Question Number 144823 by loveineq last updated on 29/Jun/21 $$\mathrm{Let}\:{a},{b}\:>\:\mathrm{0}\:\mathrm{and}\:{a}+{b}+\mathrm{1}\:=\:\mathrm{3}{ab}.\:\mathrm{Prove}\:\mathrm{that} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\frac{{a}+\mathrm{1}}{{b}+\mathrm{1}}+\frac{{b}+\mathrm{1}}{{a}+\mathrm{1}}\:\leqslant\:{a}+{b} \\ $$ Answered by ArielVyny last updated on 30/Jun/21 $${we}\:{know}\:{that}\:{a}+{b}+\mathrm{1}=\mathrm{3}{ab} \\ $$$${we}\:{suppose}\:{ab}\geqslant\mathrm{1} \\…
Question Number 79279 by jagoll last updated on 24/Jan/20 $$\mathrm{solve}\: \\ $$$$\mid\mathrm{x}\mid^{\mathrm{3}} −\mathrm{7x}^{\mathrm{2}} +\mathrm{7}\mid\mathrm{x}\mid+\mathrm{15}<\mathrm{0} \\ $$ Commented by jagoll last updated on 24/Jan/20 $$\mathrm{thanks}\: \\…
Question Number 13731 by prakash jain last updated on 22/May/17 $$\mathrm{Prove}\:\mathrm{that}\:\mathrm{if}\:{p}>{q}>\mathrm{0}\:\mathrm{and}\:{x}\geqslant\mathrm{0} \\ $$$$\frac{\mathrm{1}}{{p}}\left(\frac{{x}^{{p}} }{{p}+\mathrm{1}}−\mathrm{1}\right)\geqslant\frac{\mathrm{1}}{{q}}\left(\frac{{x}^{{q}} }{{q}+\mathrm{1}}−\mathrm{1}\right).\: \\ $$ Commented by mrW1 last updated on 25/May/17 $${function}\:{f}\left({x}\right)=\frac{\mathrm{1}}{{x}}\left(\frac{{a}^{{x}}…
Question Number 79256 by ajfour last updated on 23/Jan/20 $$\mathrm{3}{s}^{\mathrm{2}} −\mathrm{2}{ps}−\mathrm{3}{cp}−\mathrm{1}=\mathrm{0}\:\:\:{and} \\ $$$$\mathrm{3}{s}−\mathrm{2}{p}−{sp}^{\mathrm{2}} −\mathrm{3}{cp}^{\mathrm{2}} =\mathrm{0} \\ $$$${find}\:{s}\:{and}\:{p}\:{both}\:{real}\:{in}\:{terms} \\ $$$${of}\:{c}\:\in\mathbb{R}. \\ $$ Answered by MJS last…
Question Number 79249 by Pratah last updated on 23/Jan/20 Commented by mr W last updated on 23/Jan/20 $${can}\:{you}\:{tell}\:{according}\:{to}\:{the}\:{definition} \\ $$$${in}\:{your}\:{question} \\ $$$$\left[\mathrm{0}.\mathrm{8}\right]=\mathrm{0} \\ $$$$\left[−\mathrm{0}.\mathrm{8}\right]=?\:\: \\…
Question Number 144768 by ajfour last updated on 04/Jul/21 $$\:\:{x}^{\mathrm{3}} −{x}−{c}=\mathrm{0} \\ $$$${let}\:\:{x}=\sqrt{{t}+{p}}+\sqrt{{t}+{q}} \\ $$$$\left({t}+{p}\right)\sqrt{{t}+{p}}+\left({t}+{q}\right)\sqrt{{t}+{q}} \\ $$$$+\mathrm{3}\left({t}+{p}\right)\sqrt{{t}+{q}}+\mathrm{3}\left({t}+{q}\right)\sqrt{{t}+{p}} \\ $$$$−\sqrt{{t}+{p}}−\sqrt{{t}+{q}}−{c}=\mathrm{0} \\ $$$${let}\:\:\left({t}+{q}\right)+\mathrm{3}\left({t}+{p}\right)−\mathrm{1}=\mathrm{0} \\ $$$$\Rightarrow\:\mathrm{4}{t}=\mathrm{1}−\left({p}+{q}\right) \\ $$$$\Rightarrow\:\:\mathrm{4}{t}+\mathrm{4}{p}=\mathrm{3}{p}−{q}+\mathrm{1}…
