Let-p-amp-q-real-positive-number-what-the-minimum-of-p-2-q-2-q-p-3-

Question Number 142200 by liberty last updated on 27/May/21

Let p & q real positive number what the minimum of ((p^2 /q^2 ) +(q/p))^3 .

$$\:{Let}\:{p}\:\&\:{q}\:{real}\:{positive}\:{number} \\ $$$$\:\:{what}\:{the}\:{minimum}\:{of}\:\left(\frac{{p}^{\mathrm{2}} }{{q}^{\mathrm{2}} }\:+\frac{{q}}{{p}}\right)^{\mathrm{3}} . \\ $$

Answered by MJS_new last updated on 27/May/21

let q=pt∧t>0 ((p^2 /q^2 )+(q/p))^3 =(((t^3 +1)^3 )/t^6 ) (d/dt)[(((t^3 +1)^3 )/t^6 )]=0∧t>0 ⇒ t=(2)^(1/3) ⇒ min is ((27)/4)

$$\mathrm{let}\:{q}={pt}\wedge{t}>\mathrm{0} \\ $$$$\left(\frac{{p}^{\mathrm{2}} }{{q}^{\mathrm{2}} }+\frac{{q}}{{p}}\right)^{\mathrm{3}} =\frac{\left({t}^{\mathrm{3}} +\mathrm{1}\right)^{\mathrm{3}} }{{t}^{\mathrm{6}} } \\ $$$$\frac{{d}}{{dt}}\left[\frac{\left({t}^{\mathrm{3}} +\mathrm{1}\right)^{\mathrm{3}} }{{t}^{\mathrm{6}} }\right]=\mathrm{0}\wedge{t}>\mathrm{0}\:\Rightarrow\:{t}=\sqrt[{\mathrm{3}}]{\mathrm{2}}\:\Rightarrow\:\mathrm{min}\:\mathrm{is}\:\frac{\mathrm{27}}{\mathrm{4}} \\ $$

Answered by mitica last updated on 27/May/21

((p^2 /q^2 )+(q/(2p))+(q/(2p)))^3 ≥^(am−gm) (3(((p^2 /q^2 )∙(q/(2p))∙(q/(2p))))^(1/3) )^3 =((27)/4) ,,=′′ for (p^2 /q^2 )=(q/(2p))⇔p=q(2)^(1/3) ⇒min=((27)/4)

$$\left(\frac{{p}^{\mathrm{2}} }{{q}^{\mathrm{2}} }+\frac{{q}}{\mathrm{2}{p}}+\frac{{q}}{\mathrm{2}{p}}\right)^{\mathrm{3}} \overset{{am}−{gm}} {\geqslant}\:\left(\mathrm{3}\sqrt[{\mathrm{3}}]{\frac{{p}^{\mathrm{2}} }{{q}^{\mathrm{2}} }\centerdot\frac{{q}}{\mathrm{2}{p}}\centerdot\frac{{q}}{\mathrm{2}{p}}}\right)^{\mathrm{3}} =\frac{\mathrm{27}}{\mathrm{4}} \\ $$$$,,=''\:{for}\:\frac{{p}^{\mathrm{2}} }{{q}^{\mathrm{2}} }=\frac{{q}}{\mathrm{2}{p}}\Leftrightarrow{p}={q}\sqrt[{\mathrm{3}}]{\mathrm{2}}\Rightarrow{min}=\frac{\mathrm{27}}{\mathrm{4}} \\ $$

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