Áp dụng BĐT Min-côp-xki, ta có \(\sqrt{1+a^4}+\sqrt{1+b^4}\ge\sqrt{\left(1+1\right)^2+\left(a^2+b^2\right)^2}=\sqrt{4+\left(a^2+b^2\right)^2}\)
Mà \(\left(a+1\right)\left(b+1\right)=\dfrac{9}{4}\Rightarrow a+b+ab=\dfrac{5}{4}\)
Vì \(ab\le\dfrac{\left(a+b\right)^2}{4}\Rightarrow a+b+\dfrac{\left(a+b\right)^2}{4}\ge\dfrac{5}{4}\)
\(\Rightarrow4m+m^2-5\ge0\Leftrightarrow\left(m-1\right)\left(m+5\right)\ge0\Rightarrow m\ge1\)(với m=a+b)
\(\Rightarrow a^2+b^2\ge\dfrac{\left(a+b\right)^2}{2}=\dfrac{1}{2}\Rightarrow\left(a^2+b^2\right)^2\ge\dfrac{1}{4}\Rightarrow\sqrt{4+\left(a^2+b^2\right)^2}\ge\dfrac{\sqrt{17}}{2}\)
=> \(\sqrt{1+a^4}+\sqrt{1+b^4}\ge\dfrac{\sqrt{17}}{2}\)