Given that $x^4+px^3+qx^2+rx+s=0$ has four positive roots.

Question

Given that $x^4+px^3+qx^2+rx+s=0$ has four positive roots.

Prove that (1) $pr-16s\ge0$ (2) $q^2-36s\ge 0$

with equality in each case holds if and only if four roots are equal.

My Approach:

Let roots of the equation

$x^4+px^3+qx^2+rx+s=0$ be $\alpha,\beta,\eta,\delta$

$\alpha>0,\beta>0,\eta>0,\delta>0$

$\sum\alpha=-p$

$\sum\alpha\beta=q$

$\sum\alpha\beta\eta=-r$

$\alpha\beta\eta\delta=s$

I am confused , what is next step? please help me

Answer 1

We'll need the following:

If $a_1,\ldots,a_n>0$ then $$\left(\sum_{k=1}^na_k\right)\left(\sum_{k=1}^n\frac1{a_k}\right)\ge n^2$$ Proof: By AM-GM inequality, $\sum a_k\ge n\sqrt[n]{\prod a_k}$ and $\sum 1/a_k\ge n\sqrt[n]{\prod 1/a_k}$.

To ease writing, I'll use latin letters for the roots: $t,u,v,w$. $$\begin{align}pr&=(t+u+v+w)(tuv+tuw+tvw+uvw)\\ &=tuvw(t+u+v+w)\left(\frac1t+\frac1u+\frac1v+\frac1w\right)\\ &\ge 16s \end{align}$$

The second inequality can be proven similarly:

$$\begin{align}q^2&=(tu+tv+tw+uv+uw+vw)^2\\ &=tuvw(tu+tv+tw+uv+uw+vw)\left(\frac1{tu}+\frac1{tv}+\frac1{tw}+\frac1{uv}+\frac1{uw}+\frac1{vw}\right)\\ &\ge 36s \end{align}$$

Generalizing, if the polynomial $$\sum_{k=0}^na_kx^k$$ has $n$ positive real roots and $a_n=1$, then $$|a_ka_{n-k}|\ge\binom nk^2|a_0|$$ for $k\in\{1,\ldots,n-1\}$.

Answer 2

AM-GM inequality is the key.

The product $pr$ consists of $16$ terms. Four of those terms are equal to $s$. The remaining twelve are the permuted versions of $\alpha^2\beta\eta$. The product of those twelve is equal to $s^{12}$, so by AM-GM their sum is $\ge12s$.

In AM-GM we have equality iff all the involved numbers are equal. This translates easily to the requirement that $\alpha=\beta=\eta=\delta$.

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Might as well :-)

The correct version of the second inequality reads $$q^2\ge 36s.$$ This is proven as follows. The symmetric polynomial $q$ has six terms. Therefore $q^2$ has $36$ terms (some are repeated). The product of those $36$ terms is equal to $\alpha^i\beta^j\eta^k\delta^\ell$. Each of those $36$ terms is of degree four, so their product is of degree $36\cdot4$. By symmetry $i=j=k=\ell$, so we can conclude that $i=j=k=\ell=36$, and the product is thus $s^{36}$. The AM-GM inequality strikes again. Leaving the extra claim to the reader.