If-ax-2-b-x-c-for-all-x-a-b-gt-0-then-prove-minimum-value-of-27ab-2-is-4c-2-

Question Number 15869 by prakash jain last updated on 14/Jun/17

If {a x}^{2} + \frac{b}{x} ⩾ c for all x, a, b > 0 then

prove minimum value of

27 {a b}^{2} is 4 c^{2} .

Commented by prakash jain last updated on 14/Jun/17

AM ⩾ GM

\frac{{a x}^{2} + \frac{b}{2 x} + \frac{b}{2 x}}{3} ⩾ {(\frac{{a b}^{2}}{4})}^{1 / 3}

\frac{c}{3} ⩾ {(\frac{{a b}^{2}}{4})}^{1 / 3}

\frac{c^{3}}{27} ⩾ \frac{{a b}^{2}}{4}

4 c^{3} ⩾ 27 {a b}^{2}

Answered by ajfour last updated on 15/Jun/17

l e t f (x) = {a x}^{2} + \frac{b}{x}

m i n i m u m v a l u e o f f (x) = c

a s f o r x \to 0, f (x) \to \infty

a n d f o r x \to \infty, f (x) \to \infty

f (x) m u s t h a v e a p o i n t o f

m i n i m a;

f^{'} (x) = 2 a x - \frac{b}{x^{2}}

f^{'} (x) = 0 f o r x = {(\frac{b}{2 a})}^{1 / 3} = x_{m}

f (x_{m}) = {a x}_{m}^{2} + \frac{b}{x_{m}} ⩾ c (t h e n)

o r c ⩽ \frac{{a x}_{m}^{3} + b}{x_{m}}

c ⩽ \frac{a (b / 2 a) + b}{{(b / 2 a)}^{1 / 3}} = \frac{3 b}{2} {(\frac{2 a}{b})}^{1 / 3}

c^{3} ⩽ \frac{27 {a b}^{2}}{4}

27 {a b}^{2} ⩾ c^{3} i s n e c e s s a r y i f

{a x}^{2} + \frac{b}{x} ⩾ c, f o r a l l x, a, b > 0 .

Commented by prakash jain last updated on 15/Jun/17

Thanks .