đề sai nha bn ; đề zầy (cm : \(\dfrac{c}{a+b}+\dfrac{a}{b+c}+\dfrac{b}{a+c}\ge\dfrac{-3}{2}\) ) mới đúng
đặc : \(P=\dfrac{c}{a+b}+\dfrac{a}{b+c}+\dfrac{b}{a+c}\)
ta có : \(P=\dfrac{c}{a+b}+\dfrac{a}{b+c}+\dfrac{b}{a+c}\)
\(P=\dfrac{c}{a+b}+1+\dfrac{a}{b+c}+1+\dfrac{b}{a+c}+1-3\)
\(P=\dfrac{a+b+c}{a+b}+\dfrac{a+b+c}{b+c}+\dfrac{a+b+c}{a+c}-3\)
\(P=\left(a+b+c\right)\left(\dfrac{1}{a+b}+\dfrac{1}{b+c}+\dfrac{1}{a+c}\right)-3\)
\(P=\dfrac{1}{2}\left(2a+2b+2c\right)\left(\dfrac{1}{a+b}+\dfrac{1}{b+c}+\dfrac{1}{a+c}\right)-3\)
\(P=\dfrac{1}{2}\left[\left(a+b\right)+\left(b+c\right)+\left(a+c\right)\right]\left(\dfrac{1}{a+b}+\dfrac{1}{b+c}+\dfrac{1}{a+c}\right)-3\)
áp dụng bất đẳng thức Bunhiacopxki
ta có : \(\left[\left(a+b\right)+\left(b+c\right)+\left(a+c\right)\right]\left(\dfrac{1}{a+b}+\dfrac{1}{b+c}+\dfrac{1}{a+c}\right)\ge\left(\sqrt{1}+\sqrt{1}+\sqrt{1}\right)\)
\(\Rightarrow\) \(P\ge\dfrac{1}{2}.\left(\sqrt{1}+\sqrt{1}+\sqrt{1}\right)-3=\dfrac{3}{2}-3=\dfrac{-3}{2}\)
vậy \(P=\dfrac{c}{a+b}+\dfrac{a}{b+c}+\dfrac{b}{a+c}\ge\dfrac{-3}{2}\) (đpcm)
Bổ sung thêm đk: a,b,c>0
Đặt \(P=\dfrac{c}{a+b}+\dfrac{a}{b+c}+\dfrac{b}{a+c}\)
\(\Leftrightarrow P+3=\left(\dfrac{c}{a+b}+1\right)+\left(\dfrac{a}{b+c}+1\right)+\left(\dfrac{b}{c+a}+1\right)\)
\(\Leftrightarrow P+3=\dfrac{c+a+b}{a+b}+\dfrac{a+b+c}{b+c}+\dfrac{b+c+a}{c+a}\)
\(\Leftrightarrow P+3=\left(a+b+c\right)\left(\dfrac{1}{a+b}+\dfrac{1}{b+c}+\dfrac{1}{c+a}\right)\)
\(\Leftrightarrow2P+6=\left[\left(a+b\right)+\left(b+c\right)+\left(c+a\right)\right]\left(\dfrac{1}{a+b}+\dfrac{1}{b+c}+\dfrac{1}{c+a}\right)\)Đặt \(\left(a+b;b+c;c+a\right)\rightarrow\left(x;y;z\right)\)( x;y;z>0)
\(2P+6=\left(x+y+z\right)\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)\)
\(2P+6=1+1+1+\left(\dfrac{x}{y}+\dfrac{y}{x}\right)+\left(\dfrac{y}{z}+\dfrac{z}{y}\right)+\left(\dfrac{z}{x}+\dfrac{x}{z}\right)\)
\(2P+3=\left(\dfrac{x}{y}+\dfrac{y}{x}\right)+\left(\dfrac{y}{z}+\dfrac{z}{y}\right)+\left(\dfrac{z}{x}+\dfrac{x}{z}\right)\)
Áp dụng BĐT AM-GM ta có:
\(2P+3\ge2.\sqrt{\dfrac{x}{y}.\dfrac{y}{x}}+2.\sqrt{\dfrac{z}{x}.\dfrac{x}{z}}+2.\sqrt{\dfrac{y}{z}.\dfrac{z}{y}}=2+2+2=6\)
\(\Leftrightarrow2P\ge3\)
\(\Leftrightarrow P\ge\dfrac{3}{2}\)
Dấu " = " xảy ra \(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{x}{y}=\dfrac{y}{x}\\\dfrac{y}{z}=\dfrac{z}{y}\\\dfrac{x}{z}=\dfrac{z}{x}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=y\\y=z\\z=x\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a+b=b+c\\b+c=c+a\\a+c=a+b\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}a=c\\a=b\\b=c\end{matrix}\right.\Leftrightarrow a=b=c\)
phải thêm điều kiện a, b, c > 0 nữa chứ nhỉ ?
+ \(\dfrac{c}{a+b}+\dfrac{a}{b+c}+\dfrac{b}{a+c}\ge\dfrac{3}{2}\)
\(\Leftrightarrow\dfrac{c}{a+b}+1+\dfrac{a}{b+c}+1+\dfrac{b}{a+c}+1\ge3+\dfrac{3}{2}\)
\(\Leftrightarrow\dfrac{a+b+c}{a+b}+\dfrac{a+b+c}{b+c}+\dfrac{a+b+c}{a+c}\ge\dfrac{9}{2}\)
\(\Leftrightarrow\left(a+b+c\right)\left(\dfrac{1}{a+b}+\dfrac{1}{b+c}+\dfrac{1}{c+a}\right)\ge\dfrac{9}{2}\)
\(\Leftrightarrow2\left(a+b+c\right)\left(\dfrac{1}{a+b}+\dfrac{1}{b+c}+\dfrac{1}{c+a}\right)\ge9\)
\(\Leftrightarrow\left[\left(a+b\right)+\left(b+c\right)+\left(c+a\right)\right]\left(\dfrac{1}{a+b}+\dfrac{1}{b+c}+\dfrac{1}{c+a}\right)\ge9\)
+ Áp dụng bđt \(\left(x+y+z\right)\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)\ge9\forall x,y,z>0\)
Dấu "=" xảy ra <=> x = y = z ta có :
\(\left[\left(a+b\right)+\left(b+c\right)+\left(c+a\right)\right]\left(\dfrac{1}{a+b}+\dfrac{1}{b+c}+\dfrac{1}{c+a}\right)\ge9\)
=> đpcm
Dấu "=" xảy ra <=> a = b = c
* Cm bđt : \(\left(x+y+z\right)\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)\ge9\)
\(\Leftrightarrow\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\ge\dfrac{9}{x+y+z}\)
\(\Leftrightarrow\dfrac{xy+yz+zx}{xyz}\ge\dfrac{9}{x+y+z}\)
\(\Leftrightarrow\left(xy+yz+zx\right)\left(x+y+z\right)\ge9xyz\)
\(\Leftrightarrow x^2y+xy^2+x^2z+xz^2+y^2z+yz^2+3xyz\ge9xyz\)
\(\Leftrightarrow x^2y-2xyz+yz^2+xy^2-2xyz+xz^2\)
\(+x^2z-2xyz+y^2z\ge0\)
\(\Leftrightarrow y\left(x-z\right)^2+x\left(y-z\right)^2+z\left(x-y\right)^2\ge0\) luôn đúng \(\forall x,y,z>0\)
=> đpcm
