作业帮已知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]
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![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]](/uploads/image/z/10139618-2-8.jpg?t=%E5%B7%B2%E7%9F%A5a%3E0%E4%B8%94a%E2%89%A01%2C%E5%87%BD%E6%95%B0f%28X%29%3Dloga%28X%2B1%29%E5%9C%A8%E5%8C%BA%E9%97%B4%EF%BC%88-1%2C%2B%E2%88%9E%EF%BC%89%E4%B8%8A%E9%80%92%E5%87%8F%2C%E6%B1%82%E8%AF%81%EF%BC%9A%E5%AF%B9%E4%BB%BB%E6%84%8F%E5%AE%9E%E6%95%B0x1%3E0%2CX2%3E0%2C%E6%81%92%E6%9C%891%2F2%5Bf%28x1-1%29%2Bf%28x2-1%29%5D%3E%3Df%5B%28x1%2Bx2-2%EF%BC%89%2F2%5D)
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]
递减可知 0=sqrtx1x2 0