)(c+a)=(a+b)(b+c)(a+c)$. But then I'm stuck. This question is related, but a bit different.

Thank you for your help!

$\endgroup$
  • 1
    $\begingroup$ What is your source of these problems? Even the last one was brilliant about the triangle. $\endgroup$ – астон вілла олоф мэллбэрг Aug 23 '16 at 13:05
  • $\begingroup$ @астонвіллаолофмэллбэрг: I don't know the exact source of the two problems. These were told to me by a friend of mine. However, one of my previous question was apparently taken from "Aufgabe 11" here. $\endgroup$ – Alphonse Aug 23 '16 at 13:10
9
$\begingroup$

Since $\sum\limits_\text{cyc}\,\frac{a}{b+c}=1$, we have $$a+b+c=(a+b+c)\,\left(\sum_\text{cyc}\,\frac{a}{b+c}\right)=\sum_{\text{cyc}}\left(\frac{a^2}{b+c}+a\right)\,.$$ That is, $$a+b+c=\sum_{\text{cyc}}\left(\frac{a^2}{b+c}\right)+(a+b+c)\,.$$ Hence, $$\sum_{\text{cyc}}\left(\frac{a^2}{b+c}\right)=0\,.$$

$\endgroup$
  • 1
    $\begingroup$ I'm sorry, I don't get it. I get $(a+b+c)\sum_{\text{cyc}}\frac{a}{b+c}=\sum_{\text{cyc}}\left(\frac{a^2}{b+c}+a\left(\frac{b}{c+a}+\frac{c}{a+b} \right) \right)$. How is $\sum_{\text{cyc}}a\left(\frac{b}{c+a}+\frac{c}{a+b} \right) = \sum_{\text{cyc}}a$? Thanks. $\endgroup$ – Bobson Dugnutt Aug 23 '16 at 13:36
  • 1
    $\begingroup$ $\frac{a(a+b+c)}{b+c}=\frac{a^2}{b+c}+a$ $\endgroup$ – Batominovski Aug 23 '16 at 13:40
  • $\begingroup$ Thanks! .. it's a neat feature that $(a+b+c)=(b+c+a)=(c+a+b)$, so you can simply include it under the summation. $\endgroup$ – Bobson Dugnutt Aug 23 '16 at 13:45
3
$\begingroup$

Hint. You have

$$ \frac{a}{b+c} + \frac{b}{c+a} + \frac{c}{a+b} - 1 = \frac{a^3+b^3+c^3+abc}{(a+b)(b+c)(c+a)} $$

as well as

$$ \frac{a^2}{b+c} + \frac{b^2}{c+a} + \frac{c^2}{a+b} = \frac{(a^3+b^3+c^3+abc)(a+b+c)}{(a+b)(b+c)(c+a)}. $$

$\endgroup$
3
$\begingroup$

Assuming

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

we also have

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

as well as

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

and

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

These three sum together as but all terms without $(.)^2$ on the other side on sorting terms with same denominator you get $$\frac{a^{2}}{b+c} + \frac {b^2}{c+a} + \frac {c^2}{a+b} =$$ $$ a+b+c -(\frac {ac}{b+c} + \frac{ab}{b+c}) - (\frac {ab}{c+a} + \frac {bc}{c+a}) - (\frac {ac}{a+b} + \frac {bc}{a+b}) = $$ $$ a + b + c - (a) - (b)- (c) = 0$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.