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Question Number 119790 by bemath last updated on 27/Oct/20

Let x,y,z be nonnegative real numbers, which satisfy x+y+z=1 Find minimum value of Q=(√(2−x)) + (√(2−y)) + (√(2−z)) .

$${Let}\:{x},{y},{z}\:{be}\:{nonnegative}\:{real} \\ $$$${numbers},\:{which}\:{satisfy}\:{x}+{y}+{z}=\mathrm{1} \\ $$$${Find}\:{minimum}\:{value}\:{of}\: \\ $$$${Q}=\sqrt{\mathrm{2}−{x}}\:+\:\sqrt{\mathrm{2}−{y}}\:+\:\sqrt{\mathrm{2}−{z}}\:. \\ $$

Answered by 1549442205PVT last updated on 27/Oct/20

Put x+y=2a,x+z=2b,y+z=2c ⇒a+b+c=x+y+z=1;a,b,c≥0 Q=(√(1+y+z))+(√(1+x+z))+(√(1+x+y)) =(√(1+2a))+(√(1+2b))+(√(1+2c)) with a+b+c=1 Q(a,b)=(√(1+2a))+(√(1+2b))+(√(3−2(a+b))) We find extremum of { (((∂Q/∂a)=0)),(((∂Q/∂b)=0)) :} ⇔ { (((1/( (√(1+2a))))−(1/( (√(3−2(a+b)))))=0(1))),(((1/( (√(1+2b))))−(1/( (√(3−2(a+b)))))=0(2))) :} ⇒(1/( (√(1+2a))))=(1/( (√(1+2b))))⇒a=b Replace into (1) we get (√(3−4a))=(√(1+2a))⇔3−4a=1+2a ⇔6a=2⇒a=b=1/3 A=(∂^2 Q/∂a^2 )∣_(((1/3),(1/3))) =((−1)/((1+2a)(√(1+2a))))−(1/( [3−2(a+b)](√(3−2(a+b)))))∣_(((1/3),(1/3))) =((−9)/( 5(√(15))))−(9/(5(√(15))))=((−18)/(5(√(15))))<0 C=(∂^2 Q/∂b^2 )∣_(((1/3),(1/3))) =((−1)/((1+2b)(√(1+2b))))−(1/([3−2(a+b)](√(3−2(a+b)))))∣_(((1/3),(1/3))) =((−18)/( (√(515))))<0 (∂^2 Q/(∂a∂b))∣_(((1/3),(1/3))) =(∂Q/∂b)((1/( (√(1+2a))))−(1/( (√(3−2(a+b)))))) B=(1/( (√(3−2(a+b)))))∣_(((1/3),(1/3))) =(3/( (√(15)))) Δ=AC−B^2 =[((18)/( 5(√(15))))]^2 −(9/(15))=((324−225)/(25.15))>0 Since A<0,Q has maximum Q_(max) =Q((1/3),(1/3))=(√(15 )) when(x,y,z) =((1/3),(1/3),(1/3)) On the other hand, consider Q(x,y,z) at the bounded points we have Q(0,1,0)=Q(0,1,0)=Q(0,0,1)= =1+2(√2)=3.828..<(√(15)) ≈3.873 Q((1/2),(1/2),0)=(√2)+(√6)≈3.8637>1+2(√2) Comparing the above values we see Tthe smallest value of Q equal to 1+2(√2) when (x,y,z)=(0,0,1) and all the its permutation

