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)) .


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

$L e t x, y, z b e n o n n e g a t i v e r e a l$ $n u m b e r s, w h i c h s a t i s f y x + y + z = 1$ $F i n d m i n i m u m v a l u e o f$ $Q = \sqrt{2 - x} + \sqrt{2 - y} + \sqrt{2 - z} .$
	Answered by 1549442205PVT last updated on 27/Oct/20
	$Put x+y=2a,x+z=2b,y+z=2c ⇒a+b+c=x+y+z=1;a,b,c≥0 Q=(√(1+y+z))+(√(1+x+z))+(√(1+x+y)) =(√(1+2a))+(√(1+2b))+(√(1+2c)) with a+b+c=1 Q(a,b)=(√(1+2a))+(√(1+2b))+(√(3−2(a+b))) We find extremum of { (((∂Q/∂a)=0)),(((∂Q/∂b)=0)) :} ⇔ { (((1/( (√(1+2a))))−(1/( (√(3−2(a+b)))))=0(1))),(((1/( (√(1+2b))))−(1/( (√(3−2(a+b)))))=0(2))) :} ⇒(1/( (√(1+2a))))=(1/( (√(1+2b))))⇒a=b Replace into (1) we get (√(3−4a))=(√(1+2a))⇔3−4a=1+2a ⇔6a=2⇒a=b=1/3 A=(∂^2 Q/∂a^2 )∣_(((1/3),(1/3))) =((−1)/((1+2a)(√(1+2a))))−(1/( [3−2(a+b)](√(3−2(a+b)))))∣_(((1/3),(1/3))) =((−9)/( 5(√(15))))−(9/(5(√(15))))=((−18)/(5(√(15))))<0 C=(∂^2 Q/∂b^2 )∣_(((1/3),(1/3))) =((−1)/((1+2b)(√(1+2b))))−(1/([3−2(a+b)](√(3−2(a+b)))))∣_(((1/3),(1/3))) =((−18)/( (√(515))))<0 (∂^2 Q/(∂a∂b))∣_(((1/3),(1/3))) =(∂Q/∂b)((1/( (√(1+2a))))−(1/( (√(3−2(a+b)))))) B=(1/( (√(3−2(a+b)))))∣_(((1/3),(1/3))) =(3/( (√(15)))) Δ=AC−B^2 =[((18)/( 5(√(15))))]^2 −(9/(15))=((324−225)/(25.15))>0 Since A<0,Q has maximum Q_(max) =Q((1/3),(1/3))=(√(15 )) when(x,y,z) =((1/3),(1/3),(1/3)) On the other hand, consider Q(x,y,z) at the bounded points we have Q(0,1,0)=Q(0,1,0)=Q(0,0,1)= =1+2(√2)=3.828..<(√(15)) ≈3.873 Q((1/2),(1/2),0)=(√2)+(√6)≈3.8637>1+2(√2) Comparing the above values we see Tthe smallest value of Q equal to 1+2(√2) when (x,y,z)=(0,0,1) and all the its permutation$
	$Put x + y = 2 a, x + z = 2 b, y + z = 2 c$ $\Rightarrow a + b + c = x + y + z = 1; a, b, c ⩾ 0$ $Q = \sqrt{1 + y + z} + \sqrt{1 + x + z} + \sqrt{1 + x + y}$ $= \sqrt{1 + 2 a} + \sqrt{1 + 2 b} + \sqrt{1 + 2 c} with a + b + c = 1$ $Q (a, b) = \sqrt{1 + 2 a} + \sqrt{1 + 2 b} + \sqrt{3 - 2 (a + b)}$ $We find extremum of$ ${\begin{cases} \frac{\partial Q}{\partial a} = 0 \\ \frac{\partial Q}{\partial b} = 0 \end{cases}$ $\Leftrightarrow {\begin{cases} \frac{1}{\sqrt{1 + 2 a}} - \frac{1}{\sqrt{3 - 2 (a + b)}} = 0 (1) \\ \frac{1}{\sqrt{1 + 2 b}} - \frac{1}{\sqrt{3 - 2 (a + b)}} = 0 (2) \end{cases}$ $\Rightarrow \frac{1}{\sqrt{1 + 2 a}} = \frac{1}{\sqrt{1 + 2 b}} \Rightarrow a = b$ $Replace into (1) we get$ $\sqrt{3 - 4 a} = \sqrt{1 + 2 a} \Leftrightarrow 3 - 4 a = 1 + 2 a$ $\Leftrightarrow 6 a = 2 \Rightarrow a = b = 1 / 3$ $A = \frac{\partial^{2} Q}{\partial a^{2}} ∣_{(\frac{1}{3}, \frac{1}{3})} = \frac{- 1}{(1 + 2 a) \sqrt{1 + 2 a}} - \frac{1}{[3 - 2 (a + b)] \sqrt{3 - 2 (a + b)}} ∣_{(\frac{1}{3}, \frac{1}{3})}$ $= \frac{- 9}{5 \sqrt{15}} - \frac{9}{5 \sqrt{15}} = \frac{- 18}{5 \sqrt{15}} < 0$ $C = \frac{\partial^{2} Q}{\partial b^{2}} ∣_{(\frac{1}{3}, \frac{1}{3})} = \frac{- 1}{(1 + 2 b) \sqrt{1 + 2 b}} - \frac{1}{[3 - 2 (a + b)] \sqrt{3 - 2 (a + b)}} ∣_{(\frac{1}{3}, \frac{1}{3})}$ $= \frac{- 18}{\sqrt{515}} < 0$ $\frac{\partial^{2} Q}{\partial a \partial b} ∣_{(\frac{1}{3}, \frac{1}{3})} = \frac{\partial Q}{\partial b} (\frac{1}{\sqrt{1 + 2 a}} - \frac{1}{\sqrt{3 - 2 (a + b)}})$ $B = \frac{1}{\sqrt{3 - 2 (a + b)}} ∣_{(\frac{1}{3}, \frac{1}{3})} = \frac{3}{\sqrt{15}}$ $Δ = AC - B^{2} = {[\frac{18}{5 \sqrt{15}}]}^{2} - \frac{9}{15} = \frac{324 - 225}{25.15} > 0$ $Since A < 0, Q has maximum$ $Q_{\max} = Q (\frac{1}{3}, \frac{1}{3}) = \sqrt{15} when (x, y, z)$ $= (\frac{1}{3}, \frac{1}{3}, \frac{1}{3})$ $On the other hand, consider Q (x, y, z) at$ $the bounded points we have$ $Q (0, 1, 0) = Q (0, 1, 0) = Q (0, 0, 1) =$ $= 1 + 2 \sqrt{2} = 3.828 . . < \sqrt{15} \approx 3.873$ $Q (\frac{1}{2}, \frac{1}{2}, 0) = \sqrt{2} + \sqrt{6} \approx 3.8637 > 1 + 2 \sqrt{2}$ $Comparing the above values we see$ $Tthe smallest value of Q equal to 1 + 2 \sqrt{2}$ $when (x, y, z) = (0, 0, 1) and all the its$ $permutation$
	Answered by ebi last updated on 27/Oct/20

