Áp dụng BĐT Cosi với a;b;c > 0:
\(a^2+b^2\ge2ab\)
\(\Rightarrow\dfrac{a}{a^2+b^2}\le\dfrac{a}{2ab}=\dfrac{1}{2b}\)
Chứng minh tương tự: \(\dfrac{b}{b^2+c^2}\le\dfrac{1}{2c}\\ \dfrac{c}{c^2+a^2}\le\dfrac{1}{2a}\)
\(\Rightarrow\dfrac{a}{a^2+b^2}+\dfrac{b}{b^2+c^2}+\dfrac{c}{c^2+a^2}\le\dfrac{1}{2a}+\dfrac{1}{2b}+\dfrac{1}{2c}=\dfrac{1}{2}\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\)
Dấu "=" xảy ra \(\Leftrightarrow a=b=c>0\)
Ta có :
\(\dfrac{a}{a^2+b^2}\le\dfrac{a}{2ab}=\dfrac{1}{2b}\)
vì theo bđt cauchy : `a^2+b^2>=2ab`
Dấu "=" sảy ra khi `a=b`
Tương tự :
\(\dfrac{b}{b^2+c^2}\le\dfrac{b}{2bc}=\dfrac{1}{2c}\)
Dấu "=" sảy ra khi `b=c`
\(\dfrac{c}{c^2+a^2}\le\dfrac{c}{2ca}=\dfrac{2}{2a}\)
Dấu "=" sảy ra khi `c=a`
Cộng `3` đẳng thức trên vế theo vế ta có :
\(\dfrac{a}{a^2+b^2}+\dfrac{b}{b^2+c^2}+\dfrac{c}{c^2+a^2}\le \dfrac{1}{2}\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)(\text{đpcm})\)