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Chứng minh Nesbit 4 số rồi áp dụng nhé
\(\frac{a}{b+c}+\frac{b}{c+d}+\frac{c}{d+a}+\frac{d}{a+b}=\frac{a^2}{a\left(b+c\right)}+\frac{b^2}{b\left(c+d\right)}+\frac{c^2}{c\left(d+a\right)}+\frac{d^2}{d\left(a+b\right)}\) (*)
Theo Cauchy - Schwarz dạng engel , ta có
(*) \(\ge\frac{\left(a+b+c+d\right)^2}{a\left(b+c\right)+b\left(c+d\right)+c\left(d+a\right)+d\left(a+b\right)}\)
\(=\frac{2\left(a+c\right)\left(b+d\right)+\left(a+c\right)^2+\left(b+d\right)^2}{\left(a+c\right)\left(b+d\right)+2ac+2bd}\ge\frac{2\left(a+c\right)\left(b+d\right)+4ac+4bd}{\left(a+c\right)\left(b+d\right)+2ac+2bd}=2\)
Đẳng thức xảy ra <=> a = c và b = d
Áp dụng bất đẳng thức Nesbit cho 4 số ,ta có
\(\frac{2018}{x+y}+\frac{x}{y+2017}+\frac{y}{2017+2018}+\frac{2017}{x+2018}\ge2\)
Đẳng thức xảy ra <=> y = 2018 , x = 2017
\(\dfrac{2-x}{2017}-1=\dfrac{1-x}{2018}-\dfrac{x}{2019}\Leftrightarrow\left(\dfrac{2-x}{2017}+1\right)=\left(\dfrac{1-x}{2018}+1\right)+\left(1-\dfrac{x}{2019}\right)\)
\(\Leftrightarrow\dfrac{2019-x}{2017}=\dfrac{2019-x}{2018}+\dfrac{2019-x}{2019}\)\(\Leftrightarrow\left(2019-x\right)\left(\dfrac{1}{2017}-\dfrac{1}{2018}-\dfrac{1}{2019}\right)=0\)
Ta đã có: \(\dfrac{1}{2017}-\dfrac{1}{2018}-\dfrac{1}{2019}< 0\)
Vậy ta dễ dàng suy ra được \(S=\left\{2019\right\}\)
\(\dfrac{x}{2017}=\dfrac{y}{2018}=\dfrac{z}{2019}=k\\ \Rightarrow\left\{{}\begin{matrix}x=2017k\\y=2018k\\z=2019k\end{matrix}\right.\)
\(4\left(x-y\right)\left(y-z\right)=4\left(2017k-2018k\right)\left(2018k-2019k\right)=4\left(-k\right)\left(-k\right)=4k^2=\left(2k\right)^2=\left(2019k-2017k\right)^2=\left(z-x\right)^2\left(ĐPCM\right)\)
\(\frac{x+1}{2019}+\frac{x+2}{2018}=\frac{x+2017}{3}+\frac{x+2016}{4}\)
\(\Leftrightarrow\frac{x+1}{2019}+1+\frac{x+2}{2018}+1=\frac{x+2017}{3}+1+\frac{x+2016}{4}+1\)
\(\Leftrightarrow\frac{x+2020}{2019}+\frac{x+2020}{2018}-\frac{x+2020}{3}-\frac{x+2020}{4}=0\)
\(\Leftrightarrow\left(x+2020\right).\left(\frac{1}{2019}+\frac{1}{2018}-\frac{1}{3}-\frac{1}{4}\right)=0\)
Mà \(\left(\frac{1}{2019}+\frac{1}{2018}-\frac{1}{3}-\frac{1}{4}\right)\ne0\)
\(\Rightarrow x+2020=0\Leftrightarrow x=-2020\)
Vậy...
Sửa đề: \(\dfrac{x-4}{2019}+\dfrac{x-3}{2018}=\dfrac{x-2}{2017}+\dfrac{x-1}{2016}\)
\(\Leftrightarrow\dfrac{x-4}{2019}+1+\dfrac{x-3}{2018}+1=\dfrac{x-2}{2017}+1+\dfrac{x-1}{2016}+1\)
\(\Leftrightarrow\dfrac{x+2015}{2019}+\dfrac{x+2015}{2018}=\dfrac{x+2015}{2017}+\dfrac{x+2015}{2016}\)
\(\Leftrightarrow\left(x+2015\right)\left(\dfrac{1}{2019}+\dfrac{1}{2018}-\dfrac{1}{2017}-\dfrac{1}{2016}\right)=0\)
\(\Leftrightarrow x=-2015\) vì \(\left(\dfrac{1}{2019}+\dfrac{1}{2018}-\dfrac{1}{2017}-\dfrac{1}{2016}\right)\ne0\)
b) \(x,y\ge1\Rightarrow xy\ge1\)
BĐT đã cho tương đương với:
\(\left(\dfrac{1}{1+x^2}-\dfrac{1}{1+xy}\right)+\left(\dfrac{1}{1+y^2}-\dfrac{1}{1+xy}\right)\ge0\)
\(\Leftrightarrow\dfrac{xy-x^2}{\left(1+x^2\right)\left(1+xy\right)}+\dfrac{xy-y^2}{\left(1+y^2\right)\left(1+xy\right)}\ge0\)
\(\Leftrightarrow+\dfrac{x\left(y-x\right)}{\left(1+x^2\right)\left(1+xy\right)}+\dfrac{y\left(x-y\right)}{\left(1+y^2\right)\left(1+xy\right)}\ge0\)
\(\Leftrightarrow\dfrac{\left(y-x\right)^2\left(xy-1\right)}{\left(1+x^2\right)\left(1+y^2\right)\left(1+xy\right)}\ge0\)
BĐT cuối luôn đúng nên ta có đpcm
Đẳng thức xảy ra khi x=y hoặc xy=1
\(\dfrac{2-x}{2017}+1=\dfrac{x-1}{2018}-1+1-\dfrac{x}{2019}\)
\(\Leftrightarrow\dfrac{2019-x}{2017}=\dfrac{x-2019}{2018}+\dfrac{2019-x}{2019}\)
\(\Leftrightarrow\dfrac{2019-x}{2017}+\dfrac{2019-x}{2018}-\dfrac{2019-x}{2019}=0\)
\(\Leftrightarrow\left(2019-x\right)\left(\dfrac{1}{2017}+\dfrac{1}{2018}-\dfrac{1}{2019}\right)=0\)
\(\Leftrightarrow2019-x=0\) (do \(\dfrac{1}{2017}>\dfrac{1}{2019}\Rightarrow\dfrac{1}{2017}+\dfrac{1}{2018}-\dfrac{1}{2019}>0\))
\(\Rightarrow x=2019\)