Prove for x?
回答 (3)
2015-01-05 3:30 pm
Let x=1, then LHS=sqrt(1^2+1)=sqrt(2); RHS=1
sqrt(2)>=1=> the inequality holds for x=1. Now
let x=1+a, where a>0, then
sqrt[(1+a)^2+(1+a)]-(1+a)=
sqrt[(1+a)(2+a)]-sqrt[(1+a)^2]=
sqrt[(1+a)(2+a)]-sqrt[(1+a)(1+a)]>0=>
sqrt[(1+a)^2+(1+a)>(1+a)=>
sqrt(x^2+x)>=x for x>=1
sqrt(2)>=1=> the inequality holds for x=1. Now
let x=1+a, where a>0, then
sqrt[(1+a)^2+(1+a)]-(1+a)=
sqrt[(1+a)(2+a)]-sqrt[(1+a)^2]=
sqrt[(1+a)(2+a)]-sqrt[(1+a)(1+a)]>0=>
sqrt[(1+a)^2+(1+a)>(1+a)=>
sqrt(x^2+x)>=x for x>=1
2015-01-05 3:20 pm
Square root is an increasing function, so since x>0, we have
(x^2+x)^0.5 > (x^2)^0.5 = x.
This shows in fact that if x>0 then (x^2+x)^0.5 > x.
(x^2+x)^0.5 > (x^2)^0.5 = x.
This shows in fact that if x>0 then (x^2+x)^0.5 > x.
2015-01-05 3:11 pm
If you rework the equation to get x on one side, it is obvious. However you will find that it CANNOT be proved, because if the x=1, then the equation is false. It only works if x>1.
收錄日期: 2021-04-21 01:10:14
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20150105070716AAfcn87