Inequality manipulation in two variables 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 $1$.
 A: 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.
A: 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 $m>0$
$$
s\ge\tfrac56 \min(m^{4/7},m^{3/5})=\tfrac56 m^{4/7}\,\min(1,m^{1/35})
$$
