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Question Number 150719 by mathdanisur last updated on 14/Aug/21
Commented by MJS_new last updated on 15/Aug/21
((1+(√(21)))/2)≤x+y≤1+(√(11))
$$\frac{\mathrm{1}+\sqrt{\mathrm{21}}}{\mathrm{2}}\leqslant{x}+{y}\leqslant\mathrm{1}+\sqrt{\mathrm{11}} \\ $$
Commented by mathdanisur last updated on 15/Aug/21
ThankYou Ser, please the cases must be analyzd in detail
$$\mathrm{ThankYou}\:\boldsymbol{\mathrm{S}}\mathrm{er},\:\mathrm{please}\:\mathrm{the}\:\mathrm{cases}\:\mathrm{must} \\ $$$$\mathrm{be}\:\mathrm{analyzd}\:\mathrm{in}\:\mathrm{detail} \\ $$
Commented by mathdanisur last updated on 15/Aug/21
extensive solution if possible
$$\mathrm{extensive}\:\mathrm{solution}\:\mathrm{if}\:\mathrm{possible} \\ $$
Commented by mr W last updated on 15/Aug/21
typo? ((1+(√(21)))/2)≤x+y≤1+(√(11))
$${typo}? \\ $$$$\frac{\mathrm{1}+\sqrt{\mathrm{21}}}{\mathrm{2}}\leqslant{x}+{y}\leqslant\mathrm{1}+\sqrt{\mathrm{11}} \\ $$
Commented by MJS_new last updated on 15/Aug/21
yes, thank you
$$\mathrm{yes},\:\mathrm{thank}\:\mathrm{you} \\ $$
Answered by mr W last updated on 15/Aug/21
with condition x,y≥0, at x=0: −(√2)=(√(y+3))−y y−(√2)=(√(y+3)) y^2 −(2(√2)+1)y−1=0 y=((2(√2)+1+(√(14+4(√2))))/2) at y=0: x−(√(x+2))=(√3) (√(x+2))=x−(√3) x^2 −(2(√3)+1)x+1=0 x=((2(√3)+1+(√(9+4(√3))))/2) (x+y)_(min) =max(((2(√2)+1+(√(14+4(√2))))/2),((2(√3)+1+(√(9+4(√3))))/2)) (x+y)_(min) =((2(√3)+1+(√(9+4(√3))))/2)≈4.227 1−(1/(2(√(x+2))))=(1/(2(√(y+3))))y′−y′ for y′=−1: 1−(1/(2(√(x+2))))=−(1/(2(√(y+3))))+1 (1/( (√(x+2))))=(1/( (√(y+3)))) x+2=y+3 y=x−1 x−(√(x+2))=(√(x+2))−x+1 2x−1=2(√(x+2)) 4x^2 −8x−7=0 x=1+((√(11))/2) y=((√(11))/2) ⇒(x+y)_(max) =1+(√(11))≈4.316 without condition x,y≥0, at x=−2: −2=(√(y+3))−y y−2=(√(y+3)) y^2 −5y+1=0 y=((5+(√(21)))/2) x+y=−2+((5+(√(21)))/2)=((1+(√(21)))/2) at y=−3: x−(√(x+2))=3 x−3=(√(x+2)) x^2 −7x+7=0 x=((7+(√(21)))/2) x+y=((7+(√(21)))/2)−3=((1+(√(21)))/2) (x+y)_(min) =((1+(√(21)))/2)≈2.791
$$\boldsymbol{{with}}\:\boldsymbol{{condition}}\:\:\boldsymbol{{x}},\boldsymbol{{y}}\geqslant\mathrm{0},\: \\ $$$${at}\:{x}=\mathrm{0}: \\ $$$$−\sqrt{\mathrm{2}}=\sqrt{{y}+\mathrm{3}}−{y} \\ $$$${y}−\sqrt{\mathrm{2}}=\sqrt{{y}+\mathrm{3}} \\ $$$${y}^{\mathrm{2}} −\left(\mathrm{2}\sqrt{\mathrm{2}}+\mathrm{1}\right){y}−\mathrm{1}=\mathrm{0} \\ $$$${y}=\frac{\mathrm{2}\sqrt{\mathrm{2}}+\mathrm{1}+\sqrt{\mathrm{14}+\mathrm{4}\sqrt{\mathrm{2}}}}{\mathrm{2}} \\ $$$${at}\:{y}=\mathrm{0}: \\ $$$${x}−\sqrt{{x}+\mathrm{2}}=\sqrt{\mathrm{3}} \\ $$$$\sqrt{{x}+\mathrm{2}}={x}−\sqrt{\mathrm{3}} \\ $$$${x}^{\mathrm{2}} −\left(\mathrm{2}\sqrt{\mathrm{3}}+\mathrm{1}\right){x}+\mathrm{1}=\mathrm{0} \\ $$$${x}=\frac{\mathrm{2}\sqrt{\mathrm{3}}+\mathrm{1}+\sqrt{\mathrm{9}+\mathrm{4}\sqrt{\mathrm{3}}}}{\mathrm{2}} \\ $$$$\left({x}+{y}\right)_{{min}} ={max}\left(\frac{\mathrm{2}\sqrt{\mathrm{2}}+\mathrm{1}+\sqrt{\mathrm{14}+\mathrm{4}\sqrt{\mathrm{2}}}}{\mathrm{2}},\frac{\mathrm{2}\sqrt{\mathrm{3}}+\mathrm{1}+\sqrt{\mathrm{9}+\mathrm{4}\sqrt{\mathrm{3}}}}{\mathrm{2}}\right) \\ $$$$\left({x}+{y}\right)_{{min}} =\frac{\mathrm{2}\sqrt{\mathrm{3}}+\mathrm{1}+\sqrt{\mathrm{9}+\mathrm{4}\sqrt{\mathrm{3}}}}{\mathrm{2}}\approx\mathrm{4}.\mathrm{227} \\ $$$$ \\ $$$$\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}\sqrt{{x}+\mathrm{2}}}=\frac{\mathrm{1}}{\mathrm{2}\sqrt{{y}+\mathrm{3}}}{y}'−{y}' \\ $$$${for}\:{y}'=−\mathrm{1}: \\ $$$$\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}\sqrt{{x}+\mathrm{2}}}=−\frac{\mathrm{1}}{\mathrm{2}\sqrt{{y}+\mathrm{3}}}+\mathrm{1} \\ $$$$\frac{\mathrm{1}}{\:\sqrt{{x}+\mathrm{2}}}=\frac{\mathrm{1}}{\:\sqrt{{y}+\mathrm{3}}} \\ $$$${x}+\mathrm{2}={y}+\mathrm{3} \\ $$$${y}={x}−\mathrm{1} \\ $$$${x}−\sqrt{{x}+\mathrm{2}}=\sqrt{{x}+\mathrm{2}}−{x}+\mathrm{1} \\ $$$$\mathrm{2}{x}−\mathrm{1}=\mathrm{2}\sqrt{{x}+\mathrm{2}} \\ $$$$\mathrm{4}{x}^{\mathrm{2}} −\mathrm{8}{x}−\mathrm{7}=\mathrm{0} \\ $$$${x}=\mathrm{1}+\frac{\sqrt{\mathrm{11}}}{\mathrm{2}} \\ $$$${y}=\frac{\sqrt{\mathrm{11}}}{\mathrm{2}} \\ $$$$\Rightarrow\left({x}+{y}\right)_{{max}} =\mathrm{1}+\sqrt{\mathrm{11}}\approx\mathrm{4}.\mathrm{316} \\ $$$$ \\ $$$$\boldsymbol{{without}}\:\boldsymbol{{condition}}\:\boldsymbol{{x}},\boldsymbol{{y}}\geqslant\mathrm{0}, \\ $$$${at}\:{x}=−\mathrm{2}: \\ $$$$−\mathrm{2}=\sqrt{{y}+\mathrm{3}}−{y} \\ $$$${y}−\mathrm{2}=\sqrt{{y}+\mathrm{3}} \\ $$$${y}^{\mathrm{2}} −\mathrm{5}{y}+\mathrm{1}=\mathrm{0} \\ $$$${y}=\frac{\mathrm{5}+\sqrt{\mathrm{21}}}{\mathrm{2}} \\ $$$${x}+{y}=−\mathrm{2}+\frac{\mathrm{5}+\sqrt{\mathrm{21}}}{\mathrm{2}}=\frac{\mathrm{1}+\sqrt{\mathrm{21}}}{\mathrm{2}} \\ $$$${at}\:{y}=−\mathrm{3}: \\ $$$${x}−\sqrt{{x}+\mathrm{2}}=\mathrm{3} \\ $$$${x}−\mathrm{3}=\sqrt{{x}+\mathrm{2}} \\ $$$${x}^{\mathrm{2}} −\mathrm{7}{x}+\mathrm{7}=\mathrm{0} \\ $$$${x}=\frac{\mathrm{7}+\sqrt{\mathrm{21}}}{\mathrm{2}} \\ $$$${x}+{y}=\frac{\mathrm{7}+\sqrt{\mathrm{21}}}{\mathrm{2}}−\mathrm{3}=\frac{\mathrm{1}+\sqrt{\mathrm{21}}}{\mathrm{2}} \\ $$$$\left({x}+{y}\right)_{{min}} =\frac{\mathrm{1}+\sqrt{\mathrm{21}}}{\mathrm{2}}\approx\mathrm{2}.\mathrm{791} \\ $$
