Question Number 198063 by mr W last updated on 10/Oct/23
$${solve}\:{for}\:{x},\:{y}\:\in{R} \\ $$$$\sqrt{{x}^{\mathrm{2}} +\mathrm{2}{x}+\mathrm{1}}+\sqrt{{y}^{\mathrm{2}} −\mathrm{6}{y}+\mathrm{9}}+\sqrt{{x}^{\mathrm{2}} −\mathrm{4}{x}+\mathrm{4}}+\sqrt{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} −\mathrm{2}{xy}}=\mathrm{4} \\ $$
Answered by witcher3 last updated on 10/Oct/23
![∣x+1∣+∣y−3∣+∣x−2∣+∣x−y∣=4....I ∣y−3∣+∣x−y∣≥∣x−3∣,∣a+b∣≤∣a∣+∣b∣ I⇒ 4≥∣x+1∣+∣x−2∣+∣x−3∣≥∣x+1−(x−3)∣+∣x−2∣ =4+∣x−2∣ 4=4+∣x−2∣⇒x=2 x=2⇒∣y−3∣+∣2−y∣=1⇒ y∈[2,3]](https://www.tinkutara.com/question/Q198097.png)
$$\mid\mathrm{x}+\mathrm{1}\mid+\mid\mathrm{y}−\mathrm{3}\mid+\mid\mathrm{x}−\mathrm{2}\mid+\mid\mathrm{x}−\mathrm{y}\mid=\mathrm{4}….\mathrm{I} \\ $$$$\mid\mathrm{y}−\mathrm{3}\mid+\mid\mathrm{x}−\mathrm{y}\mid\geqslant\mid\mathrm{x}−\mathrm{3}\mid,\mid\mathrm{a}+\mathrm{b}\mid\leqslant\mid\mathrm{a}\mid+\mid\mathrm{b}\mid \\ $$$$\mathrm{I}\Rightarrow \\ $$$$\mathrm{4}\geqslant\mid\mathrm{x}+\mathrm{1}\mid+\mid\mathrm{x}−\mathrm{2}\mid+\mid\mathrm{x}−\mathrm{3}\mid\geqslant\mid\mathrm{x}+\mathrm{1}−\left(\mathrm{x}−\mathrm{3}\right)\mid+\mid\mathrm{x}−\mathrm{2}\mid \\ $$$$=\mathrm{4}+\mid\mathrm{x}−\mathrm{2}\mid \\ $$$$\mathrm{4}=\mathrm{4}+\mid\mathrm{x}−\mathrm{2}\mid\Rightarrow\mathrm{x}=\mathrm{2} \\ $$$$\mathrm{x}=\mathrm{2}\Rightarrow\mid\mathrm{y}−\mathrm{3}\mid+\mid\mathrm{2}−\mathrm{y}\mid=\mathrm{1}\Rightarrow \\ $$$$\mathrm{y}\in\left[\mathrm{2},\mathrm{3}\right] \\ $$$$ \\ $$
Commented by manxsol last updated on 10/Oct/23
Commented by mr W last updated on 10/Oct/23
$${thanks}\:{sirs}! \\ $$
Answered by mr W last updated on 10/Oct/23
![we have ∣a∣+∣b∣≥∣a±b∣. the equation is the same as 4=∣x+1∣+∣x−2∣+∣y−3∣+∣x−y∣ ≥∣x+1−x+2∣+∣y−3+x−y∣=3+∣x−3∣ ⇒1≤∣x−3∣ ⇒−1≤x−3≤1 ⇒2≤x≤4 ...(i) 4=∣x+1∣+∣x−2∣+∣y−3∣+∣x−y∣ ≥∣x+1∣+∣x−2∣+∣y−3+x−y∣ =∣x+1∣+∣x−2∣+∣x−3∣ ≥∣x+1∣+∣x−2−x+3∣=∣x+1∣+1 ⇒3≥∣x+1∣ ⇒−3≤x+1≤3 ⇒−4≤x≤2 ...(ii) from (i) and (ii): ⇒x=2 4=∣2+1∣+∣2−2∣+∣y−3∣+∣2−y∣ 1=∣y−3∣+∣2−y∣ ≥∣y−3−2+y∣=∣2y−5∣ ⇒−1≤2y−5≤1 ⇒2≤y≤3 ⇒the solution is x=2 ∧ y∈[2, 3]](https://www.tinkutara.com/question/Q198105.png)
