x4+16\(\ge\)2x3+8x
\(\Leftrightarrow\)x4-2x3-8x+16\(\ge\)0
\(\Leftrightarrow\)(x-2)(x3-8)\(\ge\)0
\(\Leftrightarrow\)(x-2)2(x2+x+4)\(\ge\)0 (*)
Ta có: (x-2)2\(\ge\)0
Và x2+x+4=(x+\(\dfrac{1}{2}\))2+\(\dfrac{15}{4}\)>0
Nên (*) luôn đúng
Vậy x4+16\(\ge\)2x3+8x