Đề bài đúng: \(\dfrac{1}{3a+1}+\dfrac{1}{3b+1}+\dfrac{1}{3c+1}\le\dfrac{3}{4}\)
Và cần điều kiện a;b;c dương
Ta có:
\(\dfrac{1}{3a+1}=\dfrac{1}{a+a+a+1}\le\dfrac{1}{16}\left(\dfrac{1}{a}+\dfrac{1}{a}+\dfrac{1}{a}+1\right)=\dfrac{3}{16a}+\dfrac{1}{16}\)
Tương tự:
\(\dfrac{1}{3b+1}\le\dfrac{3}{16b}+\dfrac{1}{16}\)
\(\dfrac{1}{3c+1}\le\dfrac{3}{16c}+\dfrac{1}{16}\)
Cộng vế:
\(\dfrac{1}{3a+1}+\dfrac{1}{3b+1}+\dfrac{1}{3c+1}\le\dfrac{3}{16}\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)+\dfrac{3}{16}\le\dfrac{9}{16}+\dfrac{3}{16}=\dfrac{3}{4}\)
Dấu "=" xảy ra khi \(a=b=c=1\)
tại sao \(\dfrac{1}{a+a+a+1}\le\dfrac{1}{16}\left(\dfrac{1}{a}+\dfrac{1}{a}+\dfrac{1}{a}+1\right)\) vậy?
Sử dụng BĐT sau: \(\dfrac{1}{x+y}\le\dfrac{1}{4}\left(\dfrac{1}{x}+\dfrac{1}{y}\right)\)
Chứng minh: với mọi x;y dương ta có:
\(\left(x-y\right)^2\ge0\Leftrightarrow x^2+y^2\ge2xy\)
\(\Leftrightarrow x^2+y^2+2xy\ge4xy\Leftrightarrow\left(x+y\right)^2\ge4xy\)
\(\Leftrightarrow\dfrac{x+y}{4xy}\ge\dfrac{1}{x+y}\Leftrightarrow\dfrac{1}{x+y}\le\dfrac{1}{4}\left(\dfrac{1}{x}+\dfrac{1}{y}\right)\) (đpcm)
Áp dụng 2 lần liên tiếp như sau:
\(\dfrac{1}{\left(a+a\right)+\left(a+1\right)}\le\dfrac{1}{4}\left(\dfrac{1}{a+a}+\dfrac{1}{a+1}\right)\le\dfrac{1}{4}.\left[\dfrac{1}{4}\left(\dfrac{1}{a}+\dfrac{1}{a}\right)+\dfrac{1}{4}\left(\dfrac{1}{a}+\dfrac{1}{1}\right)\right]\)
\(\Rightarrow\dfrac{1}{3a+1}\le\dfrac{1}{16}\left(\dfrac{1}{a}+\dfrac{1}{a}+\dfrac{1}{a}+1\right)\)
