Question

If the equation $x^4+p{x}^3+q{x}^2+rx+5=0$ has four positive real roots, then find the minimum value of $pr$.

Answer 1

Let $a,b,c,d$ be the roots of the polynomial.

So, $(a+b+c+d)=-p$

$(abc+abd+acd+bcd)=-r$

$abcd=5$

Now, Arithmetic Mean $\ge$ Geometric Mean

ie, $\frac{(a+b+c+d)}{4}\ge (abcd{)}^{1/4}=5^{1/4}$

$\Rightarrow -p\ge 4\times 5^{1/4}$

also, $\frac{(abc+abd+acd+bcd)}{4}\ge (abcd{)}^{3/4}=5^{3/4}$

$\Rightarrow -r\ge 4\times 5^{3/4}$

So, $pr\ge 4^2\times 5=80$

ie, minimum value of $pr=80$