Giải:

Áp dụng BĐT AM - GM ta có:

\(\dfrac{a}{1+b^2c}=a-\dfrac{ab^2c}{1+b^2c}\ge a-\dfrac{ab^2c}{2b\sqrt{c}}\) \(=a-\dfrac{ab\sqrt{c}}{2}\)

\(\ge a-\dfrac{b\sqrt{a.ac}}{2}\ge a-\dfrac{b\left(a+ac\right)}{4}\) \(\ge a-\dfrac{1}{4}\left(ab+abc\right)\)

\(\Rightarrow\dfrac{a}{1+b^2c}\ge a-\dfrac{1}{4}\left(ab+abc\right).\) Tượng tự ta cũng có:

\(\dfrac{b}{1+c^2d}\ge b-\dfrac{1}{4}\left(bc+bcd\right);\dfrac{c}{1+d^2a}\ge c-\dfrac{1}{4}\left(cd+cda\right);\dfrac{d}{1+a^2b}\ge d-\dfrac{1}{4}\left(da+dab\right)\)

Cộng theo vế 4 BĐT trên ta được:

\(\dfrac{a}{1+b^2c}+\dfrac{b}{1+c^2d}+\dfrac{c}{1+d^2a}+\dfrac{d}{1+a^2b}\)

\(\ge a+b+c+d-\dfrac{1}{4}\)\(\left(ab+bc+cd+da+abc+bcd+cda+dab\right)\)

Lại áp dụng BĐT AM - GM ta có:

\(ab+bc+cd+da\) \(\le\dfrac{1}{4}\left(a+b+c+d\right)^2=4\)

\(abc+bcd+cda+dab\) \(\le\dfrac{1}{16}\left(a+b+c+d\right)^3=4\)

Do đó:

\(\dfrac{a}{1+b^2c}+\dfrac{b}{1+c^2d}+\dfrac{c}{1+d^2a}+\dfrac{d}{1+a^2b}\)

\(\ge a+b+c+d-2=2\)

Đẳng thức xảy ra \(\Leftrightarrow a=b=c=d=1\)