已知a>0且a≠1,函数f(X)=loga(X+1)在区间（-1,+∞）上递减,求证：对任意实数x1>0,X2>0,恒有1/2[f(x1-1)+f(x2-1)]>=f[(x1+x2-2）/2]

已知a>0且a≠1,函数f(X)=loga(X+1)在区间（-1,+∞）上递减,求证：对任意实数x1>0,X2>0,恒有1/2[f(x1-1)+f(x2-1)]>=f[(x1+x2-2）/2]
已知a>0且a≠1,函数f(X)=loga(X+1)在区间（-1,+∞）上递减,求证：对任意实数x1>0,X2>0,恒有1/2[f(x1-1)+f(x2-1)]>=f[(x1+x2-2）/2]

已知a>0且a≠1,函数f(X)=loga(X+1)在区间（-1,+∞）上递减,求证：对任意实数x1>0,X2>0,恒有1/2[f(x1-1)+f(x2-1)]>=f[(x1+x2-2）/2]
∵函数f(X)=loga(X+1)在区间（-1,+∞）上递减
∴00 ∴x1-1>-1,x2-1>-1
1/2[f(x1-1)+f(x2-1)]=[loga(x1)+loga(x2)]/2
=loga(x1x2)/2=loga√(x1x2)
f[(x1+x2-2）/2]=loga[(x1+x2)/2]
∵x1>0,X2>0
根据均值定理有
√(x1x2)≤(x1+x2)/2
∵0

1/2[f(x1-1)+f(x2-1)]
=1/2[loga(x1-1)+loga(x2-1)]
=loga[(x1+1)(x2+1)]^(1/2)
√(x1-1)(x2-1)小于等于(x1-1)+(x2-1) /2
f(x)为减函数,所以恒有1/2[f(x1-1)+f(x2-1)]>=f[(x1+x2-2）/2]

E

DRESFR