Which one of the $sqrt17-sqrt12$ and $sqrt11-sqrt6$ is greater?

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Answer 1 · 2017-06-08T05:30:04

$sqrt17-sqrt12 < sqrt11-sqrt6$ and hence $sqrt11-sqrt6$ is greater.

Let us consider $sqrt17+sqrt6$ and $sqrt11+sqrt12$ for comparison.

Well we just square both of them and we get

$(sqrt17+sqrt6)^2=(sqrt17)^2+2xxsqrt17xxsqrt6+(sqrt6)^2$

= $17+6+2sqrt102=23+sqrt102$

and $(sqrt11+sqrt12)^2=(sqrt11)^2+2xxsqrt11xxsqrt12+(sqrt12)^2$

= $11+12+2sqrt132=23+2sqrt132$

As $(sqrt17+sqrt6)^2 < (sqrt11+sqrt12)^2$ ,

we have $sqrt17+sqrt6 < sqrt11+sqrt12$

and hence transposing $sqrt6$ to right and $sqrt12$ to left

$sqrt17-sqrt12 < sqrt11-sqrt6$ and hence $sqrt11-sqrt6$ is greater.

Which one of the sqrt17-sqrt12sqrt17-sqrt12 and sqrt11-sqrt6sqrt11-sqrt6 is greater?