Question

高中数学竞赛 两道不等式

xy+yz+zx=1,x,y,z>=0
求证 1/(x+y)+1/(y+z)+1/(x+z)>=5/2

abc=1

Answer 1

证明：
设f(x,y,z)=1/(x+y)+1/(y+z)+1/(z+x)
由对称性不妨设0<=x<=y<=z，先证明f(0,x+y,z')<=f(x,y,z),其中z'=1/(x+y)
即证：
1/(x+y)+1/(x+y+z')+1/z'<=1/(x+y)+1/(y+z)+1/(z+x)
由于z'=1/(x+y),z=(1-xy)/(x+y)。则上式等价于：
(x+y+2z')/[(x+y+z')z']<=(x+y+2z)/[(y+z)(z+x)]
<=>(x+y+2z')/(x+y+2z)<=(1+z'^2)/(1+z^2)
<=>(x+y+2z')/(x+y+2z)-1<=(1+z'^2)/(1+z^2)-1
<=>2(z'-z)/(x+y+2z)<=[(z'-z)(z'+z)]/(1+z^2)
<=>2/(x+y+2z)<=(z'+z)/(1+z'^2)
{注：这一步用到了z'=1/(x+y)>(1-xy)/(x+y)=z}
<=>2+2z^2<=(x+y+2z)(z'+z)
<=>xy(x+y)^2<=2(1-xy)
注意到2(1-xy)=2z(x+y)>=2(x+y)^2/2{注：这里用到了z>=y>=x}，从而，只要证明(x+y)^2>=xy(x+y)^2即可。而该式即xy<=1显然成立。
那么
f(x,y,z)>=f(0,x+y,z')=1/(x+y)+(x+y)+1/[1/(x+y)+(x+y)]
记t=1/(x+y)+(x+y)，则f(x,y,z)>=t+1/t，而由均值不等式t>=2，而我们知道函数t+1/t当x>=1时是单调递增的。所以f(x,y,z)>=t+1/t>=5/2

Answer 2

图片中的题可以用琴森不等式
构造函数 f(x)=e^x/(3e^x+1)^0.5 可以验证f``(x)>0 对所有x成立 因此f(x)是下凸函数
有f(x)+f(y)+f(z)>=3f(x+y+z/3) 令x=lna y=lnb z=lnc 则 x+y+z=0
可得结论

Answer 3

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