Đặt \(\left\{{}\begin{matrix}a^2-1=x\\b^2-1=y\\c^2-1=z\end{matrix}\right.\)(x,y,z>0)thì giả thiết trở thành \(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=1\)
chứng minh \(\dfrac{1}{\sqrt{x+1}+1}+\dfrac{1}{\sqrt{y+1}+1}+\dfrac{1}{\sqrt{z+1}+1}\le1\)
Áp dụng BĐT cauchy:(dạng \(\dfrac{1}{a+b+c}\le\dfrac{1}{9}\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\))(
\(\sum\dfrac{1}{\sqrt{x+1}+1}\le\sum\dfrac{1}{9}\left(\dfrac{4}{\sqrt{x+1}}+1\right)=\dfrac{4}{9}\left(\dfrac{1}{\sqrt{x+1}}+\dfrac{1}{\sqrt{y+1}}+\dfrac{1}{\sqrt{z+1}}\right)+\dfrac{1}{3}\)(*)
mà theo BĐT bunyakovsky:\(\left(\sum\dfrac{1}{\sqrt{x+1}}\right)^2\le3\left(\dfrac{1}{x+1}+\dfrac{1}{y+1}+\dfrac{1}{z+1}\right)\le\dfrac{3}{16}\left(\dfrac{9}{x}+\dfrac{9}{y}+\dfrac{9}{z}+3\right)=\dfrac{3}{16}\left(9+3\right)=\dfrac{9}{4}\)
\(\Leftrightarrow\sum\dfrac{1}{\sqrt{x+1}}\le\dfrac{3}{2}\)kết hợp với (*), ta có
\(VT\le\dfrac{4}{9}.\dfrac{3}{2}+\dfrac{1}{3}=\dfrac{2}{3}+\dfrac{1}{3}=1\)
Dấu = xảy ra khi x=y=z=3 hay a=b=c=2 (a,b,c>1)
Sửa đề: C/m: \(\dfrac{1}{a+1}+\dfrac{1}{b+1}-\dfrac{1}{c+1}< 1\)
Ta có: \(a,b,c>1:\)
\(\Rightarrow\left\{{}\begin{matrix}a-1>0\\b-1>0\\c-1>0\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}\dfrac{1}{a-1}>0\\\dfrac{1}{b-1}>0\\\dfrac{1}{c-1}>0\end{matrix}\right.\)
\(\Rightarrow\dfrac{1}{a-1}+\dfrac{1}{b-1}+\dfrac{1}{c-1}>0\)
Quay lại bài toán, theo giả thiết ta có:
\(\dfrac{1}{a+1}+\dfrac{1}{b+1}+\dfrac{1}{c+1}-\dfrac{1}{a-1}-\dfrac{1}{b-1}-\dfrac{1}{c-1}=1\)
\(\Leftrightarrow\dfrac{1}{a+1}+\dfrac{1}{b+1}+\dfrac{1}{c-1}=1+\dfrac{1}{a-1}+\dfrac{1}{b-1}+\dfrac{1}{c-1}\)
\(\Leftrightarrow\dfrac{1}{a+1}+\dfrac{1}{b+1}-\dfrac{1}{c+1}< 1\)(đpcm)
