help me with idea to understand: Determine the real number x for which the value of the expression $E(x)=\sqrt{x^2-4x+5}+\sqrt{x^2-6x+13}$ is minimum

Question

PROBLEM

Determine the real number x for which the value of the expression $E(x)=\sqrt{x^2-4x+5}+\sqrt{x^2-6x+13}$ is minimum.

WHAT I THOUGHT OF

So for $E(x)$ to be the minimum, we have to make $\sqrt{x^2-4x+5}$ and $\sqrt{x^2-6x+13}$ as small as possible $=>$ $x^2-4x+5$ and $x^2-6x+13$ have to be as small as possible.

I thought of writing

$x^2-4x+5=x^2-4x+2^2+1=(x-2)^2+1$

$x^2-6x+13=x^2-6x+3^2+4=(x-3)^2+4$

For $(x-2)^2+1$ to be the smallest $(x-2)^2$ has to be 0 $=>$ $x=2$

But, for $(x-3)^2+4$ to be the smallest $(x-3)^2$ has to be 0 $=>$ $x=3$

Now we can compare the numbers we get when $x=2$ and when $x=3$ and see which one is the smallest. That will be the answer. I don't know if what I thought of is correct. Can you help me solve the problem or tell me if what I thought of is good? Hope one of you can help me! Thank you!

Answer 1

Remarks: We may use Minkowski inequality.

Here is a similar problem.

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By Minkowski inequality, we have $$E(x) = \sqrt{(x - 2)^2 + 1^2} + \sqrt{(3 - x)^2 + 2^2} \ge \sqrt{(x - 2 + 3 - x)^2 + (1 + 2)^2} = \sqrt{10}.$$ (Note: Equality holds when $(x-2, 3 - x)$ and $(1, 2)$ are proportional, that is $\frac{x - 2}{1} = \frac{3 - x}{2}$ which results in $x = 7/3$. Indeed, $E(7/3) = \sqrt{10}$.)

Thus, the minimum of $E(x)$ is $\sqrt{10}$.

Answer 2

Taking the derivate: $$(x-3)\sqrt{x^2-4x+5}+(x-2)\sqrt{x^2-6x+13}=0$$ Square both sides and obtain: $$(x-3)^2\cdot(x^2-4x+5)=(x-2)^2\cdot(x^2-6x+13)\longleftrightarrow -3x^2+10x-7=0$$ So: $$x=1\,\,\,\lor \,\,\,x=\frac{7}{3}$$ Can you finish it now?

Answer 3

Here's an approach that doesn't use calculus:

$E(x)=\sqrt{(x-2)^2+1}+\sqrt{(x-3)^2+4}$ is the sum of distances of $(x,0)$ from $(2,\pm1)$ and $(3,\pm2)$.

We have that $E(x)$ is minimised when the three points $(x,0),(2,\pm1)$ and $(3,\pm2)$ are collinear.

Checking the possibilities depending on the signs of the $y$ coordinates, we find $x=\frac73$ or $1$ in order for the points to be collinear. Comparing $E(1)$ and $E\left(\frac73\right)$, we find the minimum is $E\left(\frac73\right)=\sqrt{10}$.