Subharthi's minimum

Note that , $x^2 - 4x + 7 - 2\sqrt{2} = (x - 2)^2 + (\sqrt{2} - 1)^2 ,$

$x^2 - 8x + 27 - 6\sqrt{2} = (x - 4)^2 + (\sqrt{2} - 3)^2 .$

Hence, $\sqrt{x^2 - 4x + 7 - 2\sqrt{2} } + \sqrt{x^2 - 8x + 27 - 6\sqrt{2}}$ is sum of distance from $(x, \sqrt{2})$ to $(2,1) , (4,3)$ in co-ordinate plane .

Assign name A to $(x,\sqrt{2})$ , B to $(2,1)$ , C to $(4,3) .$

_From Triangular inequality , _

$AB + AC \geq BC$ ,* Equality holds when A,B,C are co-linear *.

As , $1 <\sqrt{2} <3$ , we must can find x for which A is on line segment BC .

(of course x can be found by taking slope of AB = slope of AC, but we are not interested in *value of x ) * .

Hence, $min(AB + AC) = BC = \sqrt{8}$

& $(min(AB + AC))^4 = 64$

We will use two inequalities to solve this problem:

Cauchy-Schwarz Inequality: $x^2+y^2\geq \frac{1}{2} (x+y)^2$ (Equality occurs at $x=y$ )

Absolute Value Inequality: $|a|+|b|\geq |a+b|$ (Equality occurs when $ab \geq 0$ )

$[\sqrt{x^2-4x+7-2\sqrt{2}}+\sqrt{x^2-8x+27-6\sqrt{2}}]^4$

$=[\sqrt{(x-2)^2+(\sqrt{2}-1)^2}+\sqrt{(x-4)^2+(\sqrt{2}-3)^2}]^4$

$\geq [\frac{1}{\sqrt{2}}(|x-2+\sqrt{2}-1|+|x-4-3+\sqrt{2}|)]^4$ (Cauchy-Schwarz)

$=[\frac{1}{\sqrt{2}}(|x+\sqrt{2}-3|+|-x-\sqrt{2}+7|)]^4$ ( $|a|=|-a| \forall$ a)

$\geq [\frac{1}{\sqrt{2}}|x+\sqrt{2}-3-x-\sqrt{2}+7|]^4$ (Absolute value inequalities)

$= [\frac{1}{\sqrt{2}}|4|]^4=64$

We can easily check that equality occurs when $x=\sqrt{2}+1$ . Hence the minimum value is 64.

Rearranging the given expression

$\sqrt{\strut x^2-4x+7-2\sqrt{2}}+\sqrt{\strut x^2-8x+27-6\sqrt2}$

we get

$\sqrt{\strut (x-2)^2+(\sqrt{2}-1)^2}+\sqrt{\strut (x-4)^2+(\sqrt2-3)^2}$

Hence, we set some points on the Cartesian Plane,

$P=(x,\sqrt2)$

$A=(2,1)$

$B=(4,3)$

To get

$AP=\sqrt{\strut (x-2)^2+(\sqrt{2}-1)^2}$

$PB=\sqrt{\strut (x-4)^2+(\sqrt2-3)^2}$

Subharti'sMinimum

By Triangle Inequality ,

$AP+PB$

$\geq AB$

$\geq\sqrt{\strut (4-2)^2+(3-1)^2}$

$\geq\sqrt8$

So the minimum value

$\sqrt{\strut x^2-4x+7-2\sqrt{2}}+\sqrt{\strut x^2-8x+27-6\sqrt2}=\sqrt8$

Hence, the minimum value

$\Big [\sqrt{\strut x^2-4x+7-2\sqrt{2}}+\sqrt{\strut x^2-8x+27-6\sqrt2}\Big ]^4=64$

We have f (x) = [√(x²-4x+7-2√2) + √(x²-8x+27-6√2)]⁴

We note : f (x) attains minimum value when √(x²-4x+7-2√2) + √(x²-8x+27-6√2) attains minimum value

Re-write it as √[(x-2)²+(√2-1)²] + √[(x-4)²+(√2-3)²]

= Sum of the distances of the point (x,√2) from the points (2,1) and (4,3) which is minimum only when the three points are collinear

=> x = √2 + 1

=> min [√(x²-4x+7-2√2) + √(x²-8x+27-6√2)] = 2√2

=> f_min (x) = (2√2)⁴ = 64

Can somebody help me find solutions to this question?

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to get minima of the function, we use derivative of the function $g(x)$ by using:

let it $f(x) = (g(x))^4, g(x) = \sqrt{x^2-4x+7-2\sqrt{2}} + \sqrt{x^2-8x+27-6\sqrt{2}}$

so $\frac{d}{dx} g(x) = 0$ is $1+\sqrt{2}$ and by replacing in $g(x)$ we've $g(1+\sqrt{2}) = 2\sqrt{2}$ ,

finally $f(2\sqrt{2}) = 64$

denote $g=x^2-4x+7-2\sqrt{2}$ and $h=x^2-8x+27-6\sqrt{2}$ and $f=\sqrt{g}+\sqrt{h}$ . observe that $f>0$ for all $x$

so I will find minimum of $f$ replace $f^4$ (by calculus)

we get $\frac{df}{dx}=\frac{x-2}{\sqrt{g}}+\frac{x-4}{\sqrt{h}}$

and have to solve the equation

$\frac{x-2}{\sqrt{g}}=\frac{4-x}{\sqrt{h}}$ (observe that $x\in(2,4)$ ) $\frac{(4-x)^2}{(x-2)^2}-3=\frac{h}{g}-3 \rightarrow \frac{-2x^2+4x+4}{x^2-4x+4}=\frac{-2x^2+4x+6}{g}$ $\rightarrow \frac{-2x^2+4x+6}{-2x^2+4x+4}-1=\frac{g}{x^2-4x+4}-1 \rightarrow \frac{1}{-x^2+2x+2}=\frac{3-\sqrt{2}}{x^2-4x+4}$

to use your brute force to solve the last equation (easy to know the method) you will get $x=2-\frac{1}{\sqrt{2}},\sqrt{2}+1$ but $x\in(2,4)$ so $x=\sqrt{2}+1$ that $f(\sqrt{2}+1)=2\sqrt{2}$ and $f^4=64$