If $a+b+c=1$, then $\frac{2\sqrt{abc}}{a+bc}+\frac{2\sqrt{abc}}{b+ca}+\frac{ab-c}{ab+c} \leq \frac{3}{2}$ Someone can to help me with a hint in the following problem:
Let $a$, $b$ and $c$ be positive real numbers such that $a+b+c=1$. Prove that:
$$\frac{2\sqrt{abc}}{a+bc}+\frac{2\sqrt{abc}}{b+ca}+\frac{ab-c}{ab+c} \leq \frac{3}{2}$$

Homogenising, we have to show:
$$\frac{2\sqrt{abc(a+b+c)}}{a(a+b+c)+bc} + \frac{2\sqrt{abc(a+b+c)}}{b(a+b+c)+ca}+\frac{ab-c(a+b+c)}{ab+c(a+b+c)} \le \frac32$$
Now it motivates the triangle substitution $x=a+b, y=b+c, z = c+a$, so we have a triangle with area say  $\Delta$, to get the equivalent:
$$\frac{\Delta}{2xz}+\frac{\Delta}{2xy}+\frac{x^2-y^2-z^2}{2yz} \le \frac32$$
So with $X, Y, Z$ angles of some triangle, this is equivalent to the inequality:
$$\sin Y + \sin Z - \cos X =\sin Y + \sin Z + \cos (Y+Z) \le \frac32$$
It is sufficient to show that the unconstrained maximum of $f(A, B) = \sin A + \sin B + \cos(A+B)$ is $\frac32$.  This is easy to check as we have maximum when
$$\frac{\partial f}{\partial A} = \frac{\partial f}{\partial B} =0 \implies A=B = \sin^{-1} \tfrac12 = \frac{\pi}6$$
