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Let  \(x,y,z\) be nonnegative real numbers such that \(x + y + z = 3\) Find the maximum value of \((xy + z)(xz + y)\)
 

 Aug 7, 2019
 #1
avatar+23293 
+2

Let \(x,y,z \) be nonnegative real numbers such that \(x + y + z = 3\)

Find the maximum value of \((xy + z)(xz + y)\)

 

\(\begin{array}{|rcll|} \hline \huge{GM} &\huge{\leq}& \huge{AM} \\\\ \sqrt{(xy + z)(xz + y)} & \leq & \dfrac{(xy + z)+(xz + y)}{2} \\ \sqrt{(xy + z)(xz + y)} & \leq & \dfrac{ xy + z + xz + y }{2} \\ \sqrt{(xy + z)(xz + y)} & \leq & \dfrac{x(y+z)+(y+z)}{2} \\ \sqrt{(xy + z)(xz + y)} & \leq & \dfrac{ (y+z)(x+1)}{2} \quad | \quad \boxed{y + z = 3-x} \\ \sqrt{(xy + z)(xz + y)} & \leq & \dfrac{ (3-x)(x+1)}{2} \\ &&\begin{array}{|rcll|} \hline \large{GM} &\large{\leq}& \large{AM} \\\\ \sqrt{(3-x)(x+1)} & \leq & \dfrac{ (3-x)+(x+1)}{2} \\ \sqrt{(3-x)(x+1)} & \leq & \dfrac{ 3-x + x+1 }{2} \\ \sqrt{(3-x)(x+1)} & \leq & \dfrac{ 4 }{2} \\ \sqrt{(3-x)(x+1)} & \leq & 2 \quad | \quad \text{square both sides}\\ (3-x)(x+1) & \leq & 4 \\ \mathbf{ \dfrac{(3-x)(x+1)}{2}} & \leq & \mathbf{ 2 } \\ \hline \end{array} \\ \sqrt{(xy + z)(xz + y)} & \leq & \dfrac{ (3-x)(x+1)}{2} \leq 2 \quad | \quad \text{ The maximum value results from the equals sign} \\ \sqrt{(xy + z)(xz + y)} & = & \dfrac{ (3-x)(x+1)}{2} = 2 \quad | \quad \text{square both sides}\\ \mathbf{(xy + z)(xz + y) } &=& \mathbf{4} \\ \hline \end{array}\)

 

The maximum value of \(\mathbf{(xy + z)(xz + y)=4}\)

 

laugh

 Aug 9, 2019
edited by heureka  Aug 9, 2019
 #2
avatar+166 
+1

Thankyou

 Aug 9, 2019

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