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\(1-\dfrac{1}{1+a}\ge\dfrac{2017}{b+2017}+\dfrac{2018}{c+2018}\ge2\sqrt{\dfrac{2017.2018}{\left(b+2017\right)\left(c+2018\right)}}\)
\(1-\dfrac{2017}{b+2017}\ge\dfrac{1}{1+a}+\dfrac{2018}{b+2018}\ge2\sqrt{\dfrac{2018}{\left(1+a\right)\left(b+2018\right)}}\)
\(1-\dfrac{2018}{c+2018}\ge\dfrac{1}{1+a}+\dfrac{2017}{b+2017}\ge2\sqrt{\dfrac{2017}{\left(1+a\right)\left(b+2017\right)}}\)
Nhân vế:
\(\dfrac{abc}{\left(a+1\right)\left(b+2017\right)\left(c+2018\right)}\ge\dfrac{8.2017.2018}{\left(a+1\right)\left(b+2017\right)\left(c+2018\right)}\)
\(\Rightarrow abc\ge8.2017.2018\)
Dấu "=" xảy ra khi \(\left(a;b;c\right)=\left(2.1;2.2017;2.2018\right)=...\)
Ta có:
\(\frac{1}{1+a}+\frac{2017}{2017+b}+\frac{2018}{2018+c}\le1\)
\(\Leftrightarrow\frac{a}{1+a}\ge\frac{2017}{2017+b}+\frac{2018}{2018+c}\ge2\sqrt{\frac{2017.2018}{\left(2017+b\right)\left(2018+c\right)}}\left(1\right)\)
Tương tự ta cũng có:
\(\hept{\begin{cases}\frac{b}{2017+b}\ge2\sqrt{\frac{2018}{\left(1+a\right)\left(2018+c\right)}}\left(2\right)\\\frac{c}{2018+c}\ge2\sqrt{\frac{2017}{\left(1+a\right)\left(2017+b\right)}}\left(3\right)\end{cases}}\)
Lấy (1), (2), (3) nhân vế theo vế rút gọi ta được
\(abc\ge2\sqrt{2017.2018}.2.\sqrt{2018}.2.\sqrt{2017}=8.2017.2018\)
\(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=\dfrac{1}{2018}\)
\(\Rightarrow\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}-\dfrac{1}{a+b+c}=0\left(a+b+c=2018\right)\)
\(\Leftrightarrow\dfrac{a+b}{ab}+\dfrac{a+b+c-c}{c\left(a+b+c\right)}=0\)
\(\Leftrightarrow\left[\dfrac{1}{ab}+\dfrac{1}{c\left(a+b+c\right)}\right]\left(a+b\right)=0\)
\(\Leftrightarrow\dfrac{ac+bc+c^2+ab}{abc\left(a+b+c\right)}\times\left(a+b\right)=0\)
\(\Leftrightarrow\dfrac{\left(a+c\right)\left(b+c\right)\left(a+b\right)}{abc\left(a+b+c\right)}=0\)
\(\Leftrightarrow\left[{}\begin{matrix}a=-c\\b=-c\\a=-b\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}b=2018\\a=2018\\c=2018\end{matrix}\right.\)
\(\Rightarrow P=\dfrac{1}{2018^{2017}}\)
hình như bạn bị sai rồi
a=-c
a=-b
b=-c
=>a=-b=-(-c)=c
mà a=-c =>vô lý
