Đề sai, ngược dấu rồi.
Ta chứng minh BĐT phụ sau: \(\dfrac{x}{x+1}\le\dfrac{9}{16}x+\dfrac{1}{16}\left(\forall x\in0;1\right)\)
Thật vậy: \(\dfrac{x}{x+1}\le\dfrac{9}{16}x+\dfrac{1}{16}\)
\(\Leftrightarrow0\le\dfrac{9x+1}{16}-\dfrac{x}{x+1}\)
\(\Leftrightarrow0\le\dfrac{\left(9x+1\right)\left(x+1\right)-16x}{16\left(x+1\right)}\)
\(\Leftrightarrow0\le9x^2-6x+1=\left(3x-1\right)^2\)(Luôn đúng \(\forall x\in0;1\))
Áp dụng vào bài, ta được:
\(\dfrac{a}{a+1}\le\dfrac{9}{16}a+\dfrac{1}{16}\)
\(\dfrac{b}{b+1}\le\dfrac{9}{16}b+\dfrac{1}{16}\)
\(\dfrac{c}{c+1}\le\dfrac{9}{16}c+\dfrac{1}{16}\)
Cộng vế theo vế ta được đpcm
\(\dfrac{a}{a+1}+\dfrac{b}{b+1}+\dfrac{c}{c+1}\\ =\dfrac{a+1-1}{a+1}+\dfrac{b+1-1}{b+1}+\dfrac{c+1-1}{c+1}\\ =\dfrac{a+1}{a+1}-\dfrac{1}{a+1}+\dfrac{b+1}{b+1}-\dfrac{1}{b+1}+\dfrac{c+1}{c+1}-\dfrac{1}{c+1}\\ =\left(\dfrac{a+1}{a+1}+\dfrac{c+1}{c+1}+\dfrac{b+1}{b+1}\right)-\left(\dfrac{1}{a+1}+\dfrac{1}{b+1}+\dfrac{1}{c+1}\right)\\ =3-\left(\dfrac{1}{a+1}+\dfrac{1}{b+1}+\dfrac{1}{c+1}\right)\)
Áp dụng BDT : Cô-si
\(\Rightarrow3-\left(\dfrac{1}{a+1}+\dfrac{1}{b+1}+\dfrac{1}{c+1}\right)\le3-\dfrac{9}{a+1+b+1+c+1}\\ =3-\dfrac{9}{a+b+c+3}=3-\dfrac{9}{1+3}=\dfrac{3}{4}\left(đpcm\right)\)
Vậy \(\dfrac{a}{a+1}+\dfrac{b}{b+1}+\dfrac{c}{c+1}\le\dfrac{3}{4}\) đẳng thức xảy ra khi: \(a=b=c=\dfrac{1}{3}\)
