MHB Finding the Minimum Value of a Complex Expression

[anemone]

Determine the minimum value of $\left( \sqrt{x^2-8x+27-6\sqrt{2}}+ \sqrt{x^2-4x+7-2\sqrt{2}} \right)^4$ where $x$ is a real number.

 

[Ackbach]

Spoiler

Let
$$f(x)= \left( \sqrt{x^{2}-8x+27-6 \sqrt{2}}+ \sqrt{x^{2}-4x+7-2 \sqrt{2}} \right)^{4},$$
and let
\begin{align*}
g(x)&=x^{2}-8x+27-6 \sqrt{2} \\
h(x)&=x^{2}-4x+7-2 \sqrt{2}.
\end{align*}
By completing the square, we can determine that $g(x)>0$ and $h(x)>0$ for all $x \in \mathbb{R}$. Hence, $ \mathcal{D}(f)= \mathbb{R}$.
Since the fourth root function is monotone increasing, it follows that $f(x)$ has a minimum at the same location as $\hat{f}(x)= \sqrt[4]{f(x)}$.
Taking the derivative $\hat{f}'(x)$ and setting it equal to zero yields the equation
$$(x-4) \sqrt{x^{2}-4x+7-2 \sqrt{2}}+(x-2) \sqrt{x^{2}-8x+27-6 \sqrt{2}}=0.$$
Squaring both sides and expanding out yields the equation
$$(4 \sqrt{2}-8)x^{2}+(8 \sqrt{2}-20)x+8 \sqrt{2}-4=0.$$
The solutions are
$$x=1+ \sqrt{2} \quad \text{or} \quad x= \frac{4- \sqrt{2}}{2}.$$
Plugging these values into $\hat{f}$ yield
\begin{align*}
\hat{f}(1+ \sqrt{2})&=2 \sqrt{2} \approx 2.8284 \\
\hat{f}((4- \sqrt{2})/2)&= \frac{ \left((31-8 \sqrt{2}) \sqrt{31+8 \sqrt{2}}
+(49-28 \sqrt{2}) \sqrt{7+4 \sqrt{2}} \right) \sqrt{34}}{2 \cdot 7 \cdot 17}
\approx 3.9569.
\end{align*}
Hence, the minimum of $f$ is $(2 \sqrt{2})^{4}=(2^{3/2})^{4}=64.$

 

[anemone]

Spoiler

Let
$$f(x)= \left( \sqrt{x^{2}-8x+27-6 \sqrt{2}}+ \sqrt{x^{2}-4x+7-2 \sqrt{2}} \right)^{4},$$
and let
\begin{align*}
g(x)&=x^{2}-8x+27-6 \sqrt{2} \\
h(x)&=x^{2}-4x+7-2 \sqrt{2}.
\end{align*}
By completing the square, we can determine that $g(x)>0$ and $h(x)>0$ for all $x \in \mathbb{R}$. Hence, $ \mathcal{D}(f)= \mathbb{R}$.
Since the fourth root function is monotone increasing, it follows that $f(x)$ has a minimum at the same location as $\hat{f}(x)= \sqrt[4]{f(x)}$.
Taking the derivative $\hat{f}'(x)$ and setting it equal to zero yields the equation
$$(x-4) \sqrt{x^{2}-4x+7-2 \sqrt{2}}+(x-2) \sqrt{x^{2}-8x+27-6 \sqrt{2}}=0.$$
Squaring both sides and expanding out yields the equation
$$(4 \sqrt{2}-8)x^{2}+(8 \sqrt{2}-20)x+8 \sqrt{2}-4=0.$$
The solutions are
$$x=1+ \sqrt{2} \quad \text{or} \quad x= \frac{4- \sqrt{2}}{2}.$$
Plugging these values into $\hat{f}$ yield
\begin{align*}
\hat{f}(1+ \sqrt{2})&=2 \sqrt{2} \approx 2.8284 \\
\hat{f}((4- \sqrt{2})/2)&= \frac{ \left((31-8 \sqrt{2}) \sqrt{31+8 \sqrt{2}}
+(49-28 \sqrt{2}) \sqrt{7+4 \sqrt{2}} \right) \sqrt{34}}{2 \cdot 7 \cdot 17}
\approx 3.9569.
\end{align*}
Hence, the minimum of $f$ is $(2 \sqrt{2})^{4}=(2^{3/2})^{4}=64.$

Wow...what a brilliant way to tackle this challenge problem, well done, Ackbach!

I learned something valuable from your method, thank you for your solution and thank you for participating!:)