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1)$4\left( {\frac{{{a^5}}}{{{b^3}}} + \frac{{{b^5}}}{{{c^3}}} + \frac{{{c^5}}}{{{a^3}}}} \right) + {a^2} + {b^2} + {c^2} = \sum\limits_{cyc} {\left( {4\frac{{{a^5}}}{{{b^3}}} + {b^2}} \right)} $
2)Nhận xét: Vì cả 2 vế đề bậc 0 nên ta phải cộng thêm đa thức bậc 0
Có: $2\left( {\frac{{{a^3}}}{{{b^3}}} + \frac{{{b^3}}}{{{c^3}}} + \frac{{{c^3}}}{{{a^3}}}} \right) + 3 = \sum\limits_{cyc} {\left( {\frac{{{a^3}}}{{{b^3}}} + \frac{{{a^3}}}{{{b^3}}} + 1} \right)}  \ge \sum\limits_{cyc} {3\sqrt[3]{{\frac{{{{({a^2})}^3}}}{{{{({b^2})}^3}}}}}}  = \sum\limits_{cyc} {3\frac{{{a^2}}}{{{b^2}}} = 3\left( {\frac{{{a^2}}}{{{b^2}}} + \frac{{{b^2}}}{{{c^2}}} + \frac{{{c^2}}}{{{a^2}}}} \right)}  = $
$ = 2\left( {\frac{{{a^2}}}{{{b^2}}} + \frac{{{b^2}}}{{{c^2}}} + \frac{{{c^2}}}{{{a^2}}}} \right) + \left( {\frac{{{a^2}}}{{{b^2}}} + \frac{{{b^2}}}{{{c^2}}} + \frac{{{c^2}}}{{{a^2}}}} \right) \ge 2\left( {\frac{{{a^2}}}{{{b^2}}} + \frac{{{b^2}}}{{{c^2}}} + \frac{{{c^2}}}{{{a^2}}}} \right) + 3\sqrt[3]{{\frac{{{a^2}}}{{{b^2}}}.\frac{{{b^2}}}{{{c^2}}}.\frac{{{c^2}}}{{{a^2}}}}} = $
$ = 2\left( {\frac{{{a^2}}}{{{b^2}}} + \frac{{{b^2}}}{{{c^2}}} + \frac{{{c^2}}}{{{a^2}}}} \right) + 3 \Rightarrow DPCM$

Đẳng thức xảy ra khi a = b = c

3)Vì cả 2 vế đồng bậc (–3) nên ta cộng thêm biểu thức có bậc – 3 và sử dụng BĐT AM-GM ta được:
$\sum\limits_{cyc} {\left( {\frac{{3{a^2}}}{{{b^5}}} + \frac{2}{{{a^3}}}} \right) \ge \sum\limits_{cyc} {5\sqrt[5]{{{{\left( {\frac{{{a^2}}}{{{b^5}}}} \right)}^3}.{{\left( {\frac{1}{{{a^3}}}} \right)}^2}}} = 5\left( {\frac{1}{{{a^3}}} + \frac{1}{{{b^3}}} + \frac{1}{{{c^3}}}} \right)} } $
Có: $3\left( {\frac{1}{{{a^3}}} + \frac{1}{{{b^3}}} + \frac{1}{{{c^3}}}} \right) + 2\left( {\frac{1}{{{a^3}}} + \frac{1}{{{b^3}}} + \frac{1}{{{c^3}}}} \right) = \sum\limits_{cyc} {\left( {3.\frac{{{a^2}}}{{{b^5}}} + 2.\frac{1}{{{a^3}}}} \right)}  \Rightarrow DPCM$

Đẳng thức xảy ra khi a = b = c