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

From 
$$
 ABC = (k-a)(k-b)(k-c) \\
 = k^3 - (a+b+c)k^2 + (ab + bc + ca)k - abc \\
 = k \bigl( k^2 - (a+b+c)k + (ab + bc +ca) \bigr) - abc \\
 = k \bigl( k^2 - (a B + b C + c A ) \bigr) - abc 
$$
it follows that
$$
 a B + b C + c A = k^2 - \frac{abc + ABC}{k}
 \le k^2
$$
Equality holds if and only if
$$
 abc = ABC = 0
$$
which means that one of $a, b, c$ is  zero and another
one equal to $k$.
