Given $a + b + c = 20$, find $\max(ab + ac + bc)$ Given that $a + b + c = 20$, what's the maximum possible value for $ab +ac + bc$? 
$$\left(a, b, c \in \mathbb{N}\right)$$
I tried following this post, Evaluating  max(ab+bc+ac), which lead to:
$ab + bc + ac = 200 - (b - 10)^2 - (c - 10)^2 - bc$, but I couldn't get anywhere afterwards.
Manually trying different numbers lead me to $a = b = 7, c = 6$ with a max result of $133$, but I can't prove it. I assumed for the sum to be the maximum, $a, b, c$ have to be as big as possible, therefore they should be as close to $\frac{20}{3}$ as possible. But is that really the case? If so how can it be proven?

Here is a Solution :
$a+b+c=20$
$(a+b+c)^2=400$
$a^2+b^2+c^2+2(ab+bc+ca)=400$
$X=(ab+bc+ca)=200-(a^2+b^2+c^2)/2$
We want to maximize X which occurs when $Y=(a^2+b^2+c^2)$ is minimized.
When $a<b$ , we can increase $a$ by $\delta$ while decreasing $b$ by $\delta$ (keeping total sum Constant) , which decreases $Y$.
When $a>b$ , we can decrease $a$ by $\delta$ while increasing $b$ by $\delta$ (keeping total sum Constant) , which decreases $Y$.
Likewise , $b<c$ , $b>c$ , $c<a$ , $c>a$ can all be adjusted to "Equality" to minimize $Y$.
We can not adjust further when we have attained "Equality" for the 3 variables.
In other words , $Y$ is minimum when $a=b=c=20/3$
In that Case , $X=3(20/3)(20/3)=400/3=133.333$ is the maximum , when $a,b,c$ are real numbers. When these are Integers, this value is the upper bound.
We can take nearest Integers $(a,b,c)=(7,7,6)$ to get the Integral maximum $7\times7+7\times6+6\times7=133$.