Question Number 79233 by jagoll last updated on 23/Jan/20 $$\frac{\mathrm{1}}{\mathrm{x}\left(\mathrm{x}+\mathrm{1}\right)}+\frac{\mathrm{1}}{\left(\mathrm{x}+\mathrm{1}\right)\left(\mathrm{x}+\mathrm{2}\right)}+ \\ $$$$\frac{\mathrm{1}}{\left(\mathrm{x}+\mathrm{2}\right)\left(\mathrm{x}+\mathrm{3}\right)}\leqslant\frac{\mathrm{3}}{\mathrm{4}} \\ $$ Commented by john santu last updated on 23/Jan/20 $$\frac{\mathrm{1}}{{x}}−\frac{\mathrm{1}}{{x}+\mathrm{1}}+\frac{\mathrm{1}}{{x}+\mathrm{1}}−\frac{\mathrm{1}}{{x}+\mathrm{2}} \\ $$$$+\frac{\mathrm{1}}{{x}+\mathrm{2}}−\frac{\mathrm{1}}{{x}+\mathrm{3}}\leqslant\frac{\mathrm{3}}{\mathrm{4}}…
Question Number 144760 by mathdanisur last updated on 28/Jun/21 $$\sqrt[{\mathrm{3}}]{\mathrm{1}+\sqrt{{x}}}\:+\:\sqrt[{\mathrm{3}}]{\mathrm{1}-\sqrt{{x}}}\:=\:\sqrt[{\mathrm{3}}]{\mathrm{5}} \\ $$$${Find}\:\:\boldsymbol{{x}}=? \\ $$ Answered by Olaf_Thorendsen last updated on 28/Jun/21 $$\mathrm{Let}\:\alpha\:=\:\sqrt[{\mathrm{3}}]{\mathrm{1}+\sqrt{{x}}}\:\mathrm{and}\:\beta\:=\:\sqrt[{\mathrm{3}}]{\mathrm{1}−\sqrt{{x}}} \\ $$$$\alpha+\beta\:\:=\:\sqrt[{\mathrm{3}}]{\mathrm{5}} \\…
Question Number 144767 by mnjuly1970 last updated on 29/Jun/21 Answered by Rasheed.Sindhi last updated on 29/Jun/21 $$\left({x}^{\mathrm{3}} −\frac{\mathrm{2}}{{x}^{\mathrm{2}} }\right)^{{m}} ={a}_{\mathrm{0}} +{a}_{\mathrm{1}} {x}+{a}_{\mathrm{2}} {x}^{\mathrm{2}} +…+{a}_{{n}} {x}^{{n}}…
Question Number 144742 by loveineq last updated on 28/Jun/21 $$\mathrm{Let}\:{a},{b},{c}>\mathrm{0}\:\mathrm{and}\:{a}+{b}+{c}\:=\:\mathrm{3}.\:\mathrm{Prove}\:\mathrm{that}\:\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\frac{\frac{{ab}}{{ab}+\mathrm{1}}+\frac{{bc}}{{bc}+\mathrm{1}}+\frac{{ca}}{{ca}+\mathrm{1}}}{\frac{\mathrm{1}}{{ab}+\mathrm{1}}+\frac{\mathrm{1}}{{bc}+\mathrm{1}}+\frac{\mathrm{1}}{{ca}+\mathrm{1}}}\:\geqslant\:{abc} \\ $$$$\left(\mathrm{Found}\:\mathrm{by}\:\mathrm{WolframAlpha}\:\mathrm{and}\:\right. \\ $$$$\left.\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{inspired}\:\mathrm{by}\:\mathrm{my}\:\mathrm{old}\:\mathrm{problem}\right) \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com