Who's up to the challenge? 3

$F=a+b+{ c }+\frac { 1 }{ a } +\frac { 1 }{ b } +\frac { 1 }{ c } +\ln(abc)$ Hence, $\nabla F=\begin{pmatrix} 1-\frac { 1 }{ { a }^{ 2 } } +\frac { 1 }{ a } \\ 1-\frac { 1 }{ { b }^{ 2 } } +\frac { 1 }{ b } \\ 1-\frac { 1 }{ { c }^{ 2 } } +\frac { 1 }{ c } \end{pmatrix}$ Since the equation is symmetric and we're looking for $a,b,c$ such that $\nabla F=0$ we can solve for when $1-\frac{1}{x^2}+\frac{1}{x}=0$

There is a unique positive root to $1-\frac{1}{x^2}+\frac{1}{x}$ , hence $a,b,c$ must all be equal to this value at equilibrium.

$1-\frac { 1 }{ { a }^{ 2 } } +\frac { 1 }{ a } =0$ $a=-\phi \quad or\quad \phi -1$ Since only one value of $a$ is positive, namely $\phi-1$ , we can proceed to take the hessian. $H(a,b,c)=\begin{bmatrix} 1+\frac { 2 }{ { a }^{ 3 } } -\frac { 1 }{ { a }^{ 2 } } & 0 & 0 \\ 0 & 1+\frac { 2 }{ { b }^{ 3 } } -\frac { 1 }{ { a }^{ 2 } } & 0 \\ 0 & 0 & 1+\frac { 2 }{ { c }^{ 3 } } -\frac { 1 }{ { c }^{ 2 } } \end{bmatrix}$ $H(\phi -1,\phi -1,\phi -1)\approx \begin{pmatrix} 6.854 & 0 & 0 \\ 0 & 6.854 & 0 \\ 0 & 0 & 6.854 \end{pmatrix}$

The determinant of the first,second,and third principal minors of the matrix above are positive which implies that this is a local minimum. After plugging in we obtain $\min{F(a,b,c)} \approx 5.2$

At minima of the F, a+1/a+ln(a), b+1/b+ln(b) and c+1/c+ln(c) individually are at their minima since a b and c are independent of each other. Individually finding the minima of any of the above and then multiplying by 3 gives the answer

$A.M.\ge G.M.\\ \frac { a+b+c+\frac { 1 }{ a } +\frac { 1 }{ b } +\frac { 1 }{ c } }{ 6 } \ge \sqrt [ 6 ]{ \frac { a.b.c }{ a.b.c } } \\ =>a+b+c+\frac { 1 }{ a } +\frac { 1 }{ b } +\frac { 1 }{ c } \ge 6\quad \\ we\quad know\quad that\quad \ln { x } >0\\ =>a+b+c+\frac { 1 }{ a } +\frac { 1 }{ b } +\frac { 1 }{ c } +\ln { x } >6\quad \\ min\quad a+b+c+\frac { 1 }{ a } +\frac { 1 }{ b } +\frac { 1 }{ c } +\ln { x } =6$