Question Number 15869 by prakash jain last updated on 14/Jun/17
$$\mathrm{If}\:{ax}^{\mathrm{2}} +\frac{{b}}{{x}}\geqslant{c}\:\mathrm{for}\:\mathrm{all}\:{x},{a},{b}>\mathrm{0}\:\mathrm{then} \\ $$$$\mathrm{prove}\:\mathrm{minimum}\:\mathrm{value}\:\mathrm{of} \\ $$$$\mathrm{27}{ab}^{\mathrm{2}} \:\mathrm{is}\:\mathrm{4}{c}^{\mathrm{2}} . \\ $$
Commented by prakash jain last updated on 14/Jun/17
$$\mathrm{AM}\geqslant\mathrm{GM} \\ $$$$\frac{{ax}^{\mathrm{2}} +\frac{{b}}{\mathrm{2}{x}}+\frac{{b}}{\mathrm{2}{x}}}{\mathrm{3}}\geqslant\left(\frac{{ab}^{\mathrm{2}} }{\mathrm{4}}\right)^{\mathrm{1}/\mathrm{3}} \\ $$$$\frac{{c}}{\mathrm{3}}\geqslant\left(\frac{{ab}^{\mathrm{2}} }{\mathrm{4}}\right)^{\mathrm{1}/\mathrm{3}} \\ $$$$\frac{{c}^{\mathrm{3}} }{\mathrm{27}}\geqslant\frac{{ab}^{\mathrm{2}} }{\mathrm{4}} \\ $$$$\mathrm{4}{c}^{\mathrm{3}} \geqslant\mathrm{27}{ab}^{\mathrm{2}} \\ $$
Answered by ajfour last updated on 15/Jun/17
$$\:{let}\:\:\:{f}\left({x}\right)=\:{ax}^{\mathrm{2}} +\frac{{b}}{{x}} \\ $$$$\:{minimum}\:{value}\:{of}\:{f}\left({x}\right)\:=\:{c} \\ $$$${as}\:{for}\:{x}\rightarrow\mathrm{0}\:,\:{f}\:\left({x}\right)\rightarrow\infty \\ $$$${and}\:{for}\:{x}\rightarrow\infty\:,\:{f}\left({x}\right)\rightarrow\infty \\ $$$$\:{f}\left({x}\right)\:{must}\:{have}\:{a}\:{point}\:{of}\: \\ $$$${minima}\:; \\ $$$$\:{f}\:'\left({x}\right)=\:\mathrm{2}{ax}−\frac{{b}}{{x}^{\mathrm{2}} } \\ $$$$\:{f}\:'\left({x}\right)=\mathrm{0}\:\:\:{for}\:\:{x}=\:\left(\frac{{b}}{\mathrm{2}{a}}\right)^{\mathrm{1}/\mathrm{3}} =\:{x}_{{m}} \\ $$$$\:{f}\left({x}_{{m}} \right)=\:{ax}_{{m}} ^{\mathrm{2}} +\frac{{b}}{{x}_{{m}} }\:\geqslant\:{c}\:\:\:\:\:\left({then}\right) \\ $$$${or}\:\:\:\:{c}\leqslant\:\frac{{ax}_{{m}} ^{\mathrm{3}} +{b}}{{x}_{{m}} } \\ $$$$\:\:{c}\:\leqslant\:\frac{{a}\left({b}/\mathrm{2}{a}\right)+{b}}{\left({b}/\mathrm{2}{a}\right)^{\mathrm{1}/\mathrm{3}} }\:=\:\frac{\mathrm{3}{b}}{\mathrm{2}}\left(\frac{\mathrm{2}{a}}{{b}}\right)^{\mathrm{1}/\mathrm{3}} \\ $$$$\:\:\:{c}^{\mathrm{3}} \leqslant\:\frac{\mathrm{27}{ab}^{\mathrm{2}} }{\mathrm{4}}\: \\ $$$$\:\:\:\mathrm{27}{ab}^{\mathrm{2}} \geqslant{c}^{\mathrm{3}} \:\:\:{is}\:\:{necessary}\:{if} \\ $$$$\:\:\:{ax}^{\mathrm{2}} +\frac{{b}}{{x}}\geqslant\:{c}\:,\:{for}\:{all}\:{x},\:{a},\:{b}>\mathrm{0}\:. \\ $$
Commented by prakash jain last updated on 15/Jun/17
$$\mathrm{Thanks}. \\ $$