There must be a pattern here!

Algebra Level 4

$\left[ \sqrt{(x+y)(x+z)} + \sqrt{(y+z)(2y)} \right] + \left[ \sqrt{(x+2y)(x+y+z)} + \sqrt{yz} \right]$

Let $x, \space y,$ and $z$ be real positive number, such that $x + 2y + z = 9$ . Find the maximum value of the expression above.

Hint: Two inequalities are summed up.

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The answer is 18.

2 solutions

Fidel Simanjuntak
Jun 11, 2017

With Cauchy-Schwarz Inequality, given $\sqrt{ac} + \sqrt{bd} \leq \sqrt{(a + b)(c + d)}$ .

First, let $\color{#D61F06} a = x + y \color{#333333}; \space \color{#3D99F6} b = y + z \color{#333333}; \space \color{#20A900} c = x + z\color{#333333}; \space \color{#EC7300} d = 2y$

Then, we have

$\begin{aligned} \sqrt{(x+2y+z)(x+2y+z)} & \geq \sqrt{\color{#D61F06}(x+y)\color{#20A900}(x+z)} + \sqrt{\color{#3D99F6}(y+z)\color{#EC7300}(2y)} \\ 9 & \geq \sqrt{(x+y)(x+z)} + \sqrt{(y+z)(2y)} \space ...(1) \end{aligned}$

Second, let $\color{#D61F06} a = x + 2y \color{#333333}; \space \color{#3D99F6} b = z \color{#333333}; \space \color{#20A900} c = x + y + z\color{#333333}; \space \color{#EC7300} d = y$

Then, we have

$\begin{aligned} \sqrt{(x+2y+z)(x+2y+z)} & \geq \sqrt{\color{#D61F06}(x+2y)\color{#20A900}(x+y+z)} + \sqrt{\color{#3D99F6}(z)\color{#EC7300}(y)} \\ 9 & \geq \sqrt{(x+2y)(x+y+z)} + \sqrt{yz} \space ...(2) \end{aligned}$

$(1) + (2)$ gives

$\left[ \sqrt{(x+y)(x+z)} + \sqrt{(y+z)(2y)} \right] + \left[ \sqrt{(x+2y)(x+y+z)} + \sqrt{yz} \right] \leq 18$

Hence, the answer is $\boxed{18}$

@Fidel Simanjuntak Good solution! Anyways, it should be "There must be a pattern here!".