+0  
 
0
196
4
avatar+167 

Let x,y, and z be positive real numbers such that \(\frac{1}{x^4} + \frac{1}{y^4} + \frac{1}{z^4} = 1.\)
Find the minimum value of \(\frac{x^4 y^4 + x^4 z^4 + y^4 z^4}{x^3 y^2 z^3}.\)

 Oct 15, 2019
 #1
avatar+437 
+8

Look at this link it might help you?

 

https://web2.0calc.com/questions/1-let-x-y-and-z-be-real-numbers-such-that-find-the

 Oct 15, 2019
 #2
avatar
+1

x = 3^1/4,   y = 3^1/4,    z=3^1/4.

 

The minimum value of:  \frac{x^4 y^4 + x^4 z^4 + y^4 z^4}{x^3 y^2 z^3} = 3

 Oct 15, 2019
 #3
avatar+6046 
+1

....

Rom  Oct 16, 2019
edited by Rom  Oct 18, 2019
 #4
avatar+23903 
+3

Let x,y, and z be positive real numbers such that \(\dfrac{1}{x^4} + \dfrac{1}{y^4} + \dfrac{1}{z^4}\) = 1.
Find the minimum value of \(\dfrac{x^4 y^4 + x^4 z^4 + y^4 z^4}{x^3 y^2 z^3}\).

 

1.  \(x^4 y^4 + x^4 z^4 + y^4 z^4 = \ ?\)

\(\begin{array}{|rcll|} \hline \dfrac{1}{x^4} + \dfrac{1}{y^4} + \dfrac{1}{z^4} &=& \dfrac{x^4 y^4 + x^4 z^4 + y^4 z^4}{x^4 y^4 z^4} \\\\ 1 &=& \dfrac{x^4 y^4 + x^4 z^4 + y^4 z^4}{x^4 y^4 z^4} \\\\ \mathbf{x^4 y^4 + x^4 z^4 + y^4 z^4} &=& \mathbf{ x^4 y^4 z^4 } \\ \hline \end{array}\)

 

2. rearrange \(\dfrac{x^4 y^4 + x^4 z^4 + y^4 z^4}{x^3 y^2 z^3}\)

\(\begin{array}{|rcll|} \hline \dfrac{x^4 y^4 + x^4 z^4 + y^4 z^4}{x^3 y^2 z^3} &=& \dfrac{x^4 y^4 z^4 } {x^3 y^2 z^3} \\\\ \mathbf{\dfrac{x^4 y^4 + x^4 z^4 + y^4 z^4}{x^3 y^2 z^3} } &=& \mathbf{x y^2 z} \\ \hline \end{array}\)

 

Lagrange Multipliers

\(\begin{array}{|rcll|} \hline g(x,y,z) &=& \dfrac{1}{x^4} + \dfrac{1}{y^4} + \dfrac{1}{z^4} - 1 = 0 \\\\ f(x,y,z) &=& xy^2z \rightarrow \text{MIN.} \\\\ L(x,y,z,\lambda) &=& \underbrace{xy^2z}_{f(x,y,z)} + \lambda \left( \underbrace{\dfrac{1}{x^4} + \dfrac{1}{y^4} + \dfrac{1}{z^4} - 1 }_{g(x,y,z)}\right) \\\\ \hline \frac{\partial L(x,y,z,\lambda)}{\partial x} &=& y^2z-\dfrac{4}{x^5} \lambda \\\\ \frac{\partial L(x,y,z,\lambda)}{\partial y} &=& 2xyz-\dfrac{4}{y^5} \lambda \\\\ \frac{\partial L(x,y,z,\lambda)}{\partial z} &=& xy^2-\dfrac{4}{z^5} \lambda \\\\ \frac{\partial L(x,y,z,\lambda)}{\partial \lambda} &=& 0+ \dfrac{1}{x^4} + \dfrac{1}{y^4} + \dfrac{1}{z^4}-1 \\ \hline \end{array} \)

 

\(\begin{array}{|lrcll|} \hline & y^2z-\dfrac{4}{x^5} \lambda &=& 0 \quad | \quad \cdot x^5 \\\\ (1) & y^2x^5z-4\lambda &=& 0 \\ \hline & 2xyz-\dfrac{4}{y^5} \lambda &=& 0 \quad | \quad \cdot y^5 \\\\ (2) & 2xy^6z-4\lambda &=& 0 \\ \hline & xy^2-\dfrac{4}{z^5} \lambda &=& 0 \quad | \quad \cdot z^5 \\\\ (3) & xy^2z^5-4\lambda &=& 0 \\ \hline (4) & \dfrac{1}{x^4} + \dfrac{1}{y^4} + \dfrac{1}{z^4}-1 &=& 0 \\ \hline \end{array}\)

