a.
\(ab+bc+ca\le\dfrac{\left(a+b+c\right)^2}{3}\)
\(\Leftrightarrow3ab+3bc+3ca\le a^2+b^2+c^2+2ab+2bc+2ca\)
\(\Leftrightarrow a^2+b^2+c^2-ab-bc-ca\ge0\)
\(\Leftrightarrow2a^2+2b^2+2c^2-2ab-2bc-2ca\ge0\)
\(\Leftrightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\ge0\) (luôn đúng)
Vậy BĐT đã cho đúng
b.
Ta có: \(\dfrac{a}{1+b^2}=a-\dfrac{ab^2}{1+b^2}\ge a-\dfrac{ab^2}{2b}=a-\dfrac{1}{2}ab\)
Tương tự: \(\dfrac{b}{1+c^2}\ge b-\dfrac{1}{2}bc\) ; \(\dfrac{c}{1+a^2}\ge c-\dfrac{1}{2}ac\)
Cộng vế:
\(VT\ge a+b+c-\dfrac{1}{2}\left(ab+bc+ca\right)\ge a+b+c-\dfrac{1}{2}.\dfrac{\left(a+b+c\right)^2}{3}=\dfrac{3}{2}\) (đpcm)
Dấu "=" xảy ra khi \(a=b=c=1\)
