Đáp án:
Giải thích các bước giải:
`a/(a+1)+b/(b+1)+c/(c+1)=2`
`=>a/(a+1)=2-b/(b+1)-c/(c+1)`
`=>a/(a+1)=(1-b/(b+1))+(1-c/(c+1))`
`=>a/(a+1)=1/(b+1)+1/(c+1)>=2\sqrt{1/(b+1) . 1/(c+1)} `
`=>a/(a+1)>=2/\sqrt{(b+1) . (c+1)} `
Chứng minh tương tự
`b/(b+1)>=2/\sqrt{(a+1) . (c+1)} ,c/(c+1)>=2/\sqrt{(a+1) . (b+1)}`
`=>a/(a+1) . b/(b+1) . c/(c+1) >=2/\sqrt{(b+1) . (c+1)} . 2/\sqrt{(a+1) . (c+1)} . 2/\sqrt{(a+1) . (b+1)}`
`=>a/(a+1) . b/(b+1) . c/(c+1) >=8/(\sqrt{(a+1)(b+1) . (c+1)})^2`
`=>(abc)/[(a+1)(b+1)(c+1)] >=8/[(a+1)(b+1) . (c+1)]`
`=>abc>=8`
Do `a,b,c>0` ,Áp dụng BĐT Co-si
`ab+bc+ca>=3`$\sqrt[3]{ab.bc.ca}$
`=>ab+bc+ca>=3`$\sqrt[3]{(abc)^2}$`>=3`$\sqrt[3]{8^2}$
`=>ab+bc+ca>=3`$\sqrt[3]{8^2}$
`=>ab+bc+ca>=12`
Dấu `=` xảy ra `<=>a=b=c=2`