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Question Number 199847 by cortano12 last updated on 10/Nov/23

Find all possible value (a/(a+b+d )) +(b/(a+b+c)) + (c/(b+c+d))+(d/(a+c+d)) when a,b,c,d vary over positive reals

$$\:\mathrm{Find}\:\mathrm{all}\:\mathrm{possible}\:\mathrm{value}\: \\ $$$$\:\frac{\mathrm{a}}{\mathrm{a}+\mathrm{b}+\mathrm{d}\:}\:+\frac{\mathrm{b}}{\mathrm{a}+\mathrm{b}+\mathrm{c}}\:+\:\frac{\mathrm{c}}{\mathrm{b}+\mathrm{c}+\mathrm{d}}+\frac{\mathrm{d}}{\mathrm{a}+\mathrm{c}+\mathrm{d}}\: \\ $$$$\:\mathrm{when}\:\mathrm{a},\mathrm{b},\mathrm{c},\mathrm{d}\:\mathrm{vary}\:\mathrm{over}\:\mathrm{positive} \\ $$$$\:\mathrm{reals}\: \\ $$

Commented by Frix last updated on 10/Nov/23

No problem! I′m also trying... I think 1≤f is true but f<2?

$$\mathrm{No}\:\mathrm{problem}!\:\mathrm{I}'\mathrm{m}\:\mathrm{also}\:\mathrm{trying}... \\ $$$$\mathrm{I}\:\mathrm{think}\:\mathrm{1}\leqslant{f}\:\mathrm{is}\:\mathrm{true}\:\mathrm{but}\:{f}<\mathrm{2}? \\ $$

Commented by witcher3 last updated on 10/Nov/23

i didnt want delate Y re coment

$$\mathrm{i}\:\mathrm{didnt}\:\mathrm{want}\:\mathrm{delate}\:\mathrm{Y}\:\mathrm{re}\:\mathrm{coment} \\ $$

Commented by witcher3 last updated on 10/Nov/23

sorry frix my solution is false

$$\mathrm{sorry}\:\mathrm{frix}\:\mathrm{my}\:\mathrm{solution}\:\mathrm{is}\:\mathrm{false} \\ $$

Commented by witcher3 last updated on 12/Nov/23

f<2 i?think also

$$\mathrm{f}<\mathrm{2}\:\mathrm{i}?\mathrm{think}\:\mathrm{also} \\ $$

Commented by witcher3 last updated on 12/Nov/23

if a+b+c>d or evry sum of three ≥forthe we get f≤(4/3)

$$\mathrm{if}\:\mathrm{a}+\mathrm{b}+\mathrm{c}>\mathrm{d}\:\mathrm{or}\:\mathrm{evry}\:\mathrm{sum}\:\mathrm{of}\:\mathrm{three}\:\geqslant\mathrm{forthe} \\ $$$$\mathrm{we}\:\mathrm{get}\:\mathrm{f}\leqslant\frac{\mathrm{4}}{\mathrm{3}} \\ $$$$ \\ $$

Commented by Frix last updated on 12/Nov/23

f(a, b, c, d)=(a/(a+b+0c+d))+(b/(a+b+c+0d))+(c/(0a+b+c+d))+(d/(a+0b+c+d)) f(x, x, y, y)=((2(x^2 +4xy+y^2 ))/((x+2y)(2x+y))) ⇒ 1≤f≤(4/3) f(x, y, x, y)=((4(x^2 +xy+y^2 ))/((x+2y)(2x+y))) ⇒ (4/3)≤f≤2 This is still no proof...

$${f}\left({a},\:{b},\:{c},\:{d}\right)=\frac{{a}}{{a}+{b}+\mathrm{0}{c}+{d}}+\frac{{b}}{{a}+{b}+{c}+\mathrm{0}{d}}+\frac{{c}}{\mathrm{0}{a}+{b}+{c}+{d}}+\frac{{d}}{{a}+\mathrm{0}{b}+{c}+{d}} \\ $$$${f}\left({x},\:{x},\:{y},\:{y}\right)=\frac{\mathrm{2}\left({x}^{\mathrm{2}} +\mathrm{4}{xy}+{y}^{\mathrm{2}} \right)}{\left({x}+\mathrm{2}{y}\right)\left(\mathrm{2}{x}+{y}\right)}\:\Rightarrow\:\mathrm{1}\leqslant{f}\leqslant\frac{\mathrm{4}}{\mathrm{3}} \\ $$$${f}\left({x},\:{y},\:{x},\:{y}\right)=\frac{\mathrm{4}\left({x}^{\mathrm{2}} +{xy}+{y}^{\mathrm{2}} \right)}{\left({x}+\mathrm{2}{y}\right)\left(\mathrm{2}{x}+{y}\right)}\:\Rightarrow\:\frac{\mathrm{4}}{\mathrm{3}}\leqslant{f}\leqslant\mathrm{2} \\ $$$$\mathrm{This}\:\mathrm{is}\:\mathrm{still}\:\mathrm{no}\:\mathrm{proof}... \\ $$

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