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Question Number 215005 by Abdullahrussell last updated on 25/Dec/24
a,b,c,d∈R such that, (a+b)(c+d)=2 (a+c)(b+d)=3 (a+d)(b+c)=4 find: (a^2 +b^2 +c^2 +d^2 )_(minimum.)
$$\:{a},{b},{c},{d}\in{R}\:{such}\:{that}, \\ $$$$\:\left({a}+{b}\right)\left({c}+{d}\right)=\mathrm{2} \\ $$$$\:\left({a}+{c}\right)\left({b}+{d}\right)=\mathrm{3} \\ $$$$\:\left({a}+{d}\right)\left({b}+{c}\right)=\mathrm{4}\: \\ $$$$\:{find}:\:\:\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} +{d}^{\mathrm{2}} \right)_{{minimum}.} \\ $$
Answered by A5T last updated on 25/Dec/24
ac+ad+bc+bd=2 ab+ad+bc+cd=3 ab+ac+bd+cd=4 ⇒Σab=ab+ac+ad+bc+bd+cd=((2+3+4)/2)=4.5 a^2 +b^2 +c^2 +d^2 ≥(((a+b+c+d)^2 )/4) ⇒4(a^2 +b^2 +c^2 +d^2 )≥a^2 +b^2 +c^2 +d^2 +2Σab ⇒3(a^2 +b^2 +c^2 +d^2 )≥9 ⇒a^2 +b^2 +c^2 +d^2 ≥3
$${ac}+{ad}+{bc}+{bd}=\mathrm{2} \\ $$$${ab}+{ad}+{bc}+{cd}=\mathrm{3} \\ $$$${ab}+{ac}+{bd}+{cd}=\mathrm{4} \\ $$$$\Rightarrow\Sigma{ab}={ab}+{ac}+{ad}+{bc}+{bd}+{cd}=\frac{\mathrm{2}+\mathrm{3}+\mathrm{4}}{\mathrm{2}}=\mathrm{4}.\mathrm{5} \\ $$$${a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} +{d}^{\mathrm{2}} \geqslant\frac{\left({a}+{b}+{c}+{d}\right)^{\mathrm{2}} }{\mathrm{4}} \\ $$$$\Rightarrow\mathrm{4}\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} +{d}^{\mathrm{2}} \right)\geqslant{a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} +{d}^{\mathrm{2}} +\mathrm{2}\Sigma{ab} \\ $$$$\Rightarrow\mathrm{3}\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} +{d}^{\mathrm{2}} \right)\geqslant\mathrm{9} \\ $$$$\Rightarrow{a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} +{d}^{\mathrm{2}} \geqslant\mathrm{3} \\ $$
Commented by Abdullahrussell last updated on 25/Dec/24
Sir, equality holds with (a,b,c,d)=?
$$\:{Sir},\:{equality}\:{holds}\:{with}\:\left({a},{b},{c},{d}\right)=? \\ $$
Commented by A5T last updated on 25/Dec/24
This is a lower bound, equality may not necessarily hold.
$${This}\:{is}\:{a}\:{lower}\:{bound},\:{equality}\:{may}\:{not}\: \\ $$$${necessarily}\:{hold}. \\ $$
Commented by Abdullahrussell last updated on 25/Dec/24
Sir, if a^2 +b^2 +c^2 +d^2 =3 then (a,b,c,d)=?
$$\:{Sir},\:{if}\:{a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} +{d}^{\mathrm{2}} =\mathrm{3} \\ $$$$\:\:{then}\:\left({a},{b},{c},{d}\right)=? \\ $$
Commented by A5T last updated on 25/Dec/24
The actual minimum value is ≈7.
$${The}\:{actual}\:{minimum}\:{value}\:{is}\:\approx\mathrm{7}. \\ $$
Commented by A5T last updated on 26/Dec/24
Commented by Frix last updated on 26/Dec/24
It exactly is 7 at a=−(1/2)+((√2)/2)∧b=−(3/2)+((√2)/2)∧c=−(1/2)−((√2)/2)∧d=−(3/2)−((√2)/2)
$$\mathrm{It}\:\mathrm{exactly}\:\mathrm{is}\:\mathrm{7}\:\mathrm{at} \\ $$$${a}=−\frac{\mathrm{1}}{\mathrm{2}}+\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\wedge{b}=−\frac{\mathrm{3}}{\mathrm{2}}+\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\wedge{c}=−\frac{\mathrm{1}}{\mathrm{2}}−\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\wedge{d}=−\frac{\mathrm{3}}{\mathrm{2}}−\frac{\sqrt{\mathrm{2}}}{\mathrm{2}} \\ $$

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