Are you smarter than me? 52

The question is about the $A.M. - R.M.S$ inequality.

See that if $x,y,z,w$ are roots of this equation, then

$x+y+z+w = 8$ and $\sum xy = 24$

We know that $(x+y+z+w)^2 = \sum x^2 + 2 \sum xy$

$\therefore 8^2 = \sum x^2 + 48$

$\therefore 64 = \sum x^2 +48$

$\therefore \sum x^2 = 16$

Now as they are positive, we apply the AM-RMS inequality-

$\sqrt{\dfrac{x^2+y^2+z^2+w^2}{4}} \geq \dfrac{x+y+z+w}{4}$

and equality is only when $x=y=z=w$ .

Now using the given values, we get $RMS = AM = 2$ hence $x=y=z=w$ and hence all roots have to be $2$ .

Answer = $-(xyz+yzw+zwx+wxy) = \boxed{-32}$

This problem can also be done with the help of Calculus, here's how:

Since the given equation has $4$ real roots, $f'(x)$ will have $3$ real roots and $f''(x)$ will have $2$ real roots :

$f'(x)= 4x^3-24x^2+48x+b$

$f''(x)= 12x^2-48x+48$

The roots of $f''(x)$ comes out to be $2$ . Now putting $2$ in $f'(x)=0$ we get $b=-32$ .

Let $f(x)=x^{4}-8x^{3}+24x^{2}+bx+c$ . Then $f'(x)=4x^{3}-24x^{2}+48x+b$ and $f''(x)=12x^{2}-48x+48$ . Factoring the latter, we obtain $f''(x)=12(x-2)^{2}.$ Therefore $f''(x)\geq 0$ for all $x$ , and then the function $f'(x)$ is always increasing on $(-\infty, \infty)$ . Thus $f'(x)$ can have only one real solution.

Now we have to prove that if $f'(x)$ has only one real root and $f(x)$ has only real roots, then there will be a real number $c$ such that $f(x)=(x-c)^{4}.$ That is, $f(x)$ will has only one real solution of multiplicity 4.

Let us prove it by contradiction. Assume that $f(x)$ has more than one real solution. Of course, the number of real roots cannot be 4 or 3, because using the Rolle's Theorem this would imply that $f'(x)$ would have more than one real root. The other possibility is that $f(x)$ had 2 different real roots, let us say $\alpha$ and $\beta$ . In this case at least one these two roots would have multiplicity greater than one, let us say $\alpha$ . But this would imply that $f'(x)$ would have two different real roots: one of them would be $\alpha$ and the other one, according to Rolle's Theorem, would be a number in between $\alpha$ and $\beta$ . This contradiction proves that $f(x)$ has a unique real root and as it does not have complex roots, the given unique real root would have multiplicity 4. Then $f(x)=(x-c)^4.$ By expanding the binomial and making it equal to the given polynomial, we obtain that $c=2.$ Then from the expansion of $(x-2)^{4},$ we get that the value of $b$ would be -32.

But in this solution I never used the fact that the roots are positive numbers. I used only the fact that the polynomial does not have complex non-real roots. Did I miss something?

$x^4-8x^3+24x^2+bx+c=0$

$\Rightarrow x^2-8x+24+\dfrac {b}{x}+\dfrac {c}{x^2}=0$

$\Rightarrow \left( x^2 - 2\sqrt{c} + \dfrac {c}{x^2} \right) + 2\sqrt{c} - 8 \left( x - \dfrac {b}{8x} \right) + 24 = 0$

$\Rightarrow \left( x - \dfrac {\sqrt{c}}{x} \right)^2 - 8 \left( x - \dfrac {b}{8x} \right) + 24 + 2\sqrt{c} = 0$

Equating $\space \sqrt{c} = \dfrac {b}{8} \quad \Rightarrow x - \dfrac {\sqrt{c}}{x} = x - \dfrac {b}{8x} = y$

Then the equation has real roots when:

$8^2 - 4(24 + 2\sqrt{c} ) \ge 0\quad \Rightarrow 64 - 96 - 8\times \dfrac {b}{8} \ge 0$

$\Rightarrow y^2 - 8 y + 24 + 2\sqrt{c} = 0 \quad \Rightarrow -32 - b \ge 0$

$\Rightarrow b \le \boxed{-32}$

Looking at the graph for some values of b and c, the graph is shaped like a parabola only. How ever we have four real roots. This would be only possible if the graph touches Y=0. This points to four equal roots.
$So~ f(X)=(X + a)^4. ~Since~ there~ is~ a~ - tive~ coefficient~ of ~X^3~ term, a<0. \\ For~ m>0,~ (X - m)^4= X^4 - 4*m*X^3 + 6*m^2*X^2 - 4*m^3*X +m^4.\\ Comparing~the ~coefficients ~of~X^3~terms~with~given~eqation~we~get~m=2.\\ Coefficients ~of~X^2~terms~ confirms~ this.\\ \implies~b=-4*2^3=\Large \color{#D61F06}{ -32}.$

Let the roots be a, b, c, d and given or related values quoted as {8} and {24}:

(a + b + c + d)^2 = a^2 + b^2 + c^2 + d^2 + 2 (a b + a c + a d + b c + b d + c d)

a^2 + b^2 + c^2 + d^2 = (a + b + c + d)^2 - 2 (a b + a c + a d + b c + b d + c d)

= {8}^2 - 2 {24} = {16}

Generally, Sqrt [(a^2 + b^2 + c^2 + d^2)/ 4] >= (a + b + c + d)/ 4

Specifically, Sqrt [{16}/ 4] = {8}/ 4 => If and only if a = b = c = d.

(x - a)^4 = x^4 - 4 a x + 6 a^2 x^2 - 4 a^3 x + a^4

= x^4 - 8 x + 24 x^2 + b x + c

Quoted that:

1) 4 a = 8

2) 6 a^2 = 24

3) - 4 a^3 = b

4) a^4 = c

Therefore a = 2, b = - 4 (2^3) and c = 2^4

b = -32 and c = 16 due to the fact of only possibility of all equal roots or all repeated roots.

We can check that no other value of (b, c) other than (-32, 16) can possibly satisfy.

Answer: -32

The easiest solution relies on the observation that our given polynomial $f(x)$ can be written as $(x-2)^4$ . Some quick work with binomial expansion tells us that $b=\binom{4}{3}(-2)^3=\boxed{-32}$