Menu Close

Category: Algebra

Prove-that-0-i-lt-j-n-1-n-C-i-1-n-C-j-r-0-n-1-n-r-n-C-r-r-1-n-r-n-C-r-

Question Number 23471 by Tinkutara last updated on 31/Oct/17 Provethat$$\underset{\mathrm{0}\leqslant{i}<{j}\leqslant{n}} {\Sigma\Sigma}\left(\frac{\mathrm{1}}{\:^{{n}} {C}_{{i}} }\:+\:\frac{\mathrm{1}}{\:^{{n}} {C}_{{j}} }\right)\:=\:\underset{{r}=\mathrm{0}} {\overset{{n}−\mathrm{1}} {\sum}}\frac{{n}\:−\:{r}}{\:^{{n}} {C}_{{r}} }\:+\:\underset{{r}=\mathrm{1}} {\overset{{n}} {\sum}}\frac{{r}}{\:^{{n}} {C}_{{r}} }…

If-a-b-c-gt-0-and-n-N-then-a-2n-b-2n-c-2n-a-n-b-n-b-n-c-n-c-n-a-n-3-a-2-b-2-c-2-a-b-c-

Question Number 154493 by mathdanisur last updated on 18/Sep/21 Ifa;b;c>0andnN+then:$$\frac{\mathrm{a}^{\mathrm{2}\boldsymbol{\mathrm{n}}} \:+\:\mathrm{b}^{\mathrm{2}\boldsymbol{\mathrm{n}}} \:+\:\mathrm{c}^{\mathrm{2}\boldsymbol{\mathrm{n}}} }{\mathrm{a}^{\boldsymbol{\mathrm{n}}} \mathrm{b}^{\boldsymbol{\mathrm{n}}} \:+\:\mathrm{b}^{\boldsymbol{\mathrm{n}}} \mathrm{c}^{\boldsymbol{\mathrm{n}}} \:+\:\mathrm{c}^{\boldsymbol{\mathrm{n}}} \mathrm{a}^{\boldsymbol{\mathrm{n}}} }\:\geqslant\:\frac{\sqrt{\mathrm{3}\centerdot\left(\mathrm{a}^{\mathrm{2}} \:+\:\mathrm{b}^{\mathrm{2}} \:+\:\mathrm{c}^{\mathrm{2}} \right)}}{\mathrm{a}\:+\:\mathrm{b}\:+\:\mathrm{c}}…