xyz=1,求证:1/((x+1)^2+y^2+1)+1/((y+1)^2+z^2+1))+1/((z+1)^2+x^2+1)

Question

xyz=1,求证:1/((x+1)^2+y^2+1)+1/((y+1)^2+z^2+1))+1/((z+1)^2+x^2+1)<<1/2

Answer 1

用调和均值不等式：3/(1/x+1/y+1/z)<=(x+y+z)/3
要证明原不等式,即证(x+1)^2+y^2+1+(y+1)^2+z^2+1+(z+1)^2+x^2+1>18
整理得x^2+y^2+z^2+x+y+z>6………………(1)
又由算术-几何均值不等式得1=xyz<=[(x+y+z)/3]^3………………(2)
x+y+z>=3
x^2+y^2+z^2>=3xyz=3………………(3)
(2)式加(3)式得(1)式成立,从而原不等式成立