`\quad {6x}/1={9y}/2={12z}/3`
`=>{6x}/1={9y}/2={4z}/1`
`=>({6x}/1)^2=({9y}/2)^2=({4z}/1)^2`
`=>{36x^2}/1={81y^2}/4={16z^2}/1`
`=>{x^2}/{1/{36}}={y^2}/{4/{81}}={z^2}/{1/{16}}`
`=>{x^2}/{1/{36}}={2y^2}/{8/{81}}={3z^2}/{3/{16}}={x^2+2y^2+3z^2}/{1/{36}+8/{81}+3/{16}}=1/{{407}/{1296}}={1296}/{407}`
(Tính chất dãy tỉ số bằng nhau)
`=>{x^2}/{1/{36}}={1296}/{407}`
`=>x^2={1296}/{407}. 1/{36}={36}/{407}`
`=>x=6/{\sqrt{407}` (vì $x>0$)
$\\$
`\qquad {2y^2}/{8/{81}}={1296}/{407}`
`=>2y^2={1296}/{407}. 8/{81}={128}/{407}`
`=>y^2={64}/{407}`
`=>y={8}/{\sqrt{407}` (vì $y>0$)
$\\$
`\qquad {3z^2}/{3/{16}}={1296}/{407}`
`=>3z^2={1296}/{407}. 3/{16}`
`=>z^2={81}/{407}`
`=>z={9}/{\sqrt{407}` (vì $z>0$)
$\\$
`=>x=6/{\sqrt{407}};y=8/{\sqrt{407}};z=9/{\sqrt{407}}`
Theo giả thiết ban đầu, $P$ đạt $GTNN$ khi $a=x;b=y;c=z$
`\qquad P=2a^3+3b^3+4c^3`
`=>P_{min}=2.(6/{\sqrt{407}})^3+3.(8/{\sqrt{407}})^3+4.(9/{\sqrt{407}})^3={12}/{\sqrt{407}}`
$\\$
Vậy $GTNN$ của $P$ là `{12}/{\sqrt{407}}` khi `a=6/{\sqrt{407}};b=8/{\sqrt{407}};c=9/{\sqrt{407}}`
