theo bđt cosi ta có $\frac{a^4}{b}+\frac{b^4}{c}+\frac{c^4}{a}\\\ge 3\sqrt[3]{\frac{a^4b^4c^4}{bca}}\\=3\sqrt[3]{a^3b^3c^3}\\=3abc\\=>dpcm$
theo bđt cosi ta có $\frac{a^4}{b}+\frac{b^4}{c}+\frac{c^4}{a}\\\ge 3\sqrt[3]{\frac{a^4b^4c^2}{bca}}\\=3\sqrt[3]{a^3b^3c^3}\\=3abc\\=>dpcm$
theo bđt cosi ta có $\frac{a^4}{b}+\frac{b^4}{c}+\frac{c^4}{a}\\\ge 3\sqrt[3]{\frac{a^4b^4c^
4}{bca}}\\=3\sqrt[3]{a^3b^3c^3}\\=3abc\\=>dpcm$