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Category: Algebra

Question-227964

Question Number 227964 by mr W last updated on 05/Mar/26 Answered by Lara2440 last updated on 07/Mar/26 $$\mathrm{Let}'\mathrm{s}\:\mathrm{define}\:\mathrm{Sequence}\:{A}_{{n}} \:\mathrm{as}\:\frac{\left({n}!\right)^{\frac{\mathrm{1}}{{n}}} }{{n}}\:,\:{n}=\mathrm{1},\mathrm{2},\mathrm{3}….. \\ $$$$\mathrm{for}\:\mathrm{all}\:{n}>\mathrm{1}\:\mathrm{and}\:\mathrm{for}\:\mathrm{all}\:{x}>\mathrm{0} \\ $$$$\mathrm{1}+{nx}<\left({x}+\mathrm{1}\right)^{{n}} \:\centerdot\centerdot\centerdot\centerdot\centerdot\left(\mathrm{1}\right)\:\:\left(\mathrm{Bernoulli}\:\mathrm{Inequality}\right)…

In-ABC-holds-1-6r-h-a-h-b-h-b-h-c-h-c-h-a-1-3-3R-

Question Number 227954 by hardmath last updated on 04/Mar/26 $$\mathrm{In}\:\:\bigtriangleup\mathrm{ABC}\:\:\mathrm{holds}: \\ $$$$\left.\mathrm{1}\right)\:\mathrm{6r}\:\leqslant\:\sqrt[{\mathrm{3}}]{\left(\mathrm{h}_{\boldsymbol{\mathrm{a}}} \:+\:\mathrm{h}_{\boldsymbol{\mathrm{b}}} \right)\centerdot\left(\mathrm{h}_{\boldsymbol{\mathrm{b}}} \:+\:\mathrm{h}_{\boldsymbol{\mathrm{c}}} \right)\centerdot\left(\mathrm{h}_{\boldsymbol{\mathrm{c}}} +\:\mathrm{h}_{\boldsymbol{\mathrm{a}}} \right)}\:\leqslant\:\mathrm{3R} \\ $$ Terms of Service Privacy Policy…