<s>[INST] Let $a,b,c>0$ such that $a+b+c=1$. Prove that $\frac{a-bc}{a+bc}+\frac{b-ac}{b+ac}+\frac{c-ab}{c+ab}\le \frac32$ Let $a,b,c>0$ such that $a+b+c=1$. Prove that $\dfrac{a-bc}{a+bc}+\dfrac{b-ac}{b+ac}+\dfrac{c-ab}{c+ab}\le \dfrac32$.  
My attempt:  
First I thought if each term could be less than $\dfrac12$. That did not help. So, a little manipulation yielded  
$\dfrac{bc}{a+bc}+\dfrac{ca}{b+ca}+\dfrac{ab}{c+ab}\ge \dfrac34$.  
Again I thought that each term is greater than $\dfrac14$.  
Proceeding that way yielded $3(ab+bc+ca)\ge 1$  
Now, $a^2+b^2+c^2+2ab+2bc+2ca\ge 3(ab+bc+ca)\ge 1$.  
Here, we get $1\ge 1$, here, I am confused whether the solution is done or not. I don't think it is. So, please help. Thank you.
 [/INST] Using Cauchy-Schwarz Inequality:$$\left(\dfrac{bc}{a+bc}+\dfrac{ca}{b+ca}+\dfrac{ab}{c+ab}\right)\left(\sum\limits_{cyc} bc(a+bc)\right)\ge \left(\sum\limits_{cyc} bc\right)^2$$
So it suffices to prove: $\displaystyle 4\left(\sum\limits_{cyc} bc\right)^2 \ge 3\left(\sum\limits_{cyc} bc(a+bc)\right) = 9abc + 3\sum\limits_{cyc} b^2c^2$
i.e, $\displaystyle \sum\limits_{cyc} b^2c^2 + 8abc(a+b+c) \ge 9abc \iff \sum\limits_{cyc} b^2c^2 \ge abc(a+b+c)$ which is just Cauchy-Schwarz Inequality.
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