Giải thích các bước giải:
Áp dụng BĐT \({a^2} + {b^2} + {c^2} \ge ab + bc + ca\) ta có:
\(\begin{array}{l}
\frac{{{a^8}}}{{{b^4}}} + \frac{{{b^8}}}{{{c^4}}} + \frac{{{c^8}}}{{{a^4}}} = {\left( {\frac{{{a^4}}}{{{b^2}}}} \right)^2} + {\left( {\frac{{{b^4}}}{{{c^2}}}} \right)^2} + {\left( {\frac{{{c^4}}}{{{a^2}}}} \right)^2}\\
\ge \frac{{{a^4}}}{{{b^2}}}.\frac{{{b^4}}}{{{c^2}}} + \frac{{{b^4}}}{{{c^2}}}.\frac{{{c^4}}}{{{a^2}}} + \frac{{{c^4}}}{{{a^2}}}.\frac{{{a^4}}}{{{b^2}}}\\
= \frac{{{a^4}{b^2}}}{{{c^2}}} + \frac{{{b^4}{c^2}}}{{{a^2}}} + \frac{{{c^4}{a^2}}}{{{b^2}}}\\
= {\left( {\frac{{{a^2}b}}{c}} \right)^2} + {\left( {\frac{{{b^2}c}}{a}} \right)^2} + {\left( {\frac{{{c^2}a}}{b}} \right)^2}\\
\ge \frac{{{a^2}b}}{c}.\frac{{{b^2}c}}{a} + \frac{{{b^2}c}}{a}.\frac{{{c^2}a}}{b} + \frac{{{c^2}a}}{b}.\frac{{{a^2}b}}{c}\\
= a{b^3} + b{c^3} + c{a^3}
\end{array}\)
Dấu '=' xảy ra khi và chỉ khi a=b=c
