Question Number 193546 by York12 last updated on 16/Jun/23
$${if}\:{a}+{b}+{c}=\mathrm{1} \\ $$$${find}\:{maximum}\:{ab}\:+{bc}\:+{ca}\: \\ $$$${a}\:,\:{b}\:,\:{c}\:{are}\:{non}\:{negative}\:{integers} \\ $$$$ \\ $$
Answered by Frix last updated on 16/Jun/23
$${a}={b}=\mathrm{0}\wedge{c}=\mathrm{1}\wedge{d}=+\infty\:\Rightarrow\:\mathrm{max}\:\mathrm{is}\:+\infty \\ $$
Commented by York12 last updated on 16/Jun/23
$${yeah}\:{sir}\:{you}\:{are}\:{right} \\ $$$${but}\:{I}\:{just}\:{forgot}\:{to}\:{write}\:{that}\:{condition} \\ $$$$ \\ $$
Answered by Subhi last updated on 16/Jun/23
$${Another}\:{solution} \\ $$$$\left({a}+{b}+{c}\right)^{\mathrm{2}} ={a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} +\mathrm{2}\left({ab}+{bc}+{ac}\right) \\ $$$${a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} \geqslant{ab}+{bc}+{ac} \\ $$$$\left({a}+{b}+{c}\right)^{\mathrm{2}} \geqslant\mathrm{3}\left({ab}+{bc}+{ac}\right) \\ $$$${ab}+{bc}+{ac}\leqslant\frac{\left({a}+{b}+{c}\right)^{\mathrm{2}} }{\mathrm{3}}=\frac{\mathrm{1}}{\mathrm{3}} \\ $$
Commented by York12 last updated on 16/Jun/23
$${thnx} \\ $$
Commented by York12 last updated on 17/Jun/23
$$\sqrt{\frac{{x}_{\mathrm{1}} ^{\mathrm{2}} +{x}_{\mathrm{2}} ^{\mathrm{2}} +{x}_{\mathrm{3}} ^{\mathrm{2}} +{x}_{\mathrm{4}} ^{\mathrm{2}} +…+{x}_{{n}} ^{\mathrm{2}} }{{n}}}\geqslant\frac{{x}_{\mathrm{1}} +{x}_{\mathrm{2}} +{x}_{\mathrm{3}} +{x}_{\mathrm{4}} +…+{x}_{{n}} }{{n}} \\ $$$$\Rightarrow\left({x}_{\mathrm{1}} ^{\mathrm{2}} +{x}_{\mathrm{2}} ^{\mathrm{2}} +{x}_{\mathrm{3}} ^{\mathrm{2}} +{x}_{\mathrm{4}} ^{\mathrm{2}} +…+{x}_{{n}} ^{\mathrm{2}} \right)\geqslant\frac{\left({x}_{\mathrm{1}} +{x}_{\mathrm{2}} +{x}_{\mathrm{3}} +{x}_{\mathrm{4}} +…+{x}_{{n}} \right)^{\mathrm{2}} }{{n}} \\ $$$$\Rightarrow{n}\left({x}_{\mathrm{1}} ^{\mathrm{2}} +{x}_{\mathrm{2}} ^{\mathrm{2}} +{x}_{\mathrm{3}} ^{\mathrm{2}} +{x}_{\mathrm{4}} ^{\mathrm{2}} +…+{x}_{{n}} ^{\mathrm{2}} \right)\geqslant\left({x}_{\mathrm{1}} +{x}_{\mathrm{2}} +{x}_{\mathrm{3}} +{x}_{\mathrm{4}} +…+{x}_{{n}} \right)^{\mathrm{2}} \\ $$$$\Rightarrow\left({n}−\mathrm{1}\right)\left({x}_{\mathrm{1}} ^{\mathrm{2}} +{x}_{\mathrm{2}} ^{\mathrm{2}} +{x}_{\mathrm{3}} ^{\mathrm{2}} +{x}_{\mathrm{4}} ^{\mathrm{2}} +…+{x}_{{n}} ^{\mathrm{2}} \right)\geqslant\mathrm{2}\underset{\mathrm{1}\leqslant{i}\:\:\:<} {\sum}\underset{{j}\:\:\leqslant{n}} {\sum}{x}_{{i}} {x}_{{j}} \\ $$$$\Rightarrow\frac{\left({n}−\mathrm{1}\right)}{\mathrm{2}}\left({x}_{\mathrm{1}} ^{\mathrm{2}} +{x}_{\mathrm{2}} ^{\mathrm{2}} +{x}_{\mathrm{3}} ^{\mathrm{2}} +{x}_{\mathrm{4}} ^{\mathrm{2}} +…+{x}_{{n}} ^{\mathrm{2}} \right)\geqslant\underset{\mathrm{1}\leqslant{i}\:\:\:<} {\sum}\underset{{j}\:\:\leqslant{n}} {\sum}{x}_{{i}} {x}_{{j}} \\ $$