Commented by mathdanisur last updated on 15/Aug/21
Thank you Ser nice
$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{Ser}\:\mathrm{nice} \\ $$
Answered by MJS_new last updated on 15/Aug/21
x−(√(x+2))=(√(y+3))−y let y=m−x x−(√(x+2))=(√(m−x+3))−(m−x) m=(√(x+2))+(√(m+3−x)) ⇒ m>0 squaring and transforming m^2 −m−5=2(√(x+2))(√(m+3−x)) ⇒ m^2 +m−5≥0 ⇒ m≥((1+(√(21)))/2) ⇒ min (x+y) =((1+(√(21)))/2) squaring and transforming x^2 −(m+1)x+((m^4 −2m^3 −9m^2 +2m+1)/2)=0 D=(p^2 /4)−q=(((m+1)^2 )/4)−((m^4 −2m^3 −9m^2 +2m+1)/2)= =((m^2 (−m^2 +2m+10))/4)≥0 ⇒ m^2 −3m−10≤0∧m≥((1+(√(21)))/2) ⇒ m≤1+(√(11)) ⇒ max (x+y) =1+(√(11))
$${x}−\sqrt{{x}+\mathrm{2}}=\sqrt{{y}+\mathrm{3}}−{y} \\ $$$$\mathrm{let}\:{y}={m}−{x} \\ $$$${x}−\sqrt{{x}+\mathrm{2}}=\sqrt{{m}−{x}+\mathrm{3}}−\left({m}−{x}\right) \\ $$$${m}=\sqrt{{x}+\mathrm{2}}+\sqrt{{m}+\mathrm{3}−{x}} \\ $$$$\:\:\:\:\:\Rightarrow\:{m}>\mathrm{0} \\ $$$$\mathrm{squaring}\:\mathrm{and}\:\mathrm{transforming} \\ $$$${m}^{\mathrm{2}} −{m}−\mathrm{5}=\mathrm{2}\sqrt{{x}+\mathrm{2}}\sqrt{{m}+\mathrm{3}−{x}} \\ $$$$\:\:\:\:\:\Rightarrow\:{m}^{\mathrm{2}} +{m}−\mathrm{5}\geqslant\mathrm{0}\:\Rightarrow\:{m}\geqslant\frac{\mathrm{1}+\sqrt{\mathrm{21}}}{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\Rightarrow\:\mathrm{min}\:\left({x}+{y}\right)\:=\frac{\mathrm{1}+\sqrt{\mathrm{21}}}{\mathrm{2}} \\ $$$$\mathrm{squaring}\:\mathrm{and}\:\mathrm{transforming} \\ $$$${x}^{\mathrm{2}} −\left({m}+\mathrm{1}\right){x}+\frac{{m}^{\mathrm{4}} −\mathrm{2}{m}^{\mathrm{3}} −\mathrm{9}{m}^{\mathrm{2}} +\mathrm{2}{m}+\mathrm{1}}{\mathrm{2}}=\mathrm{0} \\ $$$${D}=\frac{{p}^{\mathrm{2}} }{\mathrm{4}}−{q}=\frac{\left({m}+\mathrm{1}\right)^{\mathrm{2}} }{\mathrm{4}}−\frac{{m}^{\mathrm{4}} −\mathrm{2}{m}^{\mathrm{3}} −\mathrm{9}{m}^{\mathrm{2}} +\mathrm{2}{m}+\mathrm{1}}{\mathrm{2}}= \\ $$$$=\frac{{m}^{\mathrm{2}} \left(−{m}^{\mathrm{2}} +\mathrm{2}{m}+\mathrm{10}\right)}{\mathrm{4}}\geqslant\mathrm{0} \\ $$$$\:\:\:\:\:\Rightarrow\:{m}^{\mathrm{2}} −\mathrm{3}{m}−\mathrm{10}\leqslant\mathrm{0}\wedge{m}\geqslant\frac{\mathrm{1}+\sqrt{\mathrm{21}}}{\mathrm{2}} \\ $$$$\:\:\:\:\:\Rightarrow\:{m}\leqslant\mathrm{1}+\sqrt{\mathrm{11}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\Rightarrow\:\mathrm{max}\:\left({x}+{y}\right)\:=\mathrm{1}+\sqrt{\mathrm{11}} \\ $$
Commented by mathdanisur last updated on 15/Aug/21
Thank you Ser cool
$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{Ser}\:\mathrm{cool} \\ $$

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