How sharp are your eyes part 2

F can be written as $F=a^2+b+c+\frac{1}{abc}=a^2+\frac{b}{2}+\frac{b}{2}+\frac{c}{2}+\frac{c}{2}+\frac{1}{2abc}+\frac{1}{2abc}$ Using AM-GM inequality we get $F\geq7\sqrt[7]{\frac{(abc)^2}{2^6(abc)^2}}=\frac{7}{\sqrt[7]{2^6}}=k$ $\Rightarrow\lceil{k}\rceil=4$ * The equality holds when $b=c=\sqrt{\frac{1}{2a^3}}$

this will be tougher than the usual algebra approach,

i used calculus because i suck at AM-GM inequalities

start with simplifying the function

$F={ a }^{ 2 }+b+c+\frac { 1 }{ abc } \\$

taking the gradient

$\nabla F=\begin{pmatrix} 2a-\frac { 1 }{ { a }^{ 2 }bc } \\ 1-\frac { 1 }{ a{ b }^{ 2 }c } \\ 1-\frac { 1 }{ ab{ c }^{ 2 } } \end{pmatrix}\\$

our goal is to find a,b,c such that

$\nabla F=0$

lets start by looking at the second and third row

$1=\frac { 1 }{ ab{ c }^{ 2 } } =\frac { 1 }{ a{ b }^{ 2 }c } \\ ab{ c }^{ 2 }=a{ b }^{ 2 }c\\ \\ c=b\\ \\ \Longrightarrow \frac { 1 }{ a{ c }^{ 3 } } =1$

from the third row we know that,

$a=\frac { 1 }{ \sqrt [ 3 ]{ 2{ c }^{ 2 } } }$

substituting we get

${ c }^{ 3 }=\sqrt [ 3 ]{ 2{ c }^{ 2 } }\\ \\$

solving for c we get

$c=\sqrt [ 7 ]{ 2 }$ i used a calculator at the end,you can definitely do it by hand but i used it to save time

you will get

$\approx 3.864\\ \\$

so the answer is $\boxed{4}$

$F=a^2+b+c+\dfrac 1{abc}. ~~~If~~ a=b=c=1., ~~ F=4.\\ \text{Let us try to reduce say c by posetive } \delta.~ So ~ F= 1^2 + 1 +1 - \delta + \dfrac 1{1*1*(1-\delta)}. \\ \therefore ~F=3+\dfrac 1{1*1*(1-\delta)} - \delta=3 + \dfrac {1-\delta + \delta}{1-\delta}=3+1+\dfrac \delta {1-\delta} - \delta \\ \therefore ~F=4+\dfrac \delta {1-\delta} - \delta > 4.~~~~~~~Note:~~\dfrac \delta {1-\delta} >\delta, \\ \text{Let us try to reduce F by increasing c by posetive } \delta,~so~~ \dfrac 1 {abc} ~ is ~~ reduced.\\ So ~ F= 1^2 + 1 +1 + \delta + \dfrac 1{1*1*(1+\delta)} =3+\dfrac{1+\delta - \delta}{1+\delta} + \delta. \\ \therefore ~F=4-\dfrac \delta {1+\delta} + \delta>4. ~~~~~~~~~Note:~~\delta>\dfrac \delta {1+\delta}.\\ \text{So any of a, b, c is increased or decreased, F>4. So F=4 is the minimum.}$

substituting a=1,b=1,c=1. answer is 4

With my own principle, let all the variables be equal to one for the minimum..