Question Number 221262 by Nicholas666 last updated on 28/May/25 $$ \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\mathrm{if}\:{a},\:{b},\:{c},\:>\:\mathrm{0}\:\:,\:\mathrm{show}\:\mathrm{that}; \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\frac{{a}^{\mathrm{5}} }{{b}^{\mathrm{2}} }\:+\:\frac{{b}}{{c}}\:+\:\frac{{c}^{\mathrm{3}} }{{a}^{\mathrm{2}} }\:>\:\mathrm{2}{a} \\ $$$$ \\ $$ Commented by mr…
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Question Number 221195 by Rojarani last updated on 26/May/25 Commented by Ghisom last updated on 26/May/25 $$\mathrm{I}\:\mathrm{get}\:\frac{\mathrm{49}}{\mathrm{34}} \\ $$ Answered by mr W last updated…
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Question Number 221185 by hardmath last updated on 26/May/25 Answered by Spillover last updated on 27/May/25 Answered by Spillover last updated on 27/May/25 Answered by…
Question Number 221151 by gregori last updated on 25/May/25 $$\:\frac{\sqrt{{x}^{\mathrm{2}} −{x}−\sqrt{{x}^{\mathrm{2}} −{x}−\sqrt{{x}^{\mathrm{2}} −{x}−\sqrt{…}}}}}{\:\sqrt[{\mathrm{3}}]{{x}^{\mathrm{2}} \:\sqrt{{x}\:\sqrt[{\mathrm{3}}]{{x}^{\mathrm{2}} \:\sqrt{{x}…}}}}}\:=\:\frac{\mathrm{3}}{\mathrm{4}} \\ $$$$\:\Rightarrow\:\frac{\mathrm{2}}{{x}}\:=?\: \\ $$ Commented by Frix last updated on…
Question Number 221168 by universe last updated on 25/May/25 Answered by Frix last updated on 26/May/25 $$\mathrm{Let}\:{b}={pa}\wedge{p}<\mathrm{0} \\ $$$$\lambda=\mathrm{min}\:\left(\frac{{p}−\mathrm{1}}{{ap}}\sqrt{\mathrm{2}{a}^{\mathrm{4}} {p}^{\mathrm{2}} +\mathrm{2}{a}^{\mathrm{2}} {p}+\mathrm{1}}\right) \\ $$$$\mathrm{Using}\:\mathrm{partial}\:\mathrm{differenciation}\:\mathrm{we}\:\mathrm{get} \\…
Question Number 221135 by MathematicalUser2357 last updated on 25/May/25 $$\mathrm{South}\:\mathrm{Korean}\:\mathrm{Grade}\:\mathrm{12}\:\mathrm{math} \\ $$$$\mathrm{Prove}\:\mathrm{log}_{{a}} {M}^{{n}} ={n}\mathrm{log}_{{a}} {M} \\ $$$$\mathrm{Using}\:\mathrm{below}: \\ $$$$\mathrm{When}\:{M}={a}^{{x}} ,\:\mathrm{log}_{{a}} {M}={x} \\ $$$$\mathrm{When}\:{N}={a}^{{y}} ,\:\mathrm{log}_{{a}} {N}={y}…
Question Number 221017 by Rojarani last updated on 22/May/25 Answered by Ghisom last updated on 23/May/25 $$\mathrm{min}\:\left(\frac{{a}^{\mathrm{2}} }{{b}}+\frac{{b}^{\mathrm{2}} }{{c}}+\frac{{c}^{\mathrm{2}} }{{a}}\right)\:=\mathrm{1}\:\mathrm{when}\:{a}={b}={c}=\frac{\mathrm{1}}{\mathrm{3}} \\ $$$$\mathrm{min}\:\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} \right)\:=\frac{\mathrm{1}}{\mathrm{3}}\:\mathrm{when}\:{a}={b}={c}=\frac{\mathrm{1}}{\mathrm{3}}…
Question Number 220987 by fantastic last updated on 21/May/25 $${Let}\:{a},{b},{c}\:{be}\:{positive}\:{reals}\:{such}\:{that}\:{abc}=\mathrm{1}.{prove}\:{that} \\ $$$$\frac{\mathrm{1}}{{a}^{\mathrm{3}} \left({b}+{c}\right)}+\frac{\mathrm{1}}{{b}^{\mathrm{3}} \left({c}+{a}\right)}+\frac{\mathrm{1}}{{c}^{\mathrm{3}} \left({a}+{b}\right)}\geqslant\frac{\mathrm{3}}{\mathrm{2}} \\ $$ Answered by mr W last updated on 21/May/25…