let-x-2-3x-p-0-has-two-positive-roots-a-and-b-then-inf-4-a-1-b-is-

Question Number 205116 by universe last updated on 09/Mar/24

let x^{2} - 3 x + p = 0 has two positive roots

^{'} a^{'} and^{'} b^{'} then \inf (\frac{4}{a} + \frac{1}{b}) is

Answered by pi314 last updated on 09/Mar/24

\frac{4 b + a}{a b} = \frac{3 + 3 b}{p}

9 - 4 p ⩾ 0; p ⩽ \frac{9}{4}

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

(\frac{4}{a} + \frac{1}{b}) = \frac{4}{3 - b} + \frac{1}{b} = f (b)

f^{'} (b) = \frac{4}{{(3 - b)}^{2}} - \frac{1}{b^{2}} = \frac{4 b^{2} - {(3 - b)}^{2}}{{(3 - b)}^{2} b^{2}} = \frac{3 b^{2} + 6 b - 9}{b^{2} {(3 - b)}^{2}}

= \frac{3 (b^{2} + 2 b - 3)}{b^{2} {(3 - b)}^{2}} = \frac{3 (b - 1) (b + 3)}{b^{2} {(3 - b)}^{2}}

i n f (f) = f (1) = \frac{4}{2} + 1 = 3

Commented by pi314 last updated on 09/Mar/24

m i s t a c k f o r m e

Answered by mr W last updated on 09/Mar/24

a, b > 0

a + b = 3

\frac{4}{a} + \frac{1}{b} = \frac{2^{2}}{a} + \frac{1^{2}}{b} ⩾ \frac{{(2 + 1)}^{2}}{a + b} = \frac{9}{3} = 3

Commented by universe last updated on 09/Mar/24

t h a n k y o u s i r

Commented by universe last updated on 09/Mar/24

s i r n a m e o f t h i s e n e q u a l i t y \frac{2^{2}}{a} + \frac{1^{2}}{b} ⩾ \frac{{(2 + 1)}^{2}}{a + b}

Commented by mr W last updated on 09/Mar/24

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