( extbf{天下事有难易乎?为之,则难者亦易矣 不为,则易者亦难矣------《为学》})
(中国第59届国际数学奥林匹克国家集训队2018.3.20日测试题)
证明:存在常数(C>0)使得对于任意的正整数(m),以及任意(m)个正整数(a_1,a_2,cdots,a_m),都有
(H(a_1)+H(a_2)+cdots+H(a_m)le Cleft(sumlimits_{k=1}^m{ka_k}
ight)^{frac{1}{2}}),其中(H(n)=sumlimits_{k=1}^{n}{dfrac{1}{k}})
证明:存在.(C=2)满足要求.记({a_1,a_2,cdots,a_m}={b_1,b_2cdots,b_m})其中(b_1ge b_2ge cdots ge b_m)
[egin{align*}
LHS&=dfrac{1}{1}+dfrac{1}{2}+cdots+dfrac{1}{b_1}+dfrac{1}{1}+dfrac{1}{2}+cdots+dfrac{1}{b_2}
+dfrac{1}{1}+dfrac{1}{2}+cdots+dfrac{1}{b_m}\
& le mleft(dfrac{1}{1}+dfrac{1}{2}+cdots+dfrac{1}{b_m}
ight) \
& le msqrt{(1^2+1^2+cdots 1^2)(dfrac{1}{1^2}+dfrac{1}{2^2}+cdots+dfrac{1}{b_m^2})}quad( extbf{此处用到柯西不等式})\
&le msqrt{b_mcdot(1+dfrac{1}{1}-dfrac{1}{2}+dfrac{1}{2}-dfrac{1}{3}cdots+dfrac{1}{b_m-1}-dfrac{1}{b_m})} quad ( extbf{此处用到}dfrac{1}{k^2}ledfrac{1}{k-1}-dfrac{1}{k})\
&=msqrt{2b_m-1}\
RHS&=Csqrt{1a_1+2a_2+cdots+ma_m}\
&ge Csqrt{(1+2+cdots m)b_m}\
&=Csqrt{dfrac{m(m+1)}{2}b_m}
end{align*}]
取(C=2)时 $ 2sqrt{dfrac{m(m+1)}{2}b_m}ge msqrt{2b_m-1}$显然成立.