Find-the-minimum-value-of-n-2-1-n-n-n-2-1-n-gt-0-

Question Number 110704 by Aina Samuel Temidayo last updated on 30/Aug/20

Find the minimum value of

\frac{n^{2} + 1}{n} + \frac{n}{n^{2} + 1}, n > 0

Answered by mr W last updated on 30/Aug/20

\frac{n^{2} + 1}{n} + \frac{n}{n^{2} + 1}

= (n + \frac{1}{n}) + \frac{1}{(n + \frac{1}{n})}

n + \frac{1}{n} ⩾ 2 \sqrt{n \times \frac{1}{n}} = 2

s o t h e q u e s t i o n i s t o f i n d t h e m i n i m u m

o f f u n c t i o n f (x) = x + \frac{1}{x} f o r x ⩾ 2 .

f^{'} (x) = 1 - \frac{1}{x^{2}} > 0 f o r x > 1

t h a t m e a n s f (x) i s s t r i c t l y i n c r e a s i n g

f o r x > 1, s o f o r x ⩾ 2 w e h a v e

f {(x)}_{m i n} = f (2) = 2 + \frac{1}{2} = \frac{5}{2} .

Answered by floor(10²Eta[1]) last updated on 30/Aug/20

with AM - GM inequality :

\frac{\frac{n^{2} + 1}{n} + \frac{n}{n^{2} + 1}}{2} ⩾ \sqrt{(\frac{n^{2} + 1}{n}) (\frac{n}{n^{2} + 1})} = 1

\Rightarrow \frac{n^{2} + 1}{n} + \frac{n}{n^{2} + 1} ⩾ 2

the minimum value is 2 .

Commented by Aina Samuel Temidayo last updated on 30/Aug/20

Thanks but the answer is 5 / 2 . I was

also surprised when I found out it was

5 / 2 .

Commented by Aina Samuel Temidayo last updated on 30/Aug/20

I {don}^{'} t know why AM - GM {doesn}^{'} t

work here .

Commented by mnjuly1970 last updated on 30/Aug/20

i f (m i n = 2) t h e n :

\frac{n^{2} + 1}{n} = \frac{n}{n^{2} + 1} \Rightarrow n^{4} + 2 n^{2} + 1 = n^{2}

n^{4} + n^{2} + 1 = 0 \Rightarrow n \notin R

s o l u t i o n : F (n) = (n \overset{[m i n]]}{+} \frac{1}{n}) + (\frac{\overset{[m a x]}{1}}{n + \frac{1}{n}})

m i n (F (n)) = 2 + \frac{1}{2} = \frac{5}{2}

You can't use 'macro parameter character #' in math mode

Commented by Aina Samuel Temidayo last updated on 30/Aug/20

Thanks .

Answered by $@y@m last updated on 30/Aug/20

L e t y = \frac{n^{2} + 1}{n} + \frac{n}{n^{2} + 1}, n > 0

\Rightarrow y = \frac{{(n^{2} + 1)}^{2} + n^{2}}{n (n^{2} + 1)}

\Rightarrow y = \frac{n^{4} + 3 n^{2} + 1}{n^{3} + n}

\frac{d y}{d n} = \frac{(n^{3} + n) (4 n^{3} + 6 n) - (n^{4} + 3 n^{2} + 1) (3 n^{2} + 1)}{{(n^{3} + n)}^{2}}

\frac{d y}{d n} = \frac{4 n^{6} + 10 n^{4} + 6 n^{2} - (3 n^{6} + 10 n^{4} + 6 n^{2} + 1)}{{(n^{3} + n)}^{2}}

\frac{d y}{d n} = \frac{n^{6} - 1}{{(n^{3} + n)}^{2}} \dots . (1)

F o r m a x i m a o r m i n i m a,

\frac{d y}{d n} = 0

\Rightarrow n^{6} - 1 = 0

\Rightarrow n = 1

N o w, \frac{d^{2} y}{{d n}^{2}} = \frac{{(n^{3} + n)}^{2} 6 n^{5} - (n^{6} - 1) . 2 (n^{3} + n) (3 n^{2} + 1)}{{(n^{3} + n)}^{4}}

{[\frac{d^{2} y}{{d n}_{}^{2}}]}_{a t n = 1} = \frac{4.6.1 - 0}{2^{4}} = \frac{3}{2} > 0

∴ y i s m i n i m u m a t n = 1

\Rightarrow y_{m i n} = \frac{1^{2} + 1}{1} + \frac{1}{1^{2} + 1} = 2 + \frac{1}{2} = \frac{5}{2}

Commented by Aina Samuel Temidayo last updated on 03/Sep/20

Thanks .