Ta có: \(0\le a\le b\le c\le1\Leftrightarrow\left\{{}\begin{matrix}1-a\ge0\\1-b\ge0\end{matrix}\right.\)
\(\Rightarrow\left(1-a\right)\left(1-b\right)\ge0\Leftrightarrow1\left(1-b\right)-a\left(1-b\right)\ge0\)
\(\Rightarrow1-b-a+ab\ge0\Leftrightarrow1+ab\ge a+b\)
Tiếp tục chứng minh ta có: \(\left\{{}\begin{matrix}1\ge c\\0\le a\le b\Leftrightarrow ab\ge0\end{matrix}\right.\)
cộng theo vế: \(1+ab+1+ab\ge a+b+c+0\)
\(\Rightarrow2\left(1+ab\right)\ge a+b+c\)
Ta có: \(\dfrac{c}{ab+1}=\dfrac{2c}{2\left(ab+1\right)}\le\dfrac{2c}{a+b+c}\) (1)
chứng minh tương tự suy ra đpcm
Ta có: 0≤a≤b≤c≤1⇔{1−a≥01−b≥00≤a≤b≤c≤1⇔{1−a≥01−b≥0
⇒(1−a)(1−b)≥0⇔1(1−b)−a(1−b)≥0⇒(1−a)(1−b)≥0⇔1(1−b)−a(1−b)≥0
⇒1−b−a+ab≥0⇔1+ab≥a+b⇒1−b−a+ab≥0⇔1+ab≥a+b
Tiếp tục chứng minh ta có: {1≥c0≤a≤b⇔ab≥0{1≥c0≤a≤b⇔ab≥0
cộng theo vế: 1+ab+1+ab≥a+b+c+01+ab+1+ab≥a+b+c+0
⇒2(1+ab)≥a+b+c⇒2(1+ab)≥a+b+c
Ta có: cab+1=2c2(ab+1)≤2ca+b+ccab+1=2c2(ab+1)≤2ca+b+c (1)
Ta có:
\(\left(1-b\right)\left(1-c\right)\ge0\)
\(\Leftrightarrow1+bc\ge b+c\)
\(\Leftrightarrow\dfrac{a}{bc+1}\le\dfrac{a}{b+c}\le\dfrac{a}{a+b}\left(1\right)\)
Tương tự ta cũng có:
\(\dfrac{b}{ac+1}\le\dfrac{b}{a+b}\left(2\right)\)
Bên cạnh đó ta lại có:
\(\dfrac{c}{ab+1}\le c\le1\left(3\right)\)
Từ (1), (2), (3) ta có:
\(\sum\dfrac{a}{bc+1}\le\dfrac{a}{a+b}+\dfrac{b}{a+b}+1=2\)
Dấu = xảy ra khi \(\left\{{}\begin{matrix}a=0\\b=c=1\end{matrix}\right.\)
\(0\le a\le b\le c\le1\\ \Rightarrow\left(a-1\right)\left(b-1\right)\ge0\\ \Rightarrow ab-a-b+1\ge0\\ \Rightarrow ab+1\ge a+b\\ \)
Chứng minh tương tự ta được :
\(bc+1\ge b+c\\ ac+1\ge a+c\)
\(\Rightarrow\dfrac{a}{bc+1}+\dfrac{b}{ac+1}+\dfrac{c}{ab+1}\\ \ge\dfrac{a}{b+c}+\dfrac{b}{a+c}+\dfrac{c}{a+b}\\ \ge\dfrac{a+a}{a+b+c}+\dfrac{b+b}{a+b+c}+\dfrac{c+c}{a+b+c}\\ =\dfrac{2\left(a+b+c\right)}{a+b+c}=2\)
:v .