cho biểu thức a=(a/a+b+c) +(b/b+c+d) +(c/c+d+a)+(d/d+a+b) (a,b,c,d>0)chứng minh rằng 1 { "@context": "https://schema.org", "@type": "QAPage", "mainEntity": { "@type": "Question", "name": " cho biểu thức a=(a/a+b+c) +(b/b+c+d) +(c/c+d+a)+(d/d+a+b) (a,b,c,d>0)chứng minh rằng 10)chứng minh rằng 1
cho biểu thức a=(a/a+b+c) +(b/b+c+d) +(c/c+d+a)+(d/d+a+b) (a,b,c,d>0)chứng minh rằng 1
By Margaret
By Margaret
Đáp án:
Giải thích các bước giải:
Ta có :
`a/[a+b+c]>a/[a+b+c+d]`
`b/[b+c+d]>b/[a+b+c+d]`
`c/[c+d+a]>c/[a+b+c+d]`
`d/[d+a+b]>d/[a+b+c+d]`
`=>a/[a+b+c]+b/[b+c+d]+c/[c+d+a]+d/[d+a+b]>a/[a+b+c+d]+b/[a+b+c+d]+c/[a+b+c+d]+d/[a+b+c+d]`
`=>a/[a+b+c]+b/[b+c+d]+c/[c+d+a]+d/[d+a+b]>[a+b+c+d]/[a+b+c+d]`
`=>a/[a+b+c]+b/[b+c+d]+c/[c+d+a]+d/[d+a+b]>1` `(1)`
Lại có :
`a/[a+b+c]<[a+d]/[a+b+c+d]`
`b/[b+c+d]<[b+a]/[a+b+c+d]`
`c/[c+d+a]<[c+b]/[a+b+c+d]`
`d/[d+a+b]<[d+c]/[a+b+c+d]`
`=>a/[a+b+c]+b/[b+c+d]+c/[c+d+a]+d/[d+a+b]<[a+d]/[a+b+c+d]+[b+a]/[a+b+c+d]+[c+b]/[a+b+c+d]+[d+c]/[a+b+c+d]`
`=>a/[a+b+c]+b/[b+c+d]+c/[c+d+a]+d/[d+a+b]<[2(a+b+c+d)]/[a+b+c+d]`
`=>a/[a+b+c]+b/[b+c+d]+c/[c+d+a]+d/[d+a+b]<2` `(2)`
Từ `(1),(2)=>1<a<2`
Đáp án :
`1<A<2`
Các bước giải tương ứng :
`+)`Ta có :
`a/(a+b+c+d)<a/(a+b+c)<a/(a+b)`
`b/(a+b+c+d)<b/(b+c+d)<b/(a+b)`
`c/(a+b+c+d)<c/(c+d+a)<c/(c+d)`
`d/(a+b+c+d)<d/(d+a+b)<d/(c+d)`
`=>a/(a+b+c+d)+b/(a+b+c+d)+c/(a+b+c+d)+d/(a+b+c+d)<a/(a+b+c)+b/(b+c+d)+c/(c+d+a)+d/(d+a+b)<a/(a+b)+b/(a+b)+c/(c+d)+d/(c+d)`
`=>(a+b+c+d)/(a+b+c+d)<A<(a+b)/(a+b)+(c+d)/(c+d)`
`=>1<A<1+1`
`=>1<A<2`
Vậy : `1<A<2`