If $a,A,b,B,c,C$ are non negative reals such that $a+A=b+B=c+C=k$ Prove that $aB+bC+cA \le k^2$ If $a,A,b,B,c,C$ are non negative reals such that $a+A=b+B=c+C=k$ Prove that $aB+bC+cA \le k^2$
I substituted $B=k-b,C=k-c,A=k-a$ and plugged them to get a quadratic of $k$ which I had to show positive .SO, the discriminant should be $<0$ but actually its not coming

Think of a equilateral triangle $\triangle ABC$ where each side is $k$, then let $$\overline {AD}=A, \overline {DB}=a, \overline {BE}=B, \overline {EC}=b,  \overline {CF}=C, \overline {FA}=c$$Where $D,E,F$ are points on $\overline {AB}, \overline {BC}, \overline {CA}$ respectively. Now, notice $$\frac{\sqrt 3}{4}Ac+\frac{\sqrt{3}}{4}Cb+\frac{\sqrt{3}}{4}Ba=\triangle {ADF}+\triangle {DBE}+\triangle {ECF} \le \triangle {ABC}=\frac{\sqrt{3}}{4}k^2$$We have our desired result. 
