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Question Number 122445 by benjo_mathlover last updated on 17/Nov/20

	Answered by TANMAY PANACEA last updated on 17/Nov/20

	$u s i n g T i t u l e m m a$ $\frac{1^{2}}{x} + \frac{3^{2}}{y} + \frac{5^{2}}{z} ⩾ \frac{{(1 + 3 + 5)}^{2}}{x + y + z}$ $\frac{1^{2}}{x} + \frac{3^{2}}{y} + \frac{5^{2}}{z} ⩾ 81$
	Commented by TANMAY PANACEA last updated on 17/Nov/20

	Commented by rs4089 last updated on 18/Nov/20

	$s i r, t h a n k y o u v e r y m u c h f o r e x p l a i n a t i o n :)$
	Answered by Olaf last updated on 17/Nov/20
	$1) 1st method : Let f(x,y) = (1/x)+(9/y)+((25)/z) with z = 1−x−y ⇒ f(x,y) = (1/x)+(9/y)+((25)/(1−x−y)) (∂f/∂x)(x,y) = −(1/x^2 )+((25)/((1−x−y)^2 )) (∂f/∂y)(x,y) = −(9/y^2 )+((25)/((1−x−y)^2 )) We solve : { (((∂f/∂x)(x,y) = 0 = −(1/x^2 )+((25)/((1−x−y)^2 )) (1))),(((∂f/∂y)(x,y) = 0 = −(9/y^2 )+((25)/((1−x−y)^2 )) (2))) :}_ (1)−(2) : y = ±3x but x>0, y>0 ⇒ y = 3x With (1) : −(1/x^2 )+((25)/((1−4x)^2 )) = 0 25x^2 −(1−4x)^2 = 0 9x^2 +8x−1 = 0 9(x+1)(x−(1/9)) = 0 x>0 ⇒ x = (1/9) ⇒ y = (1/3) ⇒ z = (5/9) ⇒ f(x,y,z) = f((1/9),(1/3),(5/9)) = (1/(1/9))+(9/(1/3))+((25)/(5/9)) = 81 2) 2nd method The function is symmetric with respect to x, y = 3x, z = 5x ⇒ x+y+z = 1 = 9x We find x = (1/9) etc... 3) 3rd method Let f(x,y,z) = (1/x)+(9/y)+((25)/z) Let the constraint c(x,y,z) = x+y+z−1 = 0 Let g(x,y,z,λ) = f(x,y,z)+λc(x,y,z) (∂g/∂x)(x,y,z,λ) = −(1/x^2 )+λ (∂g/∂y)(x,y,z,λ) = −(9/y^2 )+λ (∂g/∂z)(x,y,z,λ) = −((25)/z^2 )+λ (∂g/∂λ)(x,y,z,λ) = x+y+z−1 (∂g/∂x)(x,y,z,λ) = 0 ⇒ x = (1/( (√λ))) (∂g/∂y)(x,y,z,λ) = 0 ⇒ y = (3/( (√λ))) (∂g/∂z)(x,y,z,λ) = 0 ⇒ z = (5/( (√λ))) (∂g/∂λ)(x,y,z,λ) = 0 ⇒ (1/( (√λ)))+(3/( (√λ)))+(5/( (√λ)))−1 = 0 ⇒ (1/( (√λ))) = (1/9) ⇒ x = (1/9) ⇒ y = (1/3) ⇒z = (5/9)$
	$1) 1 st method :$ $Let f (x, y) = \frac{1}{x} + \frac{9}{y} + \frac{25}{z}$ $with z = 1 - x - y$ $\Rightarrow f (x, y) = \frac{1}{x} + \frac{9}{y} + \frac{25}{1 - x - y}$ $\frac{\partial f}{\partial x} (x, y) = - \frac{1}{x^{2}} + \frac{25}{{(1 - x - y)}^{2}}$ $\frac{\partial f}{\partial y} (x, y) = - \frac{9}{y^{2}} + \frac{25}{{(1 - x - y)}^{2}}$ $We solve :$ ${\begin{cases} \frac{\partial f}{\partial x} (x, y) = 0 = - \frac{1}{x^{2}} + \frac{25}{{(1 - x - y)}^{2}} (1) \\ \frac{\partial f}{\partial y} (x, y) = 0 = - \frac{9}{y^{2}} + \frac{25}{{(1 - x - y)}^{2}} (2) \end{cases}_{}$ $(1) - (2) : y = \pm 3 x$ $but x > 0, y > 0 \Rightarrow y = 3 x$ $With (1) :$ $- \frac{1}{x^{2}} + \frac{25}{{(1 - 4 x)}^{2}} = 0$ $25 x^{2} - {(1 - 4 x)}^{2} = 0$ $9 x^{2} + 8 x - 1 = 0$ $9 (x + 1) (x - \frac{1}{9}) = 0$ $x > 0 \Rightarrow x = \frac{1}{9} \Rightarrow y = \frac{1}{3} \Rightarrow z = \frac{5}{9}$ $\Rightarrow f (x, y, z) = f (\frac{1}{9}, \frac{1}{3}, \frac{5}{9}) = \frac{1}{1 / 9} + \frac{9}{1 / 3} + \frac{25}{5 / 9} = 81$ $2) 2 nd method$ $The function is symmetric with respect$ $to x, y = 3 x, z = 5 x$ $\Rightarrow x + y + z = 1 = 9 x$ $We find x = \frac{1}{9} etc . . .$ $3) 3 rd method$ $Let f (x, y, z) = \frac{1}{x} + \frac{9}{y} + \frac{25}{z}$ $Let the constraint c (x, y, z) = x + y + z - 1 = 0$ $Let g (x, y, z, λ) = f (x, y, z) + λ c (x, y, z)$ $\frac{\partial g}{\partial x} (x, y, z, λ) = - \frac{1}{x^{2}} + λ$ $\frac{\partial g}{\partial y} (x, y, z, λ) = - \frac{9}{y^{2}} + λ$ $\frac{\partial g}{\partial z} (x, y, z, λ) = - \frac{25}{z^{2}} + λ$ $\frac{\partial g}{\partial λ} (x, y, z, λ) = x + y + z - 1$ $\frac{\partial g}{\partial x} (x, y, z, λ) = 0 \Rightarrow x = \frac{1}{\sqrt{λ}}$ $\frac{\partial g}{\partial y} (x, y, z, λ) = 0 \Rightarrow y = \frac{3}{\sqrt{λ}}$ $\frac{\partial g}{\partial z} (x, y, z, λ) = 0 \Rightarrow z = \frac{5}{\sqrt{λ}}$ $\frac{\partial g}{\partial λ} (x, y, z, λ) = 0 \Rightarrow \frac{1}{\sqrt{λ}} + \frac{3}{\sqrt{λ}} + \frac{5}{\sqrt{λ}} - 1 = 0$ $\Rightarrow \frac{1}{\sqrt{λ}} = \frac{1}{9} \Rightarrow x = \frac{1}{9} \Rightarrow y = \frac{1}{3} \Rightarrow z = \frac{5}{9}$
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