# 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: $$\frac{2\sqrt{abc}}{a+bc}+\frac{2\sqrt{abc}}{b+ca}+\frac{ab-c}{ab+c} \leq \frac{3}{2}$$

• should all numerators be $2\sqrt{abc}$ – DeepSea Jul 21 '14 at 23:56
• Hint: After homogenisation, this inequality is the same as $$\sin Y + \sin Z - \cos X \le \frac32$$ where $X, Y, Z$ are angles of a suitably chosen triangle. – Macavity Jul 22 '14 at 7:51

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 $$\frac{\partial f}{\partial A} = \frac{\partial f}{\partial B} =0 \implies A=B = \sin^{-1} \tfrac12 = \frac{\pi}6$$