An algebra problem by Krishna Sharma

Divide numerator and denominator of f(x) by ${ x }^{ 4 }$ . therefore

$f\left( x \right) =\quad \dfrac { 1 }{ { (x }^{ 4 }+{ \frac { 16 }{ { x }^{ 4 } } ) }^{ }{ +(2x }^{ 2 }+\frac { 8 }{ { x }^{ 2 } } )\quad -4 }$ .

Now using the Fact that If a,b are positive Variables then By AM-GM inequality:

$a\quad +\quad b\quad \ge \quad 2\sqrt { ab }$ .

therefore minimum Value of denominator in f(x) ( or We can say maximum value of f(x) ) occures at when AM=GM

i.e when $\Rrightarrow \quad \quad { x }^{ 4 }=\frac { 16 }{ { x }^{ 4 } } \\ \\ \Rrightarrow \quad \quad x=\sqrt { 2 }$ .

${ f\left( x \right) }_{ max }=\frac { 1 }{ 8+8-4 } =\frac { 1 }{ 12 }$ .

And minimum Value of f(x) occurs at :

x=0

So ${ f\left( x \right) }_{ min }=0$ .

P=1 & q=12

P+q=13 Q.E.D

Let the given polynomial as y, dy/ dx will= 0 when x= 2, and 2^1/2 then put x= 0 and x= 2^1/2 to the original polynomial (y), you will get y= 0 (minimum value) and y=1/ 12 (maximum value), respectively so p/q= 0+ 1/12= 1/12, p+ q= 1+ 12= 13