两边同除以(a+b)^[(a+b)/2]
只需证明a^[(b-a)/2]*b^[(a-b)/2]≤1
只需证明:(a/b)^[(b-a)/2]]≤1
而a/b≥1 ,b-a≤0
所以命题成立!
2.
不妨设:a≥b≥c
两边同除以(abc)^[(a+b+c)/3]
只需证明:
a^[(2a-b-c)/3]*b^[(2b-a-c)/3]*c^[(2c-a-b)/3]≥1
(2a-b-c)/3+(2c-a-b)/3=-(2b-a-c)/3
只需证明:(a/b)^[(2a-b-c)/3]*(c/b)^[(2c-a-b)/3]≥1
而a/b≥1,c/b≤1,2a-b-c≥0 ,2c-a-b≤0
命题显然成立.
3.
(1+Xi)≥2√Xi
i从1到n相乘得:
(1+x1)(1+x2)*…*(1+xn)≥2^n*Sqrt(x1*x2*…xn)=2^n
4.
lg[(a+b)/2]≥lg[√ab]=(lga+lgb)/2
后面两个同样写出,三式相加得证!
5.
[loga(x)+loga(y)]/2=loga(√xy)
(x+y)/2≥√xy
当a>1时,(x+y)/2≥√xy
loga[(x+y)/2]≥loga(√xy)=[loga(x)+loga(y)]/2
当a