First, we should try to complete the squares inside the square roots. The the LHS becomes:

$\left(\sqrt{(x-3)^2+(x^2-2)^2} - \sqrt{x^2+(x^2-1)^2}\right)^4$

Notice the first square root is the distance of the point $\left(x, x^2\right)$ to the point $(3,2)$ and the second square root is the distance of the point $\left(x, x^2\right)$ to the point $(0,1)$ . Basically, we try to find the maximum difference between the points on the parabola $y=x^2$ to the two fixed points. By the triangle inequality, the absolute value of the maximum difference is simply the distance between $(0, 1)$ and $(3,2)$ , which is $\sqrt{10}$ , and this distance can be obtained on the parabola. So the answer is $\sqrt{10}^4 + 1$ which is $\boxed{101}$ .

Notice that

$x^4-3x^2-6x+13=(x^2-2)^2+(x-3)^2$ $x^4-x^2+1=(x^2-1)^2+x^2$

We let the left side of the inequality be equal to $P^4$ , then $P= \biggl| \sqrt{(x^2-2)^2+(x-3)^2} - \sqrt{(x^2-1)^2+x^2} \biggr|$

is the difference between the distances between $(3,2),(x,x^2)$ and $(0,1),(x,x^2)$ .

We let $A=(3, 2), B=(0,1), C=(x, x^2)$ . Then

$P=\bigl|\overline{AC}-\overline{BC}\bigr| \leq \overline{AB}$

by triangle inequality. Thus the maximum of $P$ is $\sqrt{(3-0)^2+(2-1)^2}=\sqrt{10}$ , which can be achieved when $x=\frac16 - \frac{\sqrt{37}}{6}$ .

As a result, $P^4$ reaches a maximum value of $100$ . Thus $N$ has to be at least $\boxed{101}$ .

One possible idea of such inequality questions with square roots is using coordinate geometry, or more specifically, using the formula of Distance Between Two Points and then Triangle Inequality.

On a side note, this question (and the following solution which I attempt to give) is similar to Subharthi's minimum , a question posted 2 weeks ago.

Step 1 Observe that we can factorise the given expression (well, for simplicity's sake, just the expression within the brackets) into the following:

$\sqrt{(x^{2} - 2)^{2} + (x-3)^{2})} - \sqrt{(x^{2} -1)^{2} + x^{2})}$ .

Step 2 Applying the Distance Formula , we easily get that the expression equals to the difference between the distance of the line connecting the points $(x^{2}, x)$ and $(2,3)$ and that of the line connecting points $(x^{2}, x)$ and $(1,0)$ .

Step 3 Now, the hardest part of the solution is to find the maximum value of the expression, which translates to the maximum distance between the 2 lines (mentioned in Step 2). Recall that $x$ ranges over all real values, so the exact value of $x$ or the length of each line does not matter. Hence, Triangle Inequality will do the trick as follows:

To enhance visualisation, lets put some labels on the 3 points: Let point $(x^{2},x)$ be $A$ , $(2,3)$ be $B$ and $(1,0)$ be $C$ .

Then, in the triangle $ABC$ , first let $AB$ be the diagonal (or longest length). Using Triangle Inequality, $AB$ $\leq AC + BC$ . From this, it is clear that $AB - AC \leq BC$ . Therefore, the maximum difference is the length of $BC$ .

Similarly, letting $AC$ be the diagonal of triangle $ABC$ , will also lead to the same conclusion that the maximum difference is the length of $BC$ .

