a-gt-0-b-gt-0-What-is-the-minimum-value-of-b-2-2-a-b-a-2-ab-1-

Question Number 39971 by math2018 last updated on 14/Jul/18

a > 0, b > 0,

W h a t i s t h e m i n i m u m v a l u e o f

\frac{b^{2} + 2}{a + b} + \frac{a^{2}}{a b + 1} ?

Answered by tanmay.chaudhury50@gmail.com last updated on 14/Jul/18

\frac{b^{2}}{a + b} + \frac{a^{2}}{a b + 1} + \frac{2}{a + b} = x (s a y)

u s i n g A M > G M

\frac{a + b}{2} > \sqrt{a b}

\frac{\frac{b^{2}}{2}}{\frac{a + b}{2}} + \frac{\frac{a^{2}}{2}}{\frac{a b + 1}{2}} + \frac{\frac{2}{2}}{\frac{a + b}{2}} < \frac{\frac{b^{2}}{2}}{\sqrt{a b}} + \frac{\frac{a^{2}}{2}}{\sqrt{a b}} + \frac{\frac{2}{2}}{\sqrt{a b}} \to y (s a y)

x < \frac{b^{2}}{2 \sqrt{a b}} + \frac{a^{2}}{2 \sqrt{a b}} + \frac{2}{2 \sqrt{a b}}

x < \frac{a^{2} + b^{2} + 2}{2 \sqrt{a b}}

x < \frac{\frac{a^{2} + b^{2}}{2} \times 2 + 2}{2 \sqrt{a b}}

u s i n g \frac{a^{2} + b^{2}}{2} > \sqrt{a^{2} b^{2}}

\frac{\frac{a^{2} + b^{2}}{2} \times 2 + 2}{2 \sqrt{a b}} > \frac{2 a b + 2}{2 \sqrt{a b}} \to z (s a y)

y > \frac{a b + 1}{\sqrt{a b}}

y > \frac{\frac{a b + 1}{2} \times 2}{\sqrt{a b}} \to z

\frac{\frac{a b + 1}{2} \times 2}{\sqrt{a b}} > \frac{\sqrt{a b} \times 2}{\sqrt{a b}}

s o z > 2

g i v e n e x p r e s s i o n a s s u m e d t o b e x

x < y

y > z

z > 2

i c a n n o t c o r e l a t e b e t w e e n x a n d z

Commented by math2018 last updated on 15/Jul/18

T h a n k y o u v e r y m u c h!

Answered by ajfour last updated on 14/Jul/18

f (a, b) = \frac{b^{2} + 2}{a + b} + \frac{a^{2}}{a b + 1}

< \frac{b^{2} + 2}{2 \sqrt{a b}} + \frac{a^{2}}{2 \sqrt{a b}} = \frac{{(a - b)}^{2} + 2 a b + 2}{2 \sqrt{a b}}

= \frac{{(a - b)}^{2}}{2 \sqrt{a b}} + \sqrt{a b} + \frac{1}{\sqrt{a b}}

t h e a b o v e e x p r e s s i o n h a s a

m i n i m u m o f v a l u e e q u a l t o 2

f o r a = b = 1

h e n c e f_{m i n} (a, b) ⩽ 2 .

Commented by math2018 last updated on 15/Jul/18

T h a n k y o u S i r!

Answered by math2018 last updated on 15/Jul/18

1 ° a + b ⩾ a b + 1

\frac{b^{2} + 2}{a + b} + \frac{a^{2}}{a b + 1} ⩾ \frac{(a^{2} + 1) + (b^{2} + 1)}{a + b} ⩾ \frac{2 a + 2 b}{a + b} = 2;

2 ° a + b < a b + 1

\frac{b^{2} + 2}{a + b} + \frac{a^{2}}{a b + 1} > \frac{a^{2} + b^{2} + 2}{a b + 1} ⩾ \frac{2 a b + 2}{a b + 1} = 2

h e n c e t h e m i n i m u m v a l u e i s 2 .

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