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
 A: 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. 
A: Here is another way.  You want to find maximum of $a(k-b)+b(k-c)+c(k-a)$.  Note that for any $k$, this is linear in $a, b, c \in [0, k]$, so its extreme values will always be when $a, b, c \in \{0, k\}$. 
Take WLOG $a \ge b \ge c$ and this leaves only the trivial cases $(a, b, c) \in \{(0, 0, 0), (k, 0, 0), (k, k, 0), (k, k, k)\}$ to check.
A: Normalize by $k$ and thus $k=1$. All other terms <=1
To show $a+b+c-ab-bc-ca\le1$.
If $a+b\ge 1$
Then, 
$a+b-ab-1<0$ (Since this is $-(1-a)(1-b)$)
Further, $c(1-a-b)\le 0$. 
Adding these, you get the result. 
If $a+b$<1, Let $a+b=d$, then $d+c-dc\le1$ and $-ab<0$, adding which gives the result.
A: 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 $k$.
