a/ Ta cần chứng minh:
[tex]a^{4}\geq 4a-3[/tex]
[tex]\Leftrightarrow (a^4-2a^2+1)+(2a^2-4a+2)\geq 0[/tex]
[tex]\Leftrightarrow (a^2-1)^2+2(a-1)^2\geq 0[/tex] đúng
Vậy ta có ĐPCM
b/ [tex]a^8+b^8=(\frac{a^8}{8}+\frac{a^8}{8}+\frac{a^8}{8}+\frac{a^8}{8}+\frac{a^8}{8}+\frac{a^8}{8}+\frac{a^8}{8}+\frac{b^8}{8})+(\frac{a^8}{8}+\frac{b^8}{8}+\frac{b^8}{8}+\frac{b^8}{8}+\frac{b^8}{8}+\frac{b^8}{8}+\frac{b^8}{8}+\frac{b^8}{8})[/tex]
[tex]\geq 8\sqrt[8]{\frac{a^{56}b^{8}}{8^{8}}}+8\sqrt[8]{\frac{b^{56}a^{8}}{8^{8}}}[/tex]
[tex]=a^7b+b^7a[/tex]