Step 4 Using the distance formula, the absolute value of $BC$ is $\sqrt{(2-1)^{2} + (3-0)^{2}} = \sqrt 10$ . Recall that we have to find $N$ , which is the smallest integer larger than the expression to the power of 4. Hence to arrive at $N$ , we just add 1 to $BC^{4}$ : $(\sqrt 10)^{4} +1 = 101$ . We are done. :)

Rearranging the given expression

$\sqrt{\strut x^4-3x^2-6x+13}- \sqrt{\strut x^4-x^2+1}$

we get

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

Hence, we can set some points on the Cartesian Plane,

$P=(x,x^2)$

$A=(3,2)$

$B=(0,1)$

To get

$AP=\sqrt{\strut (x-3)^2+(x^2-2)^2}$

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

The locus of point $P$ is the parabola

$f(x)=x^2$

$AB$ is a line segment

$g(x)=\frac{x}{3}+1$

where $x\in[0,3]$

Conclude my explanation above, we will have a graph

Smallest Inequality Bounded

Image Credit for WolframAlpha

The minimum value occurs when Point $P$ is on the line $AB$ , which means the two function gives the same output, note this as

$f(x)=g(x)$

$x^2=\frac{x}{3}+1$

$3x^2-x-3=0$

By the Quadratic Formula,

$\displaystyle x=\frac{1+\sqrt{37}}{6}$

By substituting we get the minimum value

$\sqrt{\strut x^4-3x^2-6x+13}- \sqrt{\strut x^4-x^2+1}=\sqrt{\strut 10}$

Hence, the minimum value

$\Big [\sqrt{\strut x^4-3x^2-6x+13}- \sqrt{\strut x^4-x^2+1}\Big ]^4=100$

Since $N$ is the smallest integer larger than it, so

$N=101$

END.

Here I have coded the maximum value of the function by using c++. Then I have got 100(the maximum valu of the function by substituting all real number).. as "N" must be greater than the maximum value of that function.. then "N" should be 100+1=101.. so the answer is 101... Thank you.. the code is given below:

include <bits/stdc++.h>

using namespace std;

int main()

{

ios base::sync with_stdio(0);

cin.tie(0);

double x=0,y,i,result=0,mark;

for(x=-1000;x<=1000;x=x+.0001)

{

    y=pow(sqrt(pow(x,4)-3*pow(x,2)-6*x+13)-sqrt(pow(x,4)-pow(x,2)+1),4);

    if(y>result)

        {

            result=y;

            mark=x;

        }

}

cout<<result<<" "<<mark<<endl;

return 0;

}

Let $f(x)=\sqrt{x^4-3x^2-6x+13}-\sqrt{x^4-x^2+1}$ , which can be rewritten as $f(x)=\sqrt{x^4-4x^2+4+x^2-6x+9}-\sqrt{x^4-2x^2+1+x^2}$ or $f(x)=\sqrt{(x^2-2)^2+(x-3)^2}-\sqrt{(x^2-1)^2+x^2}$ . Squaring we have $f(x)^2=2x^4-4x^2-6x+14-2\sqrt{[(x^2-2)^2+(x-3)^2][(x^2-1)^2+x^2]}$ By Cauchy Schwarz Inequality, we have $\sqrt{[(x^2-2)^2+(x-3)^2][(x^2-1)^2+x^2]}\geq (x^2-2)(x^2-1)+x(x-3)$ . Thus, $f(x)^2\leq 2x^4-4x^2-6x+14-2[(x^2-2)(x^2-1)+x(x-3)]$ , which simplifies to $f(x)^2 \leq 2x^4-4x^2-6x+14-(2x^4-4x^2-6x+4)=10$ . Therefore, $f(x)^4\leq 100$ and the smallest positive integer $N$ is $N=101$ .

This problem can be solved with help of coordinate geometry above expression can be expressed as:- {[(x^2 - 2)^2+(x-3)^2]^1/2 - [(x^2-1)^2+(x-0)^2]^1/2}^4 which is equal to difference of distances of point(x^2,x) from points(2,3)and(1,0) then by applying triangle inequality difference of two sides=or<third side thus maximum value of expression{[(x^2 - 2)^2+(x-3)^2]^1/2 - [(x^2-1)^2+(x-0)^2]^1/2} to be equal to distance between points(2,3)and(1,0)which is 10^1/2 so we get N>100 N=101