$$\mathrm{Put}\:\mathrm{x}+\mathrm{y}=\mathrm{2a},\mathrm{x}+\mathrm{z}=\mathrm{2b},\mathrm{y}+\mathrm{z}=\mathrm{2c} \\ $$$$\Rightarrow\mathrm{a}+\mathrm{b}+\mathrm{c}=\mathrm{x}+\mathrm{y}+\mathrm{z}=\mathrm{1};\mathrm{a},\mathrm{b},\mathrm{c}\geqslant\mathrm{0} \\ $$$$\mathrm{Q}=\sqrt{\mathrm{1}+\mathrm{y}+\mathrm{z}}+\sqrt{\mathrm{1}+\mathrm{x}+\mathrm{z}}+\sqrt{\mathrm{1}+\mathrm{x}+\mathrm{y}} \\ $$$$=\sqrt{\mathrm{1}+\mathrm{2a}}+\sqrt{\mathrm{1}+\mathrm{2b}}+\sqrt{\mathrm{1}+\mathrm{2c}}\:\mathrm{with}\:\mathrm{a}+\mathrm{b}+\mathrm{c}=\mathrm{1} \\ $$$$\mathrm{Q}\left(\mathrm{a},\mathrm{b}\right)=\sqrt{\mathrm{1}+\mathrm{2a}}+\sqrt{\mathrm{1}+\mathrm{2b}}+\sqrt{\mathrm{3}−\mathrm{2}\left(\mathrm{a}+\mathrm{b}\right)}\: \\ $$$$\mathrm{We}\:\mathrm{find}\:\mathrm{extremum}\:\mathrm{of}\: \\ $$$$\begin{cases}{\frac{\partial\mathrm{Q}}{\partial\mathrm{a}}=\mathrm{0}}\\{\frac{\partial\mathrm{Q}}{\partial\mathrm{b}}=\mathrm{0}}\end{cases} \\ $$$$\Leftrightarrow\begin{cases}{\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}+\mathrm{2a}}}−\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}−\mathrm{2}\left(\mathrm{a}+\mathrm{b}\right)}}=\mathrm{0}\left(\mathrm{1}\right)}\\{\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}+\mathrm{2b}}}−\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}−\mathrm{2}\left(\mathrm{a}+\mathrm{b}\right)}}=\mathrm{0}\left(\mathrm{2}\right)}\end{cases} \\ $$$$\Rightarrow\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}+\mathrm{2a}}}=\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}+\mathrm{2b}}}\Rightarrow\mathrm{a}=\mathrm{b} \\ $$$$\mathrm{Replace}\:\mathrm{into}\:\left(\mathrm{1}\right)\:\mathrm{we}\:\mathrm{get} \\ $$$$\sqrt{\mathrm{3}−\mathrm{4a}}=\sqrt{\mathrm{1}+\mathrm{2a}}\Leftrightarrow\mathrm{3}−\mathrm{4a}=\mathrm{1}+\mathrm{2a} \\ $$$$\Leftrightarrow\mathrm{6a}=\mathrm{2}\Rightarrow\mathrm{a}=\mathrm{b}=\mathrm{1}/\mathrm{3} \\ $$$$\mathrm{A}=\frac{\partial^{\mathrm{2}} \mathrm{Q}}{\partial\mathrm{a}^{\mathrm{2}} }\mid_{\left(\frac{\mathrm{1}}{\mathrm{3}},\frac{\mathrm{1}}{\mathrm{3}}\right)} =\frac{−\mathrm{1}}{\left(\mathrm{1}+\mathrm{2a}\right)\sqrt{\mathrm{1}+\mathrm{2a}}}−\frac{\mathrm{1}}{\:\left[\mathrm{3}−\mathrm{2}\left(\mathrm{a}+\mathrm{b}\right)\right]\sqrt{\mathrm{3}−\mathrm{2}\left(\mathrm{a}+\mathrm{b}\right)}}\mid_{\left(\frac{\mathrm{1}}{\mathrm{3}},\frac{\mathrm{1}}{\mathrm{3}}\right)} \\ $$$$=\frac{−\mathrm{9}}{\:\mathrm{5}\sqrt{\mathrm{15}}}−\frac{\mathrm{9}}{\mathrm{5}\sqrt{\mathrm{15}}}=\frac{−\mathrm{18}}{\mathrm{5}\sqrt{\mathrm{15}}}<\mathrm{0} \\ $$$$\mathrm{C}=\frac{\partial^{\mathrm{2}} \mathrm{Q}}{\partial\mathrm{b}^{\mathrm{2}} }\mid_{\left(\frac{\mathrm{1}}{\mathrm{3}},\frac{\mathrm{1}}{\mathrm{3}}\right)} =\frac{−\mathrm{1}}{\left(\mathrm{1}+\mathrm{2b}\right)\sqrt{\mathrm{1}+\mathrm{2b}}}−\frac{\mathrm{1}}{\left[\mathrm{3}−\mathrm{2}\left(\mathrm{a}+\mathrm{b}\right)\right]\sqrt{\mathrm{3}−\mathrm{2}\left(\mathrm{a}+\mathrm{b}\right)}}\mid_{\left(\frac{\mathrm{1}}{\mathrm{3}},\frac{\mathrm{1}}{\mathrm{3}}\right)} \\ $$$$=\frac{−\mathrm{18}}{\:\sqrt{\mathrm{515}}}<\mathrm{0} \\ $$$$\frac{\partial^{\mathrm{2}} \mathrm{Q}}{\partial\mathrm{a}\partial\mathrm{b}}\mid_{\left(\frac{\mathrm{1}}{\mathrm{3}},\frac{\mathrm{1}}{\mathrm{3}}\right)} =\frac{\partial\mathrm{Q}}{\partial\mathrm{b}}\left(\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}+\mathrm{2a}}}−\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}−\mathrm{2}\left(\mathrm{a}+\mathrm{b}\right)}}\right) \\ $$$$\mathrm{B}=\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}−\mathrm{2}\left(\mathrm{a}+\mathrm{b}\right)}}\mid_{\left(\frac{\mathrm{1}}{\mathrm{3}},\frac{\mathrm{1}}{\mathrm{3}}\right)} =\frac{\mathrm{3}}{\:\sqrt{\mathrm{15}}} \\ $$$$\Delta=\mathrm{AC}−\mathrm{B}^{\mathrm{2}} =\left[\frac{\mathrm{18}}{\:\mathrm{5}\sqrt{\mathrm{15}}}\right]^{\mathrm{2}} −\frac{\mathrm{9}}{\mathrm{15}}=\frac{\mathrm{324}−\mathrm{225}}{\mathrm{25}.\mathrm{15}}>\mathrm{0} \\ $$$$\mathrm{Since}\:\mathrm{A}<\mathrm{0},\mathrm{Q}\:\mathrm{has}\:\mathrm{maximum} \\ $$$$\mathrm{Q}_{\mathrm{max}} =\mathrm{Q}\left(\frac{\mathrm{1}}{\mathrm{3}},\frac{\mathrm{1}}{\mathrm{3}}\right)=\sqrt{\mathrm{15}\:}\:\mathrm{when}\left(\mathrm{x},\mathrm{y},\mathrm{z}\right) \\ $$$$=\left(\frac{\mathrm{1}}{\mathrm{3}},\frac{\mathrm{1}}{\mathrm{3}},\frac{\mathrm{1}}{\mathrm{3}}\right)\: \\ $$$$\mathrm{On}\:\mathrm{the}\:\mathrm{other}\:\mathrm{hand},\:\mathrm{consider}\:\mathrm{Q}\left(\mathrm{x},\mathrm{y},\mathrm{z}\right)\:\mathrm{at} \\ $$$$\mathrm{the}\:\mathrm{bounded}\:\mathrm{points}\:\mathrm{we}\:\mathrm{have} \\ $$$$\mathrm{Q}\left(\mathrm{0},\mathrm{1},\mathrm{0}\right)=\mathrm{Q}\left(\mathrm{0},\mathrm{1},\mathrm{0}\right)=\mathrm{Q}\left(\mathrm{0},\mathrm{0},\mathrm{1}\right)= \\ $$$$=\mathrm{1}+\mathrm{2}\sqrt{\mathrm{2}}=\mathrm{3}.\mathrm{828}..<\sqrt{\mathrm{15}}\:\approx\mathrm{3}.\mathrm{873} \\ $$$$\mathrm{Q}\left(\frac{\mathrm{1}}{\mathrm{2}},\frac{\mathrm{1}}{\mathrm{2}},\mathrm{0}\right)=\sqrt{\mathrm{2}}+\sqrt{\mathrm{6}}\approx\mathrm{3}.\mathrm{8637}>\mathrm{1}+\mathrm{2}\sqrt{\mathrm{2}} \\ $$$$\mathrm{Comparing}\:\mathrm{the}\:\mathrm{above}\:\mathrm{values}\:\mathrm{we}\:\mathrm{see} \\ $$$$\mathrm{Tthe}\:\mathrm{smallest}\:\mathrm{value}\:\:\mathrm{of}\:\mathrm{Q}\:\mathrm{equal}\:\mathrm{to}\:\mathrm{1}+\mathrm{2}\sqrt{\mathrm{2}} \\ $$$$\mathrm{when}\:\left(\mathrm{x},\mathrm{y},\mathrm{z}\right)=\left(\mathrm{0},\mathrm{0},\mathrm{1}\right)\:\:\mathrm{and}\:\mathrm{all}\:\mathrm{the}\:\mathrm{its} \\ $$$$\mathrm{permutation} \\ $$

