安振平老师的5909号不等式问题的证明

题目: 已知$a,b,c,d,e\in R$,求证:
$(a^2+4)(b^2+4)(c^2+4)(d^2+4)(e^2+4)\geq 125(a+b+c+d+e)^2.$
证明:先证明几个引理.
引理1.当$x_{1}\geq 1,x_{2}\geq 1,x_{3}\geq 1,x_{4}\geq 1$或者$x_{1}\leq 1,x_{2}\leq 1,x_{3}\leq 1,x_{4}\leq 1$时,有
$(x_{1}+4)(x_{2}+4)(x_{3}+4)(x_{4}+4)\geq 125(x_{1}+x_{2}+x_{3}+x_{4}+1).$
证明:(I)当$x_{1}\geq 1,x_{2}\geq 1,x_{3}\geq 1,x_{4}\geq 1$时,令
$x_{1}=1+y_{1},x_{2}=1+y_{2},x_{3}=1+y_{3},x_{4}=1+y_{4}(y_{i}\geq 0,i=1,2,3,4)$,则
$(x_{1}+4)(x_{2}+4)(x_{3}+4)(x_{4}+4)-125(x_{1}+x_{2}+x_{3}+x_{4}+1)$
$=(y_{1}+5)(y_{2}+5)(y_{3}+5)(y_{4}+5)-125(y_{1}+y_{2}+y_{3}+y_{4}+5)$
$=y_{1}y_{2}y_{3}y_{4}+5(y_{1}y_{2}y_{3}+y_{2}y_{3}y_{4}+y_{3}y_{4}y_{1}+y_{4}y_{1}y_{2})+25(y_{1}y_{2}+y_{1}y_{3}+y_{1}y_{4}+y_{2}y_{3}+y_{2}y_{4}+y_{3}y_{4})\geq 0.$
故此时原不等式成立.
(II)当$x_{1}\leq 1,x_{2}\leq 1,x_{3}\leq 1,x_{4}\leq 1$时,令
$x_{1}=1-y_{1},x_{2}=1-y_{2},x_{3}=1-y_{3},x_{4}=1-y_{4}(y_{i}\leq 0,i=1,2,3,4)$,则
$(x_{1}+4)(x_{2}+4)(x_{3}+4)(x_{4}+4)-125(x_{1}+x_{2}+x_{3}+x_{4}+1)$
$=(5-y_{1})(5-y_{2})(5-y_{3})(5-y_{4})-125[5-(y_{1}+y_{2}+y_{3}+y_{4})]$
$=y_{1}y_{2}y_{3}y_{4}-5(y_{1}y_{2}y_{3}+y_{2}y_{3}y_{4}+y_{3}y_{4}y_{1}+y_{4}y_{1}y_{2})+25(y_{1}y_{2}+y_{1}y_{3}+y_{1}y_{4}+y_{2}y_{3}+y_{2}y_{4}+y_{3}y_{4})\geq 0.$
故此时原不等式也成立.
由(I)(II)知引理1成立.

引理2.当$x_{1}\geq 1,x_{2}\geq 1,x_{3}\geq 1$或者$0\leq x_{1}\leq 1,0\leq x_{2}\leq 1,0\leq x_{3}\leq 1$时,有
$(x_{1}+4)(x_{2}+4)(x_{3}+4)\geq 25(x_{1}+x_{2}+x_{3}+2).$
证明:(I)当$x_{1}\geq 1,x_{2}\geq 1,x_{3}\geq 1$时,令
$x_{1}=1+y_{1},x_{2}=1+y_{2},x_{3}=1+y_{3}(y_{i}\geq 0,i=1,2,3)$,则
$(x_{1}+4)(x_{2}+4)(x_{3}+4)-25(x_{1}+x_{2}+x_{3}+2)$
$=(y_{1}+5)(y_{2}+5)(y_{3}+5)-25(y_{1}+y_{2}+y_{3}+5)$
$=y_{1}y_{2}y_{3}+5(y_{1}y_{2}+y_{2}y_{3}+y_{3}y_{1})\geq 0.$
故此时原不等式成立.
(II)当$x_{1}\leq 1,x_{2}\leq 1,x_{3}\leq 1$时,令
$x_{1}=1-y_{1},x_{2}=1-y_{2},x_{3}=1-y_{3}(0\leq y_{i}\leq 1,i=1,2,3)$,则
$(x_{1}+4)(x_{2}+4)(x_{3}+4)-25(x_{1}+x_{2}+x_{3}+2)$
$=(5-y_{1})(5-y_{2})(5-y_{3})-25[5-(y_{1}+y_{2}+y_{3})]$
$=5(y_{1}y_{2}+y_{2}y_{3}+y_{3}y_{1})-y_{1}y_{2}y_{3}\geq 0.$
故此时原不等式也成立.
由(I)(II)知引理2成立.

引理3.当$x_{1}\geq 1,x_{2}\geq 1$或者$x_{1}\leq 1,x_{2}\leq 1$时,有
$(x_{1}+4)(x_{2}+4)\geq 5(x_{1}+x_{2}+3).$
证明:由已知$(x_{1}-1)(x_{2}-1)\geq 0$,于是$x_{1}x_{2}\geq x_{1}+x_{2}-1$,从而
$(x_{1}+4)(x_{2}+4)=x_{1}x_{2}+4(x_{1}+x_{2})+16\geq 5(x_{1}+x_{2}+3).$
即引理3成立.

下面给出原不等式的证明.
证明:(I)若$a^2,b^2,c^2,d^2,e^2$中至少有4个不小于1或不大于1,不妨设$a^2\geq 1,b^2\geq 1,c^2\geq 1,d^2\geq 1$或$a^2\leq 1,b^2\leq 1,c^2\leq 1,d^2\leq 1$,则由引理1及柯西不等式可得
$(a^2+4)(b^2+4)(c^2+4)(d^2+4)(e^2+4)\geq 125(a^2+b^2+c^2+d^2+1)(1+1+1+1+e^2)$
$\geq 125(a+b+c+d)^2.$
故此时原不等式成立.
(II)若$a^2,b^2,c^2,d^2,e^2$中有3个不小于1,2个不大于1,不妨设$a^2\geq 1,b^2\geq 1,c^2\geq 1,d^2\leq 1,e^2\leq 1$;或$a^2,b^2,c^2,d^2,e^2$中有3个不大于1,2个不小于1,不妨设$a^2\leq 1,b^2\leq 1,c^2\leq 1,d^2\geq 1,e^2\geq 1$,则由引理2,引理3及柯西不等式可得
$(a^2+4)(b^2+4)(c^2+4)(d^2+4)(e^2+4)\geq 125(a^2+b^2+c^2+1+1)(1+1+1+d^2+e^2)$
$\geq 125(a+b+c+d)^2.$
故此时原不等式成立.
由(I)(II)知原不等式成立.