`\qquad 1/a+1/b+1/c=2=>a;b;c\ne 0`
Vì `a+b+c=abc`
`=>{a+b+c}/{abc}={abc}/{abc}`
`=>a/{abc}+b/{abc}+c/{abc}=1`
`=>1/{bc}+1/{ac}+1/{ab}=1`
$\\$
Vì ` 1/a+1/b+1/c=2`
`=>(1/a+1/b+1/c)^2=2^2`
`=>(1/a+1/b)^2+2(1/a+1/b). 1/c+1/{c^2}=4`
`=>1/{a^2}+2/{ab}+1/{b^2}+2/{ac}+2/{bc}+1/{c^2}=4`
`=>1/{a^2}+1/{b^2}+1/{c^2}+2(1/{bc}+1/{ac}+1/{ab})=4`
`=>1/{a^2}+1/{b^2}+1/{c^2}=4-2(1/{bc}+1/{ac}+1/{ab})=4-2.1=2`
Vậy: `1/{a^2}+1/{b^2}+1/{c^2}=2` (đpcm)
