Balancing An ABC Expression

The Cauchy-Schwarz inequality applied to the two vectors $(\sqrt{a},\sqrt{b},\sqrt{c})$ and $\sqrt{bc}, \sqrt{ac},\sqrt{ab})$ gives $9abc \; \le \; (a + b + c)(bc + ac + ab) \; = \; ab + ac + bc \;.$ The Cauchy-Schwarz inequality applied to the vectors $(\sqrt{a},\sqrt{b},\sqrt{c})$ and $\left(\sqrt{\frac{a}{a+bc}},\sqrt{\tfrac{b}{b+ac}},\sqrt{\tfrac{c}{c+ab}}\right)$ gives $\begin{array}{rcl} \displaystyle\left(\frac{a}{\sqrt{a+bc}} + \frac{b}{\sqrt{b + ac}} + \frac{c}{\sqrt{c + ab}}\right)^2 & \le & \displaystyle(a + b + c)\left(\frac{a}{a+bc} + \frac{b}{b+ac} + \frac{c}{c + ab}\right) \\ & = & \displaystyle\frac{a}{(1-b)(1-c)} + \frac{b}{(1-a)(1-c)} + \frac{c}{(1-a)(1-b)} \\ & = & \displaystyle \frac{a(1-a) + b(1-b) + c(1-c)}{(1-a)(1-b)(1-c)} \\ & = & \displaystyle\frac{(a+b+c) - (a+b+c)^2 + 2(ab + ac + bc)}{(b+c)(a+c)(a+b)} \\ & = &\displaystyle \frac{2(ab + ac + bc)}{(b+c)(a+c)(a+b)} \end{array}$ and so, since $(b+c)(a+c)(a+b) \; = \; (a + b + c)(ab + ac + bc) - abc \; = \; ab + ac + bc - abc$ we deduce that $\begin{array}{rcl} \displaystyle\left(\frac{a}{\sqrt{a+bc}} + \frac{b}{\sqrt{b + ac}} + \frac{c}{\sqrt{c + ab}}\right)^2 & \le &\displaystyle \frac{2(ab + ac + bc)}{ab + ac + bc - abc} \\ & \le &\displaystyle \frac{2(ab + ac + bc)}{(ab + ac + bc) - \frac19(ab + ac + bc)} \; = \; \tfrac94 \;. \end{array}$ Since the two Cauchy-Schwarz inequalities become equalities when $a=b=c$ , we deduce that $\frac{a}{\sqrt{a+bc}} + \frac{b}{\sqrt{b + ac}} + \frac{c}{\sqrt{c + ab}} \; \le \; \boxed{\frac32}$ with equality when $a=b=c=\tfrac13$ .