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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 ax^2 +(b/x)≥c for all x,a,b>0 then  prove minimum value of  27ab^2  is 4c^2 .
$$\mathrm{If}\:{ax}^{\mathrm{2}} +\frac{{b}}{{x}}\geqslant{c}\:\mathrm{for}\:\mathrm{all}\:{x},{a},{b}>\mathrm{0}\:\mathrm{then} \\ $$$$\mathrm{prove}\:\mathrm{minimum}\:\mathrm{value}\:\mathrm{of} \\ $$$$\mathrm{27}{ab}^{\mathrm{2}} \:\mathrm{is}\:\mathrm{4}{c}^{\mathrm{2}} . \\ $$
Commented by prakash jain last updated on 14/Jun/17
AM≥GM  ((ax^2 +(b/(2x))+(b/(2x)))/3)≥(((ab^2 )/4))^(1/3)   (c/3)≥(((ab^2 )/4))^(1/3)   (c^3 /(27))≥((ab^2 )/4)  4c^3 ≥27ab^2
$$\mathrm{AM}\geqslant\mathrm{GM} \\ $$$$\frac{{ax}^{\mathrm{2}} +\frac{{b}}{\mathrm{2}{x}}+\frac{{b}}{\mathrm{2}{x}}}{\mathrm{3}}\geqslant\left(\frac{{ab}^{\mathrm{2}} }{\mathrm{4}}\right)^{\mathrm{1}/\mathrm{3}} \\ $$$$\frac{{c}}{\mathrm{3}}\geqslant\left(\frac{{ab}^{\mathrm{2}} }{\mathrm{4}}\right)^{\mathrm{1}/\mathrm{3}} \\ $$$$\frac{{c}^{\mathrm{3}} }{\mathrm{27}}\geqslant\frac{{ab}^{\mathrm{2}} }{\mathrm{4}} \\ $$$$\mathrm{4}{c}^{\mathrm{3}} \geqslant\mathrm{27}{ab}^{\mathrm{2}} \\ $$
Answered by ajfour last updated on 15/Jun/17
 let   f(x)= ax^2 +(b/x)   minimum value of f(x) = c  as for x→0 , f (x)→∞  and for x→∞ , f(x)→∞   f(x) must have a point of   minima ;   f ′(x)= 2ax−(b/x^2 )   f ′(x)=0   for  x= ((b/(2a)))^(1/3) = x_m    f(x_m )= ax_m ^2 +(b/x_m ) ≥ c     (then)  or    c≤ ((ax_m ^3 +b)/x_m )    c ≤ ((a(b/2a)+b)/((b/2a)^(1/3) )) = ((3b)/2)(((2a)/b))^(1/3)      c^3 ≤ ((27ab^2 )/4)      27ab^2 ≥c^3    is  necessary if     ax^2 +(b/x)≥ c , for all x, a, b>0 .
$$\:{let}\:\:\:{f}\left({x}\right)=\:{ax}^{\mathrm{2}} +\frac{{b}}{{x}} \\ $$$$\:{minimum}\:{value}\:{of}\:{f}\left({x}\right)\:=\:{c} \\ $$$${as}\:{for}\:{x}\rightarrow\mathrm{0}\:,\:{f}\:\left({x}\right)\rightarrow\infty \\ $$$${and}\:{for}\:{x}\rightarrow\infty\:,\:{f}\left({x}\right)\rightarrow\infty \\ $$$$\:{f}\left({x}\right)\:{must}\:{have}\:{a}\:{point}\:{of}\: \\ $$$${minima}\:; \\ $$$$\:{f}\:'\left({x}\right)=\:\mathrm{2}{ax}−\frac{{b}}{{x}^{\mathrm{2}} } \\ $$$$\:{f}\:'\left({x}\right)=\mathrm{0}\:\:\:{for}\:\:{x}=\:\left(\frac{{b}}{\mathrm{2}{a}}\right)^{\mathrm{1}/\mathrm{3}} =\:{x}_{{m}} \\ $$$$\:{f}\left({x}_{{m}} \right)=\:{ax}_{{m}} ^{\mathrm{2}} +\frac{{b}}{{x}_{{m}} }\:\geqslant\:{c}\:\:\:\:\:\left({then}\right) \\ $$$${or}\:\:\:\:{c}\leqslant\:\frac{{ax}_{{m}} ^{\mathrm{3}} +{b}}{{x}_{{m}} } \\ $$$$\:\:{c}\:\leqslant\:\frac{{a}\left({b}/\mathrm{2}{a}\right)+{b}}{\left({b}/\mathrm{2}{a}\right)^{\mathrm{1}/\mathrm{3}} }\:=\:\frac{\mathrm{3}{b}}{\mathrm{2}}\left(\frac{\mathrm{2}{a}}{{b}}\right)^{\mathrm{1}/\mathrm{3}} \\ $$$$\:\:\:{c}^{\mathrm{3}} \leqslant\:\frac{\mathrm{27}{ab}^{\mathrm{2}} }{\mathrm{4}}\: \\ $$$$\:\:\:\mathrm{27}{ab}^{\mathrm{2}} \geqslant{c}^{\mathrm{3}} \:\:\:{is}\:\:{necessary}\:{if} \\ $$$$\:\:\:{ax}^{\mathrm{2}} +\frac{{b}}{{x}}\geqslant\:{c}\:,\:{for}\:{all}\:{x},\:{a},\:{b}>\mathrm{0}\:. \\ $$
Commented by prakash jain last updated on 15/Jun/17
Thanks.
$$\mathrm{Thanks}. \\ $$

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