设 a、b、ca、b、ca、b、c 是某三角形得三条边的边长,且 a+b+c=1a+b+c=1a+b+c=1 .若整数 n≥2,n\geq2,n≥2, 证明:
an+bnn+bn+cnn+cn+ann<1+2n2.\sqrt[n]{a^n+b^n}+\sqrt[n]{b^n+c^n}+\sqrt[n]{c^n+a^n}<1+\frac{\sqrt[n]2}2.nan+bn+nbn+cn+ncn+an<1+2n2.
\qquad证明\qquad 设 a≥b≥c>0a\geq b\geq c>0a≥b≥c>0 .
\qquad因为 a+b+c=1a+b+c=1a+b+c=1,所以,
\qquad (b+c2)n=bn+c2Cn1bn−1+c24Cn2bn−2+⋯+cn2nCnn(b+\frac c2)^n=b^n+\frac c2C_n^1b^{n-1}+\frac{c^2}4C_n^2b^{n-2}+\cdots+\frac{c^n}{2^n}C_n^n(b+2c)n=bn+2cCn1bn−1+4c2Cn2bn−2+⋯+2ncnCnn
≥bn+[12Cn1+(12)2Cn2+⋯+(12)nCnn]cn\qquad\qquad\qquad\geq b^n+[\frac12C_n^1+(\frac12)^2C_n^2+\cdots+(\frac12)^nC_n^n]c^n≥bn+[21Cn1+(21)2Cn2+⋯+(21)nCnn]cn
=bn+[(12+1)n−1]cn\qquad\qquad\qquad=b^n+[(\dfrac12+1)^n-1]c^n=bn+[(21+1)n−1]cn.
\qquad因为 n≥2n\geq 2n≥2,所以,(12+1)n−1>1.(\dfrac12+1)^n-1>1.(21+1)n−1>1.从而, (b+c2)n>bn+cn(b+\frac c2)^n>b^n+c^n(b+2c)n>bn+cn.
\qquad故 bn+cnn<b+c2.⋯①\sqrt[n]{b^n+c^n}<b+\dfrac c2.\cdots①nbn+cn<b+2c.⋯①
\qquad同理, an+cnn<a+c2⋯②\sqrt[n]{a^n+c^n}<a+\dfrac c2\cdots②nan+cn<a+2c⋯②.
\qquad又因为 a<12,b<12a<\dfrac12,b<\dfrac12a<21,b<21 , 所以,
\qquad an+bnn<(12)n+(12)nn=2n2⋯③\sqrt[n]{a^n+b^n}<\sqrt[n]{(\frac12)^n+(\frac12)^n}=\frac{\sqrt[n]2}2\cdots③nan+bn<n(21)n+(21)n=2n2⋯③
\qquad①+②+③得 an+bnn+bn+cnn+cn+ann<1+2n2.\sqrt[n]{a^n+b^n}+\sqrt[n]{b^n+c^n}+\sqrt[n]{c^n+a^n}<1+\frac{\sqrt[n]2}2.nan+bn+nbn+cn+ncn+an<1+2n2.