	$Q = {(2 - x)}^{\frac{1}{2}} + {(2 - y)}^{\frac{1}{2}} + {(2 - z)}^{\frac{1}{2}}$ $c o n s t r a i n t, x + y + z = 1 = F$ $t h e s o l u t i o n l i e s o n 0 ⩽ x, y, z ⩽ 1$ $▽ Q = λ ▽ F$ $⟨ Q_{x}, Q_{y}, Q_{z} ⟩ = λ ⟨ F_{x}, F_{y}, F_{z} ⟩$ $λ = l a g r a n g e m u l t i p l i e r$ $Q_{x} = λ F_{x}$ $- \frac{1}{2 \sqrt{2 - x}} = λ$ ${(- \frac{1}{2 \sqrt{2 - x}})}^{2} = λ^{2}$ $λ^{2} = \frac{1}{4 (2 - x)}$ $x = 2 - \frac{1}{4 λ^{2}} . . . . . . (1)$ $Q_{y} = λ F_{y}$ $- \frac{1}{2 \sqrt{2 - y}} = λ$ $λ^{2} = \frac{1}{4 (2 - y)}$ $y = 2 - \frac{1}{4 λ^{2}} . . . . . . (2)$ $Q_{z} = λ F_{z}$ $- \frac{1}{2 \sqrt{2 - z}} = λ$ $λ^{2} = \frac{1}{4 (2 - z)}$ $z = 2 - \frac{1}{4 λ^{2}} . . . . . . (3)$ $x + y + z = 1 . . . . . . (4)$ $s u b s t i t u t e (1), (2), (3) i n t o (4)$ $(2 - \frac{1}{4 λ^{2}}) + (2 - \frac{1}{4 λ^{2}}) + (2 - \frac{1}{4 λ^{2}}) = 1$ $- \frac{3}{4 λ^{2}} = - 5$ $λ^{2} = \frac{3}{20}$ $x = 2 - \frac{1}{4 (\frac{3}{20})} = \frac{1}{3}$ $y = 2 - \frac{1}{4 (\frac{3}{20})} = \frac{1}{3}$ $z = 2 - \frac{1}{4 (\frac{3}{20})} = \frac{1}{3}$ $∴ t h e s o l u t i o n i s (\frac{1}{3}, \frac{1}{3}, \frac{1}{3})$ $G (\frac{1}{3}, \frac{1}{3}, \frac{1}{3}) = \sqrt{2 - \frac{1}{3}} + \sqrt{2 - \frac{1}{3}} + \sqrt{2 - \frac{1}{3}}$ $= 3 \sqrt{\frac{5}{3}} = \sqrt{15} \approx 3.873$ $b e c a u s e o f t h i s p o i n t i s t h e o n l y$ $s t a t i o n a r y p o i n t, w e h a v e t o c o n s i d e r$ $o t h e r p o i n t s f o r m a x i m a o r m i n i m a$ $c o n s i d e r t h e s e 3 p o i n t s :$ $(1, 0, 0), (0, 1, 0), (0, 0, 1)$ $G (1, 0, 0) = G (0, 1, 0) = G (0, 0, 1)$ $= 1 + 2 \sqrt{2} \approx 3.828$ $t h u s,$ $G (\frac{1}{3}, \frac{1}{3}, \frac{1}{3}) = 3.873 \to m a x i m u m$ $G (1, 0, 0) = G (0, 1, 0) = G (0, 0, 1) = 3.828 \to m i n i m u m$
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