$${we}\:{have}\:\mid{a}\mid+\mid{b}\mid\geqslant\mid{a}\pm{b}\mid. \\ $$$$ \\ $$$${the}\:{equation}\:{is}\:{the}\:{same}\:{as} \\ $$$$\mathrm{4}=\mid{x}+\mathrm{1}\mid+\mid{x}−\mathrm{2}\mid+\mid{y}−\mathrm{3}\mid+\mid{x}−{y}\mid \\ $$$$\:\:\:\geqslant\mid{x}+\mathrm{1}−{x}+\mathrm{2}\mid+\mid{y}−\mathrm{3}+{x}−{y}\mid=\mathrm{3}+\mid{x}−\mathrm{3}\mid \\ $$$$\Rightarrow\mathrm{1}\leqslant\mid{x}−\mathrm{3}\mid \\ $$$$\Rightarrow−\mathrm{1}\leqslant{x}−\mathrm{3}\leqslant\mathrm{1}\:\Rightarrow\mathrm{2}\leqslant{x}\leqslant\mathrm{4}\:\:\:…\left({i}\right) \\ $$$$ \\ $$$$\mathrm{4}=\mid{x}+\mathrm{1}\mid+\mid{x}−\mathrm{2}\mid+\mid{y}−\mathrm{3}\mid+\mid{x}−{y}\mid \\ $$$$\:\:\geqslant\mid{x}+\mathrm{1}\mid+\mid{x}−\mathrm{2}\mid+\mid{y}−\mathrm{3}+{x}−{y}\mid \\ $$$$\:\:=\mid{x}+\mathrm{1}\mid+\mid{x}−\mathrm{2}\mid+\mid{x}−\mathrm{3}\mid \\ $$$$\:\:\geqslant\mid{x}+\mathrm{1}\mid+\mid{x}−\mathrm{2}−{x}+\mathrm{3}\mid=\mid{x}+\mathrm{1}\mid+\mathrm{1} \\ $$$$\Rightarrow\mathrm{3}\geqslant\mid{x}+\mathrm{1}\mid \\ $$$$\Rightarrow−\mathrm{3}\leqslant{x}+\mathrm{1}\leqslant\mathrm{3}\:\Rightarrow−\mathrm{4}\leqslant{x}\leqslant\mathrm{2}\:\:\:…\left({ii}\right) \\ $$$$ \\ $$$${from}\:\left({i}\right)\:{and}\:\left({ii}\right): \\ $$$$\Rightarrow{x}=\mathrm{2} \\ $$$$ \\ $$$$\mathrm{4}=\mid\mathrm{2}+\mathrm{1}\mid+\mid\mathrm{2}−\mathrm{2}\mid+\mid{y}−\mathrm{3}\mid+\mid\mathrm{2}−{y}\mid \\ $$$$\mathrm{1}=\mid{y}−\mathrm{3}\mid+\mid\mathrm{2}−{y}\mid \\ $$$$\:\:\:\geqslant\mid{y}−\mathrm{3}−\mathrm{2}+{y}\mid=\mid\mathrm{2}{y}−\mathrm{5}\mid \\ $$$$\Rightarrow−\mathrm{1}\leqslant\mathrm{2}{y}−\mathrm{5}\leqslant\mathrm{1}\:\Rightarrow\mathrm{2}\leqslant{y}\leqslant\mathrm{3} \\ $$$$ \\ $$$$\Rightarrow{the}\:{solution}\:{is}\:{x}=\mathrm{2}\:\wedge\:{y}\in\left[\mathrm{2},\:\mathrm{3}\right] \\ $$