An algebra problem by mohamed aboalamayem

Algebra Level 2

when a > 1 a > 1 , arrange the following three expressions by magnitude : 1 , a a − 1 , a + 1 a 1 , \quad \dfrac a{a-1} , \quad \dfrac{ a+1}a

1 > a / ( a - 1 ) > ( a + 1 ) / a a / ( a - 1 ) > 1 > ( a + 1 ) / a ( a +1 ) / a > 1 > a / ( a - 1 ) a / (a - 1) > (a + 1 ) / a > 1

Mohamed Aboalamayem by mohamed aboalamayem , 57, Egypt

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2 solutions

Hung Woei Neoh
Apr 21, 2016

a a − 1 = a − 1 + 1 a − 1 = a − 1 a − 1 + 1 a − 1 = 1 + 1 a − 1 \dfrac{a}{a-1} = \dfrac{a-1+1}{a-1} = \dfrac{a-1}{a-1} + \dfrac{1}{a-1} = 1 + \dfrac{1}{a-1}

a + 1 a = a a + 1 a = 1 + 1 a \dfrac{a+1}{a} = \dfrac{a}{a} + \dfrac{1}{a} = 1 + \dfrac{1}{a}

Since a > 1 a>1 , we know that

1 a > 0 \dfrac{1}{a} > 0 and 1 a − 1 > 0 \dfrac{1}{a-1} > 0

and

a − 1 < a ⟹ 1 a − 1 > 1 a a-1 < a \implies \dfrac{1}{a-1} > \dfrac{1}{a}

Therefore:

0 < 1 a < 1 a − 1 ⟹ 1 < 1 + 1 a < 1 + 1 a − 1 ⟹ 1 < a + 1 a < a a − 1 0 < \dfrac{1}{a} < \dfrac{1}{a-1}\\ \implies 1 < 1 + \dfrac{1}{a} < 1 + \dfrac{1}{a-1}\\ \implies 1 < \dfrac{a+1}{a} < \dfrac{a}{a-1}

Turn the inequality around, and we get the answer: a a − 1 > a + 1 a > 1 \boxed{\dfrac{a}{a-1} > \dfrac{a+1}{a} > 1}

1 Helpful 0 Interesting 0 Brilliant 0 Confused

For : a > 1 we see that : a / ( a - 1 ) > ( a + 1 ) / a > 1 , You can let : a = 5 , so : 5 / 4 > 6 / 5 > 1 .

1 Helpful 0 Interesting 0 Brilliant 0 Confused

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