Câu 1:$P\le\sqrt[3]{\frac{(a+b+c+d)^4}{256}}+\sqrt[3]{\frac{(4-a-b-c-d)^4}{256}}=\sqrt[3]{\frac{t^4}{256}}+\sqrt[3]{\frac{(4-t)^4}{256}}$$Do: 0\le a,b,c,d\le 1\Rightarrow 0\le t\le 4$
$P\le\sqrt[3]{\frac{t^4}{256}}+\sqrt[3]{\frac{(t-4)^4}{256}}\le\sqrt[3]{\frac{4^4}{256}}+\sqrt[3]{\frac{(4-4)^4}{256}}=1$
$P\le \sqrt[3]{\frac{(-t)^4}{256}}+\sqrt[3]{\frac{(4-t)^4}{256}}\le 1$
Dấu bằng có khi a=b=c=0 hoặc =1