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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.

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.
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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 \(x\sqrt{m}=18>\sqrt{(m + x + 1)}=4\).
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Re: If x > 1 and m > 1 , is x m < (m + x + 1) ? [#permalink]

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28 May 2015, 06:10

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Re: If x > 1 and m > 1 , is x m < (m + x + 1) ? [#permalink]

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13 Mar 2016, 06:28

Also the first part of the statement can be proven insufficient by taking examples as x=1.0000000001 and m=100 and x=20 and m = 400 Remember => Do not discard statement 1 as it does not satisfies the inequality directly => rather look for values and prove it both sufficient and insufficient
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Re: If x > 1 and m > 1 , is x m < (m + x + 1) ?
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13 Mar 2016, 06:28

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