Palindromes in math!

I will only give hint ! try to kill further , by own ! and ask me freely if you still have any problem

Here answer is " 4/5 " .

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Edited

Method-1)- (Geometrical Approach)

$\displaystyle{{ ({ ({ x }^{ 2 }+\cfrac { 1 }{ { x }^{ 2 } } )+a(x+\cfrac { 1 }{ x } )+b=0\\ x+\cfrac { 1 }{ x } =t\quad (say)\quad \quad (t>2)\\ { t }^{ 2 }+at+b-2=0\\ (say)\quad b=Y\quad \& \quad a=X\\ Xt+Y-2+{ t }^{ 2 }=0\quad (straight\quad Line\quad (a,b)\quad lies\quad on\quad it) }) }}$

Now see what is asked in problem . Think ........ ! Got something ........ ?

Yes we are asked to find square value of shortest distance P(a,b) from origin

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Method-2)- (Algebraic approach)

What about AM-HM ...........? Think ......!!!!

and then what about use of Cauchy schz. inequality ? Think .....! !! Got it .......?

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I'am waiting's for other's to post all such possible method's for this problem!

Enjoy maths ! cheer's $\ddot\smile$

Let $P(x)=x^4+ax^3+bx^2+ax+1.$ If $x=1$ is a root of $P(x)$ , it can be proved that its multiplicity is at least 2 . Indeed, if you divide the given 4-th degree polynomial by $(x-1)$ twice, then we get the remainder $2P(1)=0.$ In a similar way, it can be proved that if $x=-1$ is a root of $P(x),$ then it has multiplicity at least 2. On the other hand, since $P(x)=x^4P(1/x)$ , if $r$ is a real root of $P(x)$ then $1/r$ is another real root.

Therefore, if $P(x)$ has at least one real root, then $P(x)$ has the factor $(x^2-(r+\frac{1}{r})x+1)$ , where $r$ can be any real number, including the case when $r=1$ or $r=-1.$ Then the other quadratic factor has to have the form $x^2+px+1$ where $p$ is an arbitrary real number.

From the fact that $P(x)=(x^2-(r+\frac{1}{r})x+1)(x^2+px+1),$ we obtain that $a=p-(r+\frac{1}{r})$ and $b=2-p(r+\frac{1}{r}).$ Therefore, $a^2+b^2=(1+(r+\frac{1}{r})^2)p^2- 6(r+\frac{1}{r})p+(r+\frac{1}{r})^2+4.$

Since $r$ is a non-zero real number and $\frac{1}{r}$ has the same sign as $r,$ $|r+\frac{1}{r}|=|r|+|\frac{1}{r}|=\frac{|r|^2+1}{|r|}\geq \frac{|r|^2-2|r|+1+2|r|}{|r|}\geq \frac{(|r|-1)^2+2|r|}{|r|}\geq \frac{2|r|}{2}\geq 2.$

From the previous inequalities we also obtain that the equality is reached when $r=1$ or $r=-1.$ Now, making $t=r+\frac{1}{r}$ the problem of finding the minimum value of $a^2+b^2$ can be rephrased in the following way: we need to find the absolute minimum of the form $(1+t^2)x^2- 6t x+t^2+4$ where $t$ and $x$ are any real numbers and $|t|\geq 2.$

By factoring $(1+t^2)$ out and completing the square we get $(1+t^2)x^2- 6t x+t^2+4=(1+t^2)(x-\frac{3t}{1+t^2})^2+t^2+4-\frac{9t^2}{1+t^2}.\:\:\:\:\:\:\:\:(*)$

Since the term $(1+t^2)(x-\frac{3t}{1+t^2})^2$ is always greater than or equal to zero, we can find the minimum value for the given expression by finding a value of $t$ , such that $|t|\geq 2,$ that minimizes the function $f(t)=t^2+4-\frac{9t^2}{1+t^2},$ and then making $x=\frac{3t}{1+t^2}$ for that value of $t.$ We can verify that $f '(t)= \frac{2 t (t^2-2) (t^2+4)}{(t^2+1)^2},$ so the only roots of the derivative of $f(t)$ are $-\sqrt{2}, 0,$ and $\sqrt{2}.$ and it is never undefined at any number. Now, it is easy to determine by using test points that $f '(t)$ is negative on $(-\infty, -2),$ and positive on $(2 ,\infty ).$ Therefore the minimum value of $f(t)$ for all values of $t$ such that $|t|\geq 2$ will be $f(-2)=f(2)=\frac{4}{5}=0.8.$ Thus we can conclude that the expression $(*)$ gets its minimum value a $t=2$ or $-2$ and taking $x=\frac{3(2)}{1+2^2}=\frac{6}{5}$ or $x=\frac{3(-2)}{1+(-2)^2}=-\frac{6}{5},$ respectively, and the minimum values of $(*)$ will be 0.8.

A polynomial $f(x)$ of this form with minimal $a^2+b^2$ must have a multiple root, $c$ . Given the required format of the polynomial, we must have $f(x)=(x\pm 1)^2(x^2+px+1)$ or $f(x)=(x-c)^2\left(x-\frac{1}{c}\right)^2$ . In the first case, with the factor $(x-1)^2$ , the minimum is $a^2+b^2=(p-2)^2+(2-2p)^2$ $=5p^2-12p+8= \boxed{0.8}$ when $p=1.2$ . In the second case we have $b>4$ so $a^2+b^2>16$ .