Problem 14

Relevant wiki: Cauchy-Schwarz Inequality

I am not sure if what I did is completely fine. Denote $a+b+c=3u , ab+bc+ca=3v^2, abc=w^3$ . We have $u=tv^2, t\geq1$ . Now the expression we try to minimize is:

$\frac{9u^2-6v^2+3u}{3v^2}=\frac{9t^2v^4-6v^2+3tv^2}{3v^2}=3v^2t^2+t-2$ . Now since $3v^2>0$ we have in the RHS an increasing function of the variable t, which gets a minimum when t is minimal, thus, $t=1$ or $a+b+c=ab+bc+ca$ . In this point, the result of the expression will be $3v^2-1=3u-1$ . Thus, we want to minimize $v^2=u$ . Now we look on the constraint. It can be re-written as:

$3w+w^3\leq3u+w^3\leq3v^2+w^3\leq4$ , or $3w+w^3\leq4$ . Now since we want to maximize $w$ in order to minimze $u$ , we have $3w+w^3=4$ . The one real solution is $w=1$ , and since the equality happens when $a=b=c$ we have $a=b=c=1$ , substituting will give us the result, 2.