145 followers Problems #2

since $a^3b^3c^3=27$ , $b^3c+ac^3+a^3b=\dfrac{b^3a^3c^3}{a^3c^2}+\dfrac{b^3a^3c^3}{a^2b^3}+\dfrac{b^3a^3c^3}{b^2c^3}=\dfrac{27}{a^3c^2}+\dfrac{27}{a^2b^3}+\dfrac{27}{b^2c^3}$ by cauchy sarwaz: $({ a }^{ 3 }{ c }^{ 2 }+{ a }^{ 2 }{ b }^{ 3 }+{ b }^{ 2 }{ c }^{ 3 })({ b }^{ 3 }c+a{ c }^{ 3 }+{ a }^{ 3 }b)=({ a }^{ 3 }{ c }^{ 2 }+{ a }^{ 2 }{ b }^{ 3 }+{ b }^{ 2 }{ c }^{ 3 })(\dfrac{27}{a^3c^2}+\dfrac{27}{a^2b^3}+\dfrac{27}{b^2c^3})\geq (\sqrt{27}+\sqrt{27}+\sqrt{27})^2=\boxed{243}$

My LATEX game is weak (or nonexistent) so I'll try to verbally explain this. If you factor (abc)^2 from the left term and (abc) from the right and then apply AM-GM inequality to each you obtain a nice cancelation and are left with 3^5=243 once you apply the constraint that (abc)^3=27. Equality when a=b=c=3^(1/3).