Question

设x1>=x2>=x3>=x4>=2,x2+x3+x4>=x1,求证：（x1+x2+x3+x4）^2

Answer 1

令x1=k(x2+x3+x4)
1/3(x2+x3+x4)<=x1<=x2+x3+x4
则1/3<=k<=1
原不等式变形为
(1+k)^2(x2+x3+x4)^2<=4k(x2+x3+x4)x2x3x4
[(1+k)^2/4k](x2+x3+x4)<=x2x3x4①
[(1+k)^2/4k](x2+x3+x4)<=[(1+k)^2/4k](x2+x2+x2)=[(1+k)^2/4k]*3x2
x2x3x4>=2*2*x2=4x2
证①成立 只需证明
[(1+k)^2/4k]*3x2<=4x2
1/4(k+1/k+2)*3<=4
因为f(x)=x+1/x在[1/3,1]上是减函数
所以
1/4(k+1/k+2)*3<=(1/4)*(1/3+3+2)*3=4
因此(x1+x2+x3+x4)^2<=4x1x2x3x4