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Given the inequality $m^4/s^5 - m \leq 2s^2$, the paper I am reading says that this implies that $s \geq 1/10\cdot m^{4/7}$. I tried manipulating the inequality to try and reproduce their result, but the best that I can get is $s \geq \sqrt[7]{m^4/(2+m)}$. Does anyone know how they managed to obtain their bound?

Edit: Both $s$ and $m$ are greater than or equal to $1$.

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  • $\begingroup$ I've edited the original inequality as I made a typo. $\endgroup$ – Nizbel99 May 30 '14 at 20:15
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    $\begingroup$ Do you know anything else about $s$ and $m$? For example, are they both positive? $\endgroup$ – Peter Woolfitt May 30 '14 at 20:42
  • $\begingroup$ (the derived inequality given is incorrect if they are allowed to be negative - choosing $m=1$, $s=-1$ gives $-2=-1-1\leq (2)1^2$, but $-1\not\geq 1/10$) $\endgroup$ – Peter Woolfitt May 30 '14 at 20:45
  • $\begingroup$ Yes, they are both positive. I will edit the question to reflect that. $\endgroup$ – Nizbel99 May 30 '14 at 20:45
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Set $s=xm^{4/7}$, then $$ 2x^7m^4\ge m^4-x^5m^{27/7}\iff 2x^7+x^5m^{-1/7}\ge 1 $$ For any polynomial $p(z)=a_nz^n+...+a_1z+a_0$ with $a_0\ne 0$, all of the roots are bounded below by $$ |z|\ge\frac{|a_0|}{|a_0|+\max_{k=1..n}|a_k|} $$ which follows from the upper Cauchy bound for the reverse polynomial $q(t)=a_n+a_{n-1}t+\dots+a_1t^{n-1}+a_0t^n$.

In the case above, the inequality is wrong for $x=0$, any change can only happen at a root of the polynomial that is the difference of both sides, so any $x$ satisfying the inequality also satisfies $$ x\ge\frac{1}{1+\max(2,m^{-1/7})}, $$ and for $m\ge 1$, the denominator has the value $3$. So in fact $$ s=xm^{4/7}\ge \tfrac13 m^{4/7}. $$


There is another (also named after) Cauchy bound that, applied to the reverse polynomial, results in a lower bound $B$ for the positive roots of the polynomial, $$ B^{-1}=\max\left\{\left(|N|\cdot \frac{|a_i|}{|a_0|}\right)^{\frac{1}{i}}\middle| i\in N\right\} \text{ where } N=\left\{k\in\{0,1,\dotsc,n-1\}\middle| a_0\cdot a_k < 0\right\}. $$ From the original inequality we have $(a_0,a_5,a_7)=(m^4,-m,-2)$, so one gets the lower bound for positive roots $$ B=\min\left\{\left(\frac{m^3}2\right)^{1/5},\left(\frac{m^4}{4}\right)^{1/7}\right\} $$


Or one can use the inequality directly, only using monotonicity and roots of the simplified polynomials. If $s$ and $x$ satisfy the inequalities above, then $x$ also satisfies $$ 2x^7+x^5\ge 1, $$ and since this inequality is not true for $x<0.8388629$, $x$ necessarily is greater than $0.8388629>\frac56=0.833333...$, so $$ s\ge \tfrac56 m^{4/7} $$


For the excluded case $m<1$ use $s=ym^{3/5}$ so that the inequality translates into $$ m^4-m^4y^5\le 2 m^{21/5}y^7\iff 2m^{1/5}y^7+y^5\ge 1 $$ which again implies the weaker inequality $2y^7+y^5\ge 1$ and thus $$ s=ym^{3/5}\ge \tfrac56m^{3/5} $$ Thus for any $m>0$ $$ s\ge\tfrac56 \min(m^{4/7},m^{3/5})=\tfrac56 m^{4/7}\,\min(1,m^{1/35}) $$

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  • $\begingroup$ What is the name of the Theorem/Result that you have used to bound the roots of the polynomial? $\endgroup$ – Nizbel99 Jun 4 '14 at 16:38
  • $\begingroup$ There is no specific theorem, the first, general root bound is named after Cauchy, the bound for positive roots too. The slightly better bounds below use rescaling and monotonicity of the power function for positive arguments. Links to wikipedia articles containing the bounds added. $\endgroup$ – LutzL Jun 4 '14 at 17:35
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I've found a counterexample to the paper's deduction $s\geq\frac1{10}m^{4/7}$ for very small values of $s$ and $m$, namely $s=(1.158)10^{-24}$ and $m=10^{-40}$

Here are the wolframalpha evaluations: first inequality evaluates to True, second inequality evaluates to False.

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  • $\begingroup$ Interesting! What happens if the values are greater than or equal to one instead? I don't think the values of $s$ and $m$ will be that small (although not explicitly stated). $\endgroup$ – Nizbel99 May 30 '14 at 22:06
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    $\begingroup$ I've read over parts of the paper again, and I think that in this context, both $m$ and $s$ should be greater than or equal to one. Something I didn't catch right away since it isn't stated explicitly. I will adjust the question accordingly. $\endgroup$ – Nizbel99 May 30 '14 at 22:15

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