Question

Answer 1

\(a,b,c,d>0;a+b+c+d=8\)

\(\sum\dfrac{a^3}{a^2+b+c}=\sum\left[a-\dfrac{a\left(b+c\right)}{a^2+b+c}\right]\ge\sum\left[a-\dfrac{a\left(b+c\right)}{2a\sqrt{b+c}}\right]=\sum\left(a-\dfrac{\sqrt{b+c}}{2}\right)\)

\(=\sum\left(a-\dfrac{\sqrt{\left(b+c\right).4}}{4}\right)\ge\sum\left(a-\dfrac{b+c+4}{8}\right)=\left(a+b+c+d\right)-\dfrac{2\left(a+b+c+d\right)+4.4}{8}=8-\dfrac{2.8+16}{8}=4\left(dpcm\right)\)

Dấu "=" xảy ra khi \(a=b=c=d=2\)

Note: 2 bài anh đã làm cho em, anh dùng phương pháp Cauchy ngược dấu để làm nhé.