If $x,y,z$ are real numbers satisfying$x/(y+z) +y/(z+x) +z/(x+y) =1$ then $x^2/(y+z) +y^2/(z+x)+z^2/(x+y)=$

Question

I have tried it a few times but I am not making any progress. Please help.

Closely related: https://math.stackexchange.com/q/1082415/42969. — Martin R, Jul 10 '19 at 08:21
Welcome to MSE. Please include your question in the body of the question, instead of putting it only in the title. — José Carlos Santos, Jul 10 '19 at 08:23
This identity https://math.stackexchange.com/a/1082469/42969 is all you need — Martin R, Jul 10 '19 at 08:39

score 3 · Accepted Answer · answered Jul 10 '19 at 08:37

3

$$\dfrac{x^2}{y+z}+\dfrac{y^2}{z+x}+\dfrac{z^2}{x+y}$$

$$=\dfrac{x^2}{y+z}+x+\dfrac{y^2}{z+x}+y+\dfrac{z^2}{x+y}+z-(x+y+z)$$

$$=(x+y+z)\left(\dfrac x{y+z}+\cdots\right)-(x+y+z)$$

$$=?$$

answered Jul 10 '19 at 08:37

lab bhattacharjee

274,582

Whoa man thanks a lot. I had tried the same method before but due to an error while doing calculation i was not able to get the answer.thanks – Okabe Jul 10 '19 at 11:52

score 1 · Answer 2 · answered Jul 10 '19 at 08:36

Your first equation simplifies to $$x^3+y^3+z^3+xyz=0$$. The left-hand side of your given term is (factorized) $${\frac { \left( x+y+z \right) \left( {x}^{3}+xyz+{y}^{3}+{z}^{3} \right) }{ \left( y+z \right) \left( z+x \right) \left( x+y \right) }} $$ Using your equation above so we get $$0$$

If $x,y,z$ are real numbers satisfying$x/(y+z) +y/(z+x) +z/(x+y) =1$ then $x^2/(y+z) +y^2/(z+x)+z^2/(x+y)=$

2 Answers2