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Question-225599

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Question Number 225599 by Jubr last updated on 04/Nov/25
Commented by Frix last updated on 04/Nov/25
If a, b, c, d ∈R no maximum exists. Let a=b=−r; c=1; d=2r (1−r)^2 (1+2r)=1−3r^2 +2r^3 lim_(r→+∞) (1−3r^2 +2r^3 ) =+∞ If a, b, c, d >0 we′ve got a 4D “cube” which has the max “volume” at a=b=c=d=(1/4) ⇒ maximum is ((5/4))^4
$$\mathrm{If}\:{a},\:{b},\:{c},\:{d}\:\in\mathbb{R}\:\mathrm{no}\:\mathrm{maximum}\:\mathrm{exists}. \\ $$$$\mathrm{Let}\:{a}={b}=−{r};\:{c}=\mathrm{1};\:{d}=\mathrm{2}{r} \\ $$$$\left(\mathrm{1}−{r}\right)^{\mathrm{2}} \left(\mathrm{1}+\mathrm{2}{r}\right)=\mathrm{1}−\mathrm{3}{r}^{\mathrm{2}} +\mathrm{2}{r}^{\mathrm{3}} \\ $$$$\underset{{r}\rightarrow+\infty} {\mathrm{lim}}\left(\mathrm{1}−\mathrm{3}{r}^{\mathrm{2}} +\mathrm{2}{r}^{\mathrm{3}} \right)\:=+\infty \\ $$$$\mathrm{If}\:{a},\:{b},\:{c},\:{d}\:>\mathrm{0}\:\mathrm{we}'\mathrm{ve}\:\mathrm{got}\:\mathrm{a}\:\mathrm{4D}\:“\mathrm{cube}''\:\mathrm{which} \\ $$$$\mathrm{has}\:\mathrm{the}\:\mathrm{max}\:“\mathrm{volume}''\:\mathrm{at}\:{a}={b}={c}={d}=\frac{\mathrm{1}}{\mathrm{4}} \\ $$$$\Rightarrow\:\mathrm{maximum}\:\mathrm{is}\:\left(\frac{\mathrm{5}}{\mathrm{4}}\right)^{\mathrm{4}} \\ $$
Commented by Jubr last updated on 04/Nov/25
I appreciate sir
$${I}\:{appreciate}\:{sir} \\ $$
Answered by mr W last updated on 04/Nov/25
assume a,b,c,d>−1 let A=a+1, B=b+1, C=c+1, D=d+1 >0 a+b+c+d=1 (a+1)+(b+1)+(c+1)+(d+1)=5 A+B+C+D=5 (1+a)(1+b)(1+c)(1+d) =ABCD =(((ABCD))^(1/4) )^4 ≤(((A+B+C+D)/4))^4 =((5/4))^4 i.e. maximum is ((5/4))^4 =((625)/(256)).
$${assume}\:{a},{b},{c},{d}>−\mathrm{1} \\ $$$${let}\:{A}={a}+\mathrm{1},\:{B}={b}+\mathrm{1},\:{C}={c}+\mathrm{1},\:{D}={d}+\mathrm{1}\:>\mathrm{0} \\ $$$${a}+{b}+{c}+{d}=\mathrm{1} \\ $$$$\left({a}+\mathrm{1}\right)+\left({b}+\mathrm{1}\right)+\left({c}+\mathrm{1}\right)+\left({d}+\mathrm{1}\right)=\mathrm{5} \\ $$$${A}+{B}+{C}+{D}=\mathrm{5} \\ $$$$\left(\mathrm{1}+{a}\right)\left(\mathrm{1}+{b}\right)\left(\mathrm{1}+{c}\right)\left(\mathrm{1}+{d}\right) \\ $$$$={ABCD} \\ $$$$=\left(\sqrt[{\mathrm{4}}]{{ABCD}}\right)^{\mathrm{4}} \leqslant\left(\frac{{A}+{B}+{C}+{D}}{\mathrm{4}}\right)^{\mathrm{4}} =\left(\frac{\mathrm{5}}{\mathrm{4}}\right)^{\mathrm{4}} \\ $$$${i}.{e}.\:{maximum}\:{is}\:\left(\frac{\mathrm{5}}{\mathrm{4}}\right)^{\mathrm{4}} =\frac{\mathrm{625}}{\mathrm{256}}. \\ $$
Commented by mr W last updated on 04/Nov/25
you could also put the question as: find the maximum from (1+a)(2+b)(3+c)(4+d). then (a+1)+(b+2)+(c+3)+(d+4)=11 A+B+C+D=11 (a+1)(b+2)(c+3)(d+4) =ABCD =(((ABCD))^(1/4) )^4 ≤(((A+B+C+D)/4))^4 =(((11)/4))^4 i.e. maximum is (((11)/4))^4 .
$${you}\:{could}\:{also}\:{put}\:{the}\:{question}\:{as}: \\ $$$${find}\:{the}\:{maximum}\:{from} \\ $$$$\left(\mathrm{1}+{a}\right)\left(\mathrm{2}+{b}\right)\left(\mathrm{3}+{c}\right)\left(\mathrm{4}+{d}\right). \\ $$$${then} \\ $$$$\left({a}+\mathrm{1}\right)+\left({b}+\mathrm{2}\right)+\left({c}+\mathrm{3}\right)+\left({d}+\mathrm{4}\right)=\mathrm{11} \\ $$$${A}+{B}+{C}+{D}=\mathrm{11} \\ $$$$\left({a}+\mathrm{1}\right)\left({b}+\mathrm{2}\right)\left({c}+\mathrm{3}\right)\left({d}+\mathrm{4}\right) \\ $$$$={ABCD} \\ $$$$=\left(\sqrt[{\mathrm{4}}]{{ABCD}}\right)^{\mathrm{4}} \leqslant\left(\frac{{A}+{B}+{C}+{D}}{\mathrm{4}}\right)^{\mathrm{4}} =\left(\frac{\mathrm{11}}{\mathrm{4}}\right)^{\mathrm{4}} \\ $$$${i}.{e}.\:{maximum}\:{is}\:\left(\frac{\mathrm{11}}{\mathrm{4}}\right)^{\mathrm{4}} . \\ $$
Commented by Jubr last updated on 04/Nov/25
I appreciate sir.
$${I}\:{appreciate}\:{sir}. \\ $$

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