Áp dụng BĐT cauchy ngược dấu ta có:
\(\dfrac{1}{a^2+1}=1-\dfrac{a^2}{a^2+1}\ge1-\dfrac{a^2}{2a}=1-\dfrac{a}{2}\)
Chứng minh tương tự ta có:
\(\dfrac{1}{b^2+1}\ge1-\dfrac{b}{2};\dfrac{1}{c^2+1}\ge1-\dfrac{c}{2}\)
Từ đó ta có: \(\dfrac{1}{a^2+1}+\dfrac{1}{b^2+1}+\dfrac{1}{c^2+1}\ge1-\dfrac{a}{2}+1-\dfrac{b}{2}+1-\dfrac{c}{2}=\)\(=3-\dfrac{a+b+c}{2}=3-\dfrac{3}{2}=\dfrac{3}{2}\left(đpcm\right)\)
Áp dụng BĐT Cauchy dạng Engel , ta có :
\(\dfrac{1}{a^2+1}+\dfrac{1}{b^2+1}+\dfrac{1}{c^2+1}\) ≥ \(\dfrac{\left(1+1+1\right)^2}{a^2+b^2+c^2+3}=\dfrac{9}{a^2+b^2+c^2+3}\left(1\right)\)
Ta có BĐT : \(a^2+b^2+c^2\text{≥}ab+bc+ac\)
⇔ \(3\left(a^2+b^2+c^2\right)\text{≥}\left(a+b+c\right)^2\)
⇔ \(a^2+b^2+c^2\text{≥}\dfrac{9}{3}=3\left(2\right)\)
Từ ( 1 ; 2 ) ⇒ đpcm .
"=" ⇔ \(a=b=c=\dfrac{1}{3}\)
Ta cần chứng minh: \(a^2+b^2+c^2+3\ge2\left(a+b+c\right)\)
\(\Rightarrow\left(a^2-2a+1\right)+\left(b^2-2b+1\right)+\left(c^2-2c+1\right)\ge0\)
\(\Rightarrow\left(a-1\right)^2+\left(b-1\right)^2+\left(c-1\right)^2\ge0\) ( luôn đúng)
Suy ra: ta có BĐT: \(a^2+b^2+c^2+3\ge2\left(a+b+c\right)\)
\(a^2+b^2+c^2+3\ge6\)
\(\Rightarrow a^2+b^2+c^2\ge3\)
Áp dụng BĐT AM - GM cho 2 số dương, ta có:
\(\dfrac{1}{a^2+1}+\dfrac{a^2+1}{4}\ge2\sqrt{\dfrac{1}{a^2+1}.\dfrac{a^2+1}{4}}=1\)
\(\dfrac{1}{b^2+1}+\dfrac{b^2+1}{4}\ge2\sqrt{\dfrac{1}{b^2+1}.\dfrac{b^2+1}{4}}=1\)
\(\dfrac{1}{c^2+1}+\dfrac{c^2+1}{4}\ge2\sqrt{\dfrac{1}{c^2+1}.\dfrac{c^2+1}{4}}=1\)
Cộng vế theo vế BĐT, ta được:
\(\dfrac{1}{a^2+1}+\dfrac{1}{b^2+1}+\dfrac{1}{c^2+1}+\dfrac{a^2+b^2+c^2+3}{4}\ge1+1+1\)
\(\Rightarrow\dfrac{1}{a^2+1}+\dfrac{1}{b^2+1}+\dfrac{1}{c^2+1}\ge3-\dfrac{a^2+b^2+c^2+3}{4}\ge3-\dfrac{3+3}{4}=\dfrac{3}{2}\)
\(\Rightarrowđpcm\)
Dấu "=" xảy ra khi: \(a=b=c=1\)
