Question

Cho ba số dương \(0\le a\le b\le c\le1\) chứng minh rằng \(\dfrac{a}{bc+1}+\dfrac{b}{ac+1}+\dfrac{c}{ab+1}\le2\)

Answer 1

Vào đây đi:

https://hoc24.vn/hoi-dap/question/32718.html

Answer 2

0≤a≤b≤c≤1 suy ra \(A=\dfrac{a}{bc+1}+\dfrac{a}{ac+1}+\dfrac{c}{ab+1}\le\dfrac{a+b+c}{abc+1}\) vì a;b;c <1 suy ra \(\left\{{}\begin{matrix}\left(a-1\right)\left(bc-1\right)\ge0\\\left(b-1\right)\left(c-1\right)\ge0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}2abc+1\ge abc+1\ge bc+a\\bc+1\ge b+c\end{matrix}\right.\) 2abc + 2 \(\ge a+bc+1\ge a+b+c\) dấu bằng xảy ra khi (a;b;c) = (0;1;1)

Answer 3

Vậy cũng được tick???

Answer 4

Giải:

Từ giả thiết ta có:

\(\left(1-b\right)\left(1-c\right)\ge0\)

\(\Leftrightarrow1-\left(b+c\right)+bc\ge0\)

\(\Leftrightarrow bc+1\ge b+c\)

\(\Rightarrow\dfrac{a}{bc+1}\le\dfrac{a}{b+c}\le\dfrac{a}{a+b}\left(1\right)\)

Tương tự ta cũng có:

\(\dfrac{b}{ac+1}\le\dfrac{b}{a+c}\le\dfrac{b}{a+b}\left(2\right)\)

\(\dfrac{c}{ab+1}\le c\le1\left(3\right)\)

Cộng theo vế \(\left(1\right);\left(2\right);\left(3\right)\) ta đươc:

\(\dfrac{a}{bc+1}+\dfrac{b}{ac+1}+\dfrac{c}{ab+1}\le\dfrac{a+b}{a+b}+1=2\)

Vậy \(\dfrac{a}{bc+1}+\dfrac{b}{ac+1}+\dfrac{c}{ab+1}\le2\) (Đpcm)

Answer 5

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