Answered by ebi last updated on 27/Oct/20

Q=(2−x)^(1/2) +(2−y)^(1/2) +(2−z)^(1/2) constraint, x+y+z=1=F the solution lies on 0≤x,y,z≤1 ▽Q=λ▽F ⟨Q_x ,Q_y ,Q_z ⟩=λ⟨F_x ,F_y ,F_z ⟩ λ=lagrange multiplier Q_x =λF_x −(1/(2(√(2−x))))=λ (−(1/(2(√(2−x)))))^2 =λ^2 λ^2 =(1/(4(2−x))) x=2−(1/(4λ^2 ))......(1) Q_y =λF_y −(1/(2(√(2−y))))=λ λ^2 =(1/(4(2−y))) y=2−(1/(4λ^2 ))......(2) Q_z =λF_z −(1/(2(√(2−z))))=λ λ^2 =(1/(4(2−z))) z=2−(1/(4λ^2 ))......(3) x+y+z=1......(4) substitute (1),(2),(3) into (4) (2−(1/(4λ^2 )))+(2−(1/(4λ^2 )))+(2−(1/(4λ^2 )))=1 −(3/(4λ^2 ))=−5 λ^2 =(3/(20)) x=2−(1/(4((3/(20)))))=(1/3) y=2−(1/(4((3/(20)))))=(1/3) z=2−(1/(4((3/(20)))))=(1/3) ∴the solution is ((1/3),(1/3),(1/3)) G((1/3),(1/3),(1/3))=(√(2−(1/3)))+(√(2−(1/3)))+(√(2−(1/3))) =3(√(5/3))=(√(15))≈3.873 because of this point is the only stationary point, we have to consider other points for maxima or minima consider these 3 points: (1,0,0),(0,1,0),(0,0,1) G(1,0,0)=G(0,1,0)=G(0,0,1) =1+2(√2)≈3.828 thus, G((1/3),(1/3),(1/3))=3.873→maximum G(1,0,0)=G(0,1,0)=G(0,0,1)=3.828→minimum

