Menu Close

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

Tinku Tara 0 Comments
Pin




Question Number 205116 by universe last updated on 09/Mar/24
let x^2 −3x+p = 0 has two positive roots ′a′ and ′b′ then inf((4/a)+(1/b)) is
$$\:\:\mathrm{let}\:\mathrm{x}^{\mathrm{2}} −\mathrm{3x}+\mathrm{p}\:=\:\mathrm{0}\:\mathrm{has}\:\mathrm{two}\:\mathrm{positive}\:\mathrm{roots} \\ $$$$\:'\mathrm{a}'\:\mathrm{and}\:'\mathrm{b}'\:\mathrm{then}\:\:\mathrm{inf}\left(\frac{\mathrm{4}}{\mathrm{a}}+\frac{\mathrm{1}}{\mathrm{b}}\right)\:\mathrm{is}\: \\ $$
Answered by pi314 last updated on 09/Mar/24
((4b+a)/(ab))=((3+3b)/p) 9−4p≥0;p≤(9/4) p#0;p>0 ;0<p≤(9/4) ((4/a)+(1/b))=(4/(3−b))+(1/b)=f(b) f′(b)=(4/((3−b)^2 ))−(1/b^2 )=((4b^2 −(3−b)^2 )/((3−b)^2 b^2 ))=((3b^2 +6b−9)/(b^2 (3−b)^2 )) =((3(b^2 +2b−3))/(b^2 (3−b)^2 ))=((3(b−1)(b+3))/(b^2 (3−b)^2 )) inf (f)=f(1)=(4/2)+1=3
$$\frac{\mathrm{4}{b}+{a}}{{ab}}=\frac{\mathrm{3}+\mathrm{3}{b}}{{p}} \\ $$$$\mathrm{9}−\mathrm{4}{p}\geqslant\mathrm{0};{p}\leqslant\frac{\mathrm{9}}{\mathrm{4}} \\ $$$${p}#\mathrm{0};{p}>\mathrm{0}\:;\mathrm{0}<{p}\leqslant\frac{\mathrm{9}}{\mathrm{4}} \\ $$$$\left(\frac{\mathrm{4}}{{a}}+\frac{\mathrm{1}}{{b}}\right)=\frac{\mathrm{4}}{\mathrm{3}−{b}}+\frac{\mathrm{1}}{{b}}={f}\left({b}\right) \\ $$$${f}'\left({b}\right)=\frac{\mathrm{4}}{\left(\mathrm{3}−{b}\right)^{\mathrm{2}} }−\frac{\mathrm{1}}{{b}^{\mathrm{2}} }=\frac{\mathrm{4}{b}^{\mathrm{2}} −\left(\mathrm{3}−{b}\right)^{\mathrm{2}} }{\left(\mathrm{3}−{b}\right)^{\mathrm{2}} {b}^{\mathrm{2}} }=\frac{\mathrm{3}{b}^{\mathrm{2}} +\mathrm{6}{b}−\mathrm{9}}{{b}^{\mathrm{2}} \left(\mathrm{3}−{b}\right)^{\mathrm{2}} } \\ $$$$=\frac{\mathrm{3}\left({b}^{\mathrm{2}} +\mathrm{2}{b}−\mathrm{3}\right)}{{b}^{\mathrm{2}} \left(\mathrm{3}−{b}\right)^{\mathrm{2}} }=\frac{\mathrm{3}\left({b}−\mathrm{1}\right)\left({b}+\mathrm{3}\right)}{{b}^{\mathrm{2}} \left(\mathrm{3}−{b}\right)^{\mathrm{2}} } \\ $$$${inf}\:\left({f}\right)={f}\left(\mathrm{1}\right)=\frac{\mathrm{4}}{\mathrm{2}}+\mathrm{1}=\mathrm{3}\: \\ $$
Commented by pi314 last updated on 09/Mar/24
mistack for me
$${mistack}\:{for}\:{me}\: \\ $$
Answered by mr W last updated on 09/Mar/24
a, b>0 a+b=3 (4/a)+(1/b)=(2^2 /a)+(1^2 /b)≥(((2+1)^2 )/(a+b))=(9/3)=3
$${a},\:{b}>\mathrm{0} \\ $$$${a}+{b}=\mathrm{3} \\ $$$$\frac{\mathrm{4}}{{a}}+\frac{\mathrm{1}}{{b}}=\frac{\mathrm{2}^{\mathrm{2}} }{{a}}+\frac{\mathrm{1}^{\mathrm{2}} }{{b}}\geqslant\frac{\left(\mathrm{2}+\mathrm{1}\right)^{\mathrm{2}} }{{a}+{b}}=\frac{\mathrm{9}}{\mathrm{3}}=\mathrm{3}\: \\ $$
Commented by universe last updated on 09/Mar/24
thank you sir
$${thank}\:{you}\:{sir} \\ $$
Commented by universe last updated on 09/Mar/24
sir name of this enequality (2^2 /a)+(1^2 /b)≥(((2+1)^2 )/(a+b))
$$\:{sir}\:{name}\:{of}\:{this}\:{enequality}\:\frac{\mathrm{2}^{\mathrm{2}} }{{a}}+\frac{\mathrm{1}^{\mathrm{2}} }{{b}}\geqslant\frac{\left(\mathrm{2}+\mathrm{1}\right)^{\mathrm{2}} }{{a}+{b}} \\ $$
Commented by mr W last updated on 09/Mar/24

Pin

Leave a Reply

Your email address will not be published. Required fields are marked *