[TEX]n\ge 1, \ \ \ \ a,b,c>0[/TEX] ta cần chứng minh [TEX]\sum_{cyclic}\frac{a^n}{b^n+c^n}\ge \sum_{cyclic}\frac{a}{b+c}[/TEX]
Giả sử rằng [TEX]a\ge b\ge c[/TEX]
[TEX]\frac{a^n}{b^n+c^n} -\frac{a}{b+c}:=\frac{ab(a^{n-1}-b^{n-1})+ac(a^{n-1}-c^{n-1})}{(b^n+c^n)(b+c)}=\frac{TS_1+TS_2}{MS_{bc}}[/TEX]
[TEX]\frac{b^n}{c^n+a^n} -\frac{b}{c+a}:=\frac{bc(b^{n-1}-c^{n-1})+ba(b^{n-1}-a^{n-1})}{(c^n+a^n)(c+a)}=\frac{TS_3+TS_4}{MS_{ca}}[/TEX]
[TEX]\frac{c^n}{a^n+b^n} -\frac{c}{a+b}:=\frac{ca(c^{n-1}-a^{n-1})+cb(c^{n-1}-b^{n-1})}{(a^n+b^n)(a+b)}=\frac{TS_5+TS_6}{MS_{ab}}[/TEX]
Và hãy chú ý như sau :
[TEX]\frac{TS_1}{MS_{bc}}+\frac{TS_4}{MS_{ca}}=ab(a^{n-1}-b^{n-1})\left[\frac{1}{(b^n+c^n)(b+c)}-\frac{1}{(c^n+a^n)(c+a)}\right]\ge 0[/TEX]
[TEX]\frac{TS_2}{MS_{bc}}+\frac{TS_5}{MS_{ab}}=ac(a^{n-1}-c^{n-1})\left[\frac{1}{(b^n+c^n)(b+c)}-\frac{1}{(a^n+b^n)(a+b)}\right]\ge 0[/TEX]
[TEX]\frac{TS_3}{MS_{ca}}+\frac{TS_6}{MS_{ab}}=bc(b^{n-1}-c^{n-1})\left[\frac{1}{(c^n+a^n)(c+a)}-\frac{1}{(a^n+b^n)(a+b)}\right]\ge 0[/TEX]
Vậy bài toán chứng minh xong