$$ \\ $$$${Q}=\left(\mathrm{2}−{x}\right)^{\frac{\mathrm{1}}{\mathrm{2}}} +\left(\mathrm{2}−{y}\right)^{\frac{\mathrm{1}}{\mathrm{2}}} +\left(\mathrm{2}−{z}\right)^{\frac{\mathrm{1}}{\mathrm{2}}} \\ $$$${constraint},\:{x}+{y}+{z}=\mathrm{1}={F} \\ $$$${the}\:{solution}\:{lies}\:{on}\:\mathrm{0}\leqslant{x},{y},{z}\leqslant\mathrm{1} \\ $$$$ \\ $$$$\bigtriangledown{Q}=\lambda\bigtriangledown{F} \\ $$$$\langle{Q}_{{x}} ,{Q}_{{y}} ,{Q}_{{z}} \rangle=\lambda\langle{F}_{{x}} ,{F}_{{y}} ,{F}_{{z}} \rangle \\ $$$$\lambda={lagrange}\:{multiplier} \\ $$$$ \\ $$$${Q}_{{x}} =\lambda{F}_{{x}} \\ $$$$−\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{2}−{x}}}=\lambda \\ $$$$\left(−\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{2}−{x}}}\right)^{\mathrm{2}} =\lambda^{\mathrm{2}} \\ $$$$\lambda^{\mathrm{2}} =\frac{\mathrm{1}}{\mathrm{4}\left(\mathrm{2}−{x}\right)} \\ $$$${x}=\mathrm{2}−\frac{\mathrm{1}}{\mathrm{4}\lambda^{\mathrm{2}} }......\left(\mathrm{1}\right) \\ $$$${Q}_{{y}} =\lambda{F}_{{y}} \\ $$$$−\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{2}−{y}}}=\lambda \\ $$$$\lambda^{\mathrm{2}} =\frac{\mathrm{1}}{\mathrm{4}\left(\mathrm{2}−{y}\right)} \\ $$$${y}=\mathrm{2}−\frac{\mathrm{1}}{\mathrm{4}\lambda^{\mathrm{2}} }......\left(\mathrm{2}\right) \\ $$$${Q}_{{z}} =\lambda{F}_{{z}} \\ $$$$−\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{2}−{z}}}=\lambda \\ $$$$\lambda^{\mathrm{2}} =\frac{\mathrm{1}}{\mathrm{4}\left(\mathrm{2}−{z}\right)} \\ $$$${z}=\mathrm{2}−\frac{\mathrm{1}}{\mathrm{4}\lambda^{\mathrm{2}} }......\left(\mathrm{3}\right) \\ $$$${x}+{y}+{z}=\mathrm{1}......\left(\mathrm{4}\right) \\ $$$$ \\ $$$${substitute}\:\left(\mathrm{1}\right),\left(\mathrm{2}\right),\left(\mathrm{3}\right)\:{into}\:\left(\mathrm{4}\right) \\ $$$$\left(\mathrm{2}−\frac{\mathrm{1}}{\mathrm{4}\lambda^{\mathrm{2}} }\right)+\left(\mathrm{2}−\frac{\mathrm{1}}{\mathrm{4}\lambda^{\mathrm{2}} }\right)+\left(\mathrm{2}−\frac{\mathrm{1}}{\mathrm{4}\lambda^{\mathrm{2}} }\right)=\mathrm{1} \\ $$$$−\frac{\mathrm{3}}{\mathrm{4}\lambda^{\mathrm{2}} }=−\mathrm{5} \\ $$$$\lambda^{\mathrm{2}} =\frac{\mathrm{3}}{\mathrm{20}} \\ $$$$ \\ $$$${x}=\mathrm{2}−\frac{\mathrm{1}}{\mathrm{4}\left(\frac{\mathrm{3}}{\mathrm{20}}\right)}=\frac{\mathrm{1}}{\mathrm{3}} \\ $$$${y}=\mathrm{2}−\frac{\mathrm{1}}{\mathrm{4}\left(\frac{\mathrm{3}}{\mathrm{20}}\right)}=\frac{\mathrm{1}}{\mathrm{3}} \\ $$$${z}=\mathrm{2}−\frac{\mathrm{1}}{\mathrm{4}\left(\frac{\mathrm{3}}{\mathrm{20}}\right)}=\frac{\mathrm{1}}{\mathrm{3}} \\ $$$$\therefore{the}\:{solution}\:{is}\:\left(\frac{\mathrm{1}}{\mathrm{3}},\frac{\mathrm{1}}{\mathrm{3}},\frac{\mathrm{1}}{\mathrm{3}}\right) \\ $$$${G}\left(\frac{\mathrm{1}}{\mathrm{3}},\frac{\mathrm{1}}{\mathrm{3}},\frac{\mathrm{1}}{\mathrm{3}}\right)=\sqrt{\mathrm{2}−\frac{\mathrm{1}}{\mathrm{3}}}+\sqrt{\mathrm{2}−\frac{\mathrm{1}}{\mathrm{3}}}+\sqrt{\mathrm{2}−\frac{\mathrm{1}}{\mathrm{3}}} \\ $$$$=\mathrm{3}\sqrt{\frac{\mathrm{5}}{\mathrm{3}}}=\sqrt{\mathrm{15}}\approx\mathrm{3}.\mathrm{873} \\ $$$${because}\:{of}\:{this}\:{point}\:{is}\:{the}\:{only} \\ $$$${stationary}\:{point},\:{we}\:{have}\:{to}\:{consider} \\ $$$${other}\:{points}\:{for}\:{maxima}\:{or}\:{minima} \\ $$$$ \\ $$$${consider}\:{these}\:\mathrm{3}\:{points}: \\ $$$$\left(\mathrm{1},\mathrm{0},\mathrm{0}\right),\left(\mathrm{0},\mathrm{1},\mathrm{0}\right),\left(\mathrm{0},\mathrm{0},\mathrm{1}\right) \\ $$$${G}\left(\mathrm{1},\mathrm{0},\mathrm{0}\right)={G}\left(\mathrm{0},\mathrm{1},\mathrm{0}\right)={G}\left(\mathrm{0},\mathrm{0},\mathrm{1}\right) \\ $$$$=\mathrm{1}+\mathrm{2}\sqrt{\mathrm{2}}\approx\mathrm{3}.\mathrm{828} \\ $$$$ \\ $$$${thus}, \\ $$$${G}\left(\frac{\mathrm{1}}{\mathrm{3}},\frac{\mathrm{1}}{\mathrm{3}},\frac{\mathrm{1}}{\mathrm{3}}\right)=\mathrm{3}.\mathrm{873}\rightarrow{maximum} \\ $$$${G}\left(\mathrm{1},\mathrm{0},\mathrm{0}\right)={G}\left(\mathrm{0},\mathrm{1},\mathrm{0}\right)={G}\left(\mathrm{0},\mathrm{0},\mathrm{1}\right)=\mathrm{3}.\mathrm{828}\rightarrow{minimum} \\ $$

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