cho a.b.c =1 . chứng minh rằng 1/ab+a +1 . 1/bc+b+1 + 1/ca+c+1 > 2019/2020 13/11/2021 Bởi Caroline cho a.b.c =1 . chứng minh rằng 1/ab+a +1 . 1/bc+b+1 + 1/ca+c+1 > 2019/2020
Đáp án: Giải thích các bước giải: abc=1 ⇒1/ab+a +1 . 1/bc+b+1 + 1/ca+c+1 =1/(ab+a+1) +abc/(bc+b+1) +abc/(abc+ac+c ) =1/(ab+b+1 ) +abc/b(c+1+ac) + abc/(c.(ab+a+1) =1/(ab+b+1) + ac/(c(1+ab+a) + ab/(a+1+ab) =1/(ab+b+1) + a/(1+ab+a) + ab/(a+1+ab)=1>2019/2020 Bình luận
Thay $1=abc$ vào biểu thức ta có :$\dfrac{1}{ab+a+1} + \dfrac{1}{bc+b+1}+\dfrac{1}{ca+c+1}$$ = \dfrac{1}{ab+a+abc}+\dfrac{1}{bc+b+1}+\dfrac{1}{ca+c+abc}$$ = \dfrac{abc}{a.(b+1+bc)}+\dfrac{1}{bc+b+1} + \dfrac{1}{c.(a+1+ab)}$$ = \dfrac{bc}{b+1+bc}+\dfrac{1}{bc+b+1}+\dfrac{abc}{c.(a+abc+ab)}$$ = \dfrac{bc+1}{b+1+bc}+\dfrac{abc}{ca.(1+b+bc)}$$= \dfrac{bc+1}{b+1+bc}+\dfrac{b}{1+b+bc}$$ = \dfrac{bc+1+b}{b+1+bc} = 1 > \dfrac{2019}{2020}$ Vậy $\dfrac{1}{ab+a+1} + \dfrac{1}{bc+b+1}+\dfrac{1}{ca+c+1} ≥ \dfrac{2019}{2020}$ Bình luận
Đáp án:
Giải thích các bước giải:
abc=1 ⇒1/ab+a +1 . 1/bc+b+1 + 1/ca+c+1
=1/(ab+a+1) +abc/(bc+b+1) +abc/(abc+ac+c )
=1/(ab+b+1 ) +abc/b(c+1+ac) + abc/(c.(ab+a+1)
=1/(ab+b+1) + ac/(c(1+ab+a) + ab/(a+1+ab)
=1/(ab+b+1) + a/(1+ab+a) + ab/(a+1+ab)=1>2019/2020
Thay $1=abc$ vào biểu thức ta có :
$\dfrac{1}{ab+a+1} + \dfrac{1}{bc+b+1}+\dfrac{1}{ca+c+1}$
$ = \dfrac{1}{ab+a+abc}+\dfrac{1}{bc+b+1}+\dfrac{1}{ca+c+abc}$
$ = \dfrac{abc}{a.(b+1+bc)}+\dfrac{1}{bc+b+1} + \dfrac{1}{c.(a+1+ab)}$
$ = \dfrac{bc}{b+1+bc}+\dfrac{1}{bc+b+1}+\dfrac{abc}{c.(a+abc+ab)}$
$ = \dfrac{bc+1}{b+1+bc}+\dfrac{abc}{ca.(1+b+bc)}$
$= \dfrac{bc+1}{b+1+bc}+\dfrac{b}{1+b+bc}$
$ = \dfrac{bc+1+b}{b+1+bc} = 1 > \dfrac{2019}{2020}$
Vậy $\dfrac{1}{ab+a+1} + \dfrac{1}{bc+b+1}+\dfrac{1}{ca+c+1} ≥ \dfrac{2019}{2020}$