Question

(1)證明

(2)若x、y、z∈(0,1),求證:x(1-y)+y(1-z)+z(1-x)＜1.

Answer 1

解析:(1)由于待定的不等式可整理為關(guān)于變量x的二次方程的形式,可用判別式證明.?

(2)由于待定的不等式關(guān)于x最高次數(shù)為一次,可整理成關(guān)于x的一次函數(shù),由一次函數(shù)的單調(diào)性進(jìn)行證明.

證明:(1)設(shè)則(y-1)x²+(3y+3)x+4y-4=0.

①當(dāng)y-1≠0時(shí),則x∈R,Δ≥0,?

即(3y+3)²-4(y-1)(4y-4)≥0.?

解得≤y≤7(y≠1).?

②當(dāng)y-1=0時(shí),x=0∈R.?

綜上,≤y≤7.∴

(2)構(gòu)造函數(shù):?

f(x)=x(1-y)+y(1-z)+z(1-x)-1=(1-y-z)x+y(1-z)+z-1(0＜x＜1),

①當(dāng)1-y-z=0,即y+z=1時(shí),f(x)=y(1-z)+z-1=y+z-yz-1=-yz＜0;?

②當(dāng)1-y-z≠0時(shí),f(x)為一次函數(shù),由一次函數(shù)的單調(diào)性,只要證明f(0)＜0,f(1)＜0,?

∵f(0)=y-yz+z-1=(y-1)(1-z)＜0,f(1)=-yz＜0,

∴對(duì)任意x∈(0,1)都有f(x)＜0,?

即x(1-y)+y(1-z)+z(1-x)＜1.