\(a,b,c,d>0;abcd=1\)
Ta có: \(\sum\dfrac{1}{a^3+b+c+d}=\sum\dfrac{\dfrac{1}{a}+b+c+d}{\left(a^3+b+c+d\right)\left(\dfrac{1}{a}+b+c+d\right)}\le^{CBS}\sum\dfrac{\dfrac{1}{a}+b+c+d}{\left(a+b+c+d\right)^2}\)
\(=\sum\dfrac{bcd}{\left(a+b+c+d\right)^2}+\dfrac{3\left(a+b+c+d\right)}{\left(a+b+c+d\right)^2}=\sum\dfrac{bcd}{\left(a+b+c+d\right)^2}+\dfrac{3}{a+b+c+d}\)
Lại có: \(a+b+c+d\ge4\sqrt[4]{abcd}=4\Rightarrow\dfrac{3\left(a+b+c+d\right)}{16}\ge\dfrac{3}{a+b+c+d}\)
Do đó: \(\sum\dfrac{1}{a^3+b+c+d}\le\dfrac{3\left(a+b+c+d\right)}{16}+\sum\dfrac{bcd}{\left(a+b+c+d\right)^2}\)
Ta có: \(bcd+cda+dab+abc=cd\left(a+b\right)+ab\left(c+d\right)\le\dfrac{\left(c+d\right)^2}{4}\left(a+b\right)+\dfrac{\left(a+b\right)^2}{4}\left(c+d\right)\)
\(=\dfrac{\left(a+b\right)\left(c+d\right)}{4}\left(a+b+c+d\right)\le\dfrac{\dfrac{\left(a+b+c+d\right)^2}{4}}{4}\left(a+b+c+d\right)=\dfrac{\left(a+b+c+d\right)^3}{16}\)
Do đó: \(\sum\dfrac{1}{a^3+b+c+d}\le\dfrac{3\left(a+b+c+d\right)}{16}+\dfrac{\left(a+b+c+d\right)^3}{16\left(a+b+c+d\right)^2}=\dfrac{a+b+c+d}{4}\left(đpcm\right)\)
Đẳng thức xảy ra khi \(a=b=c=d=1\)
