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Re: Interesting Inequality [#permalink]
20 Jun 2010, 08:09

2

This post received KUDOS

Expert's post

Hussain15 wrote:

If x > 1 and m > 1, is x\sqrt{m} < \sqrt{(m + x + 1)} ?

(1) m > x + 1

(2) m > \frac{1}{x-1}

Is x\sqrt{m}<\sqrt{(m+x+1)} --> square both sides (we can safely do this as both sides are positive) --> mx^2<m+x+1 --> mx^2-m<x+1 --> m(x^2-1)<x+1 --> m(x-1)(x+1)<x+1 --> reduce by x+1 (we can safely do this as x+1>0) --> m(x-1)<1.

So, finally the question becomes: is m(x-1)<1?

(1) m>x+1 --> Not sufficient.

(2) m>\frac{1}{x-1} --> cross multiply (again we can safely doth is as x-1>0) --> m(x-1)>1. Sufficient.

Re: Interesting Inequality [#permalink]
20 Jun 2010, 08:48

Bunuel, another feather in your hat. Good explanation. I was eagerly waiting for your answer as I was not able to solve the same Thanks once again. _________________

------------------------------------- Please give kudos, if my post is helpful.

Re: Interesting Inequality [#permalink]
21 Jun 2010, 02:30

Expert's post

ParvezDhamani wrote:

Can you explain why statement 1 alone doesnt suffice, it is given both x and m are >1

Because if m is some number, let's say m=100 and x is very small (not violating the condition x>1), let's say x=1.000001, then: x\sqrt{m}=10.00001<\sqrt{(m + x + 1)}\approx{10.01}.

Bit if m=9 and x=6, then x\sqrt{m}=18>\sqrt{(m + x + 1)}=4. _________________