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Category: Algebra

2-3-2-3-2-

Question Number 145224 by mathdanisur last updated on 03/Jul/21 $$\sqrt{\mathrm{2}\sqrt{\mathrm{3}}\:+\:\mathrm{2}}\:-\:\sqrt{\sqrt{\mathrm{3}}\:-\:\sqrt{\mathrm{2}}}\:=\:? \\ $$ Answered by Olaf_Thorendsen last updated on 03/Jul/21 $$\left(\sqrt{\mathrm{3}}+\sqrt{\mathrm{2}}+\mathrm{1}\right)^{\mathrm{2}} \:=\:\mathrm{6}+\mathrm{2}\sqrt{\mathrm{6}}+\mathrm{2}\sqrt{\mathrm{3}}+\mathrm{2}\sqrt{\mathrm{2}} \\ $$$$ \\ $$$${x}\:=\:\sqrt{\mathrm{2}\sqrt{\mathrm{3}}+\mathrm{2}}−\sqrt{\sqrt{\mathrm{3}}−\sqrt{\mathrm{2}}}…

Let-a-b-c-0-c-3-a-b-3-0-and-a-2-b-2-c-2-3-Prove-that-1-2-a-3-b-c-3-b-3-c-a-3-c-3-a-b-3-2-Determine-when-equality-holds-

Question Number 145198 by loveineq last updated on 03/Jul/21 $$\mathrm{Let}\:{a}\geqslant{b}\geqslant{c}\geqslant\mathrm{0}\:,\:{c}^{\mathrm{3}} +\left({a}+{b}\right)^{\mathrm{3}} \neq\mathrm{0}\:\mathrm{and}\:{a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} \:=\:\mathrm{3}.\:\mathrm{Prove}\:\mathrm{that} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\frac{\mathrm{1}}{\mathrm{2}}\:\leqslant\:\frac{{a}^{\mathrm{3}} +\left({b}+{c}\right)^{\mathrm{3}} +{b}^{\mathrm{3}} +\left({c}+{a}\right)^{\mathrm{3}} }{{c}^{\mathrm{3}} +\left({a}+{b}\right)^{\mathrm{3}} }\:\leqslant\:\mathrm{2} \\ $$$$\mathrm{Determine}\:\mathrm{when}\:\mathrm{equality}\:\mathrm{holds}.…

if-q-1-and-x-gt-1-then-1-x-q-1-x-q-1-x-1-qx-

Question Number 145191 by mathdanisur last updated on 03/Jul/21 $${if}\:\:{q}\geqslant\mathrm{1}\:\:{and}\:\:{x}>−\mathrm{1}\:\:{then}: \\ $$$$\left(\mathrm{1}+{x}\right)^{\boldsymbol{{q}}} \:\geqslant\:\left(\mathrm{1}+{x}\right)^{\boldsymbol{{q}}−\mathrm{1}} \:+\:{x}\:\geqslant\:\mathrm{1}+{qx} \\ $$ Answered by Olaf_Thorendsen last updated on 03/Jul/21 $$\left(\mathrm{1}+{x}\right)^{{q}} \:=\:\left(\mathrm{1}+{x}\right)^{{q}−\mathrm{1}}…