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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 byprakash 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 byprakash jain last updated on 15/Jun/17

Thanks.

$$\mathrm{Thanks}. \\ $$

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