Max and min part 4

Call the expression P, we see that $P=(a+b+c)^2-(ab+bc+ac)=81-(ab+bc+ac)$ We can prove that $ab+bc+ac\leq \frac{(a+b+c)^2}{3}=27 \therefore P\geq 54$ , the equality holds when $a=b=c=3$

Now we assume that $a\geq b\geq c\geq 0$ we have $f(a,a,c)=3a^2+c^2+2ac$ $P-f(a,a,c)=a^2+b^2+c^2+ab+bc+ac-3a^2-c^2-2ac$ $= -2a^2+b^2+ab+bc-ac=(b-a)(b+a)+a(b-a)+c(b-a)\leq 0$ $\Rightarrow P\leq f(a,a,c)$ Now we have $f(a,a,c)=3a^2+c^2+2ac$ and $2a+c=9$ . Since $a,c\in[0;4]$ and $a\geq c$ , we have $a\geq 3$ and $c\leq 3$

Subtituting $c=9-2a$ into $f(a,a,c)$ , we get $f(a,a,c)=f(a)=3a^2+(9-2a)^2+2a(9-2a)=3a^2-18a+81$ We see that $f(a,a,c)$ increases in $[3;4]$ so $f(a,a,c)\leq f(4)=57$ The equality holds when $(a,b,c)=(4,4,1)$ and its permutations