 

\(\begin{array}{|lrcll|} \hline & y^2x^5z-4\lambda &=& 0 \qquad (1) \\ (5) & \mathbf{z} &=& \mathbf{ \dfrac{4\lambda}{y^2x^5} } \\ \hline \hline & 2xy^6z-4\lambda &=& 0 \qquad (2) \\ (7) & \mathbf{x} &=& \mathbf{ \dfrac{4\lambda}{2y^6z} } \\ \hline & 2xy^6z-4\lambda &=& 0 \qquad (2) \quad | \quad \mathbf{z=\dfrac{4\lambda}{y^2x^5} } \\ & 2xy^6\dfrac{4\lambda}{y^2x^5}-4\lambda &=& 0 \\ & 2y^4\dfrac{4\lambda}{x^4} &=& 4\lambda \\ & 2y^4 &=& x^4 \\ (8)& \mathbf{y^4} &=& \mathbf{\dfrac{x^4}{2}} \\ \hline & xy^2z^5-4\lambda &=& 0 \qquad (3) \quad | \quad \mathbf{x=\dfrac{4\lambda}{2y^6z} } \\ & \dfrac{4\lambda}{2y^6z}y^2z^5-4\lambda &=& 0 \\ & \dfrac{4\lambda}{2y^4}z^4 &=& 4\lambda \\ & z^4 &=& 2y^4 \\ (9) & \mathbf{y^4} &=& \mathbf{ \dfrac{z^4}{2} } \\ \hline (8)=(9): & y^4 = \dfrac{x^4}{2} &=& \dfrac{z^4}{2} \\\\ (10) & \mathbf{x^4} &=& \mathbf{z^4} \\ \hline \end{array} \)

 

\(\mathbf{z=\ ?}\)

\(\begin{array}{|rcll|} \hline \dfrac{1}{x^4} + \dfrac{1}{y^4} + \dfrac{1}{z^4}-1 &=& 0 \qquad (4) \quad | \quad \mathbf{x^4=z^4} \\\\ \dfrac{1}{z^4} + \dfrac{1}{y^4} + \dfrac{1}{z^4}-1 &=& 0 \quad | \quad \mathbf{y^4=\dfrac{z^4}{2}} \\\\ \dfrac{1}{z^4} + \dfrac{1}{\dfrac{z^4}{2}} + \dfrac{1}{z^4}-1 &=& 0 \\\\ \dfrac{1}{z^4} + \dfrac{2}{z^4} + \dfrac{1}{z^4} &=& 1 \\\\ \dfrac{4}{z^4} &=& 1 \\\\ z^4 &=& 4 \\ z^2 &=& 2 \\ \mathbf{z} &=& \mathbf{\sqrt{2}} \\ \hline \end{array} \)

 

\(\mathbf{x=\ ?} \)

\(\begin{array}{|rcll|} \hline \mathbf{x^4} &=& \mathbf{z^4} \qquad (10) \quad | \quad \mathbf{z^4 = 4} \\\\ x^4 &=& 4 \\ x^2 &=& 2 \\ \mathbf{x} &=& \mathbf{\sqrt{2}} \\ \hline \end{array}\)

 

\(\mathbf{y=\ ?}\)

\(\begin{array}{|rcll|} \hline \mathbf{y^4} &=& \mathbf{\dfrac{z^4}{2} } \qquad (9) \quad | \quad \mathbf{z^4 = 4} \\ y^4 &=& \dfrac{4}{2} \\ y^4 &=& 2 \\ y &=& \sqrt[4]{2} \\ \hline \end{array}\)

 

\(\begin{array}{|rcll|} \hline \mathbf{\dfrac{x^4 y^4 + x^4 z^4 + y^4 z^4}{x^3 y^2 z^3}} &=& xy^2z \\\\ &=& \sqrt{2} \left( \sqrt[4]{2} \right)^2 \sqrt{2} \\ &=& 2 \cdot \left( \sqrt[4]{2} \right)^2 \\ &=& 2\cdot 2^{\frac{2}{4}} \\ &=& 2\cdot 2^{\frac{1}{2}} \\ &=& \mathbf{2 \sqrt{2}} \\ \hline \end{array}\)

 

 

laugh

 Oct 17, 2019

21 Online Users

avatar