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Category: Algebra

Let-a-b-c-0-and-1-a-1-b-1-c-8-Prove-that-a-2b-1-a-b-1-b-2c-1-b-c-1-c-2a-1-c-a-1-8-

Question Number 143229 by loveineq last updated on 11/Jun/21 $$\mathrm{Let}\:{a},{b},{c}\:\geqslant\:\mathrm{0}\:\mathrm{and}\:\left(\mathrm{1}+{a}\right)\left(\mathrm{1}+{b}\right)\left(\mathrm{1}+{c}\right)\:=\:\mathrm{8}. \\ $$$$\mathrm{Prove}\:\mathrm{that} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left({a}+\frac{\mathrm{2}{b}+\mathrm{1}}{{a}+{b}+\mathrm{1}}\right)\left({b}+\frac{\mathrm{2}{c}+\mathrm{1}}{{b}+{c}+\mathrm{1}}\right)\left({c}+\frac{\mathrm{2}{a}+\mathrm{1}}{{c}+{a}+\mathrm{1}}\right)\:\geqslant\:\mathrm{8}\:\:\:\:\:\:\:\:\:\:\:\: \\ $$$$ \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com

1-1-1-2-1-4-2-1-2-2-2-4-3-1-3-2-3-4-

Question Number 12144 by rish@bh last updated on 14/Apr/17 $$\frac{\mathrm{1}}{\mathrm{1}+\mathrm{1}^{\mathrm{2}} +\mathrm{1}^{\mathrm{4}} }+\frac{\mathrm{2}}{\mathrm{1}+\mathrm{2}^{\mathrm{2}} +\mathrm{2}^{\mathrm{4}} }+\frac{\mathrm{3}}{\mathrm{1}+\mathrm{3}^{\mathrm{2}} +\mathrm{3}^{\mathrm{4}} }+…..\infty=? \\ $$ Answered by b.e.h.i.8.3.4.1.7@gmail.com last updated on 14/Apr/17…

Question-12126

Question Number 12126 by tawa last updated on 13/Apr/17 Answered by b.e.h.i.8.3.4.1.7@gmail.com last updated on 16/Apr/17 $${a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} +\mathrm{2}{ab}+\mathrm{2}{ac}+\mathrm{2}{bc}=\mathrm{0} \\ $$$$\Rightarrow{ab}+{bc}+{ca}=−\mathrm{21} \\ $$$${A}=\left({a}−{b}\right)\left({b}−{c}\right)\left({c}−{a}\right) \\…

10-25-28-125-82-625-

Question Number 143184 by bramlexs22 last updated on 11/Jun/21 $$\:\:\:\:\frac{\mathrm{10}}{\mathrm{25}}+\frac{\mathrm{28}}{\mathrm{125}}+\frac{\mathrm{82}}{\mathrm{625}}+…\:=\:? \\ $$ Answered by Canebulok last updated on 11/Jun/21 $$\boldsymbol{{Solution}}: \\ $$$${In}\:{terms}\:{of}\:{summation}, \\ $$$$\Rightarrow\:\underset{{n}=\mathrm{1}} {\overset{\infty}…

Question-12080

Question Number 12080 by sin (x) last updated on 12/Apr/17 Answered by ajfour last updated on 12/Apr/17 $${f}\left(−\mathrm{1},\mathrm{3}\right)=\mathrm{max}\left(−\mathrm{4},−\frac{\mathrm{1}}{\mathrm{3}}\right)=\frac{−\mathrm{1}}{\mathrm{3}} \\ $$$${g}\left(\mathrm{4},\mathrm{4}\right)={min}\left(\mathrm{8},\:\mathrm{16}\right)\:=\mathrm{8} \\ $$$${f}\left[{f}\left(−\mathrm{1},\mathrm{3}\right),\:{g}\left(\mathrm{4},\mathrm{4}\right)\right] \\ $$$$\:\:\:=\:{f}\left(−\frac{\mathrm{1}}{\mathrm{3}},\:\mathrm{8}\right)={max}\left(−\frac{\mathrm{25}}{\mathrm{3}},\:−\frac{\mathrm{1}}{\mathrm{24}}\right) \\…