Question Number 215883 by CrispyXYZ last updated on 20/Jan/25
$$\mathrm{Given}\:{m}>\mathrm{0},\:{n}>\mathrm{0},\:{m}+{n}=\sqrt{{a}}. \\ $$$$\mathrm{Find}\:\mathrm{the}\:\mathrm{range}\:\mathrm{of}\:{a}\:\mathrm{such}\:\mathrm{that} \\ $$$$``\left({m}+\frac{\mathrm{1}}{{m}}\right)\left({n}+\frac{\mathrm{1}}{{n}}\right)\:\mathrm{gets}\:\mathrm{its}\:\mathrm{minimum}\:\mathrm{iff}\:{m}={n}''. \\ $$
Commented by MathematicalUser2357 last updated on 21/Jan/25
$$``{iff}''?\:{does}\:{it}\:{mean}\:{ifh}.{cc}? \\ $$
Commented by CrispyXYZ last updated on 21/Jan/25
$$``\mathrm{if}\:\mathrm{and}\:\mathrm{only}\:\mathrm{if}''. \\ $$
Answered by Eze120 last updated on 23/Jan/25
$${m}={n} \\ $$$$\left({m}+{m}^{−\mathrm{1}} \right)\left({m}+{m}^{−\mathrm{1}} \right)={m}^{\mathrm{2}} +\left({m}^{−\mathrm{1}} \right)^{\mathrm{2}} \\ $$$$\mathrm{2}{m}=\sqrt{{a}} \\ $$$$\mathrm{2}{m}^{\mathrm{2}} ={a} \\ $$$${m}^{\mathrm{2}} =\frac{{a}}{\mathrm{2}} \\ $$$$\left({m}^{−\mathrm{1}} \right)^{\mathrm{2}} =\frac{\mathrm{1}}{{m}^{\mathrm{2}} }=\frac{\mathrm{1}}{{a}/\mathrm{2}} \\ $$$$\frac{{a}}{\mathrm{2}}+\frac{\mathrm{1}}{{a}/\mathrm{2}}=\left(\mathrm{1}/\mathrm{2}\right){a}+\frac{\mathrm{2}}{{a}} \\ $$$${PD}:{It}'{s}\:{a}\:{guess},\:{it}'{s}\:{not}\:{accurate}. \\ $$