A lower bound of a particular convex function Hello,
I suspect this reduces to a homework problem, but I've been a bit hung up on it for the last few hours.  I'm trying to minimize the (convex) function $f(x) = 1/x + ax + bx^2$ , where $x,a,b>0$.  Specifically, I'm interested in the minimal objective function value as a function of $a$ and $b$.  Since finding the minimizer $x^*$ is tricky (requires solving a cubic), I figured I'd try and find a lower bound using the following argument:  if $b=0$, the minimizer is $x=1/\sqrt{a}$ and the minimal value is $2\sqrt{a}$.  If $a=0$, the minimizer is $x=(2b)^{-1/3}$ and the minimal value is $\frac{3\cdot2^{1/3}}{2}b^{1/3}$.  Therefore, one possible approximate solution is the convex combination
$(\frac{a}{a+b})\cdot2\sqrt{a} + (\frac{b}{a+b})\cdot\frac{3\cdot2^{1/3}}{2}b^{1/3}$.
Numerical simulations suggest that the above expression is a lower bound for the minimal value.  Does this follow from some nice result about parameterized convex functions?  It seems like it shouldn't be hard to prove.  I guess in a nutshell I just want to prove that for all $x,a,b>0$ we have
$(\frac{a}{a+b})\cdot2\sqrt{a} + (\frac{b}{a+b})\cdot\frac{3\cdot2^{1/3}}{2}b^{1/3} \leq 1/x + ax + bx^2$.  Thanks!
EDIT:  It also appears that if I take the convex combination
$(\frac{a^{3/5}}{a^{3/5}+b^{2/5}})\cdot2\sqrt{a} + (\frac{b^{2/5}}{a^{3/5}+b^{2/5}})\cdot\frac{3\cdot2^{1/3}}{2}b^{1/3}$
then I get a tighter lower bound, and in fact the lower bound is within a factor of something like $3/2$ of the true minimal solution.

The first inequality is true. Write
$$f=\frac{a}{a+b}f_0+\frac{b}{a+b}f_1,$$
where $f_0$ and $f_1$ correspond to the case $b=0$ and $a=0$, respectively. You know that $f_0\ge2\sqrt a$ and $f_1\ge\frac{3\cdot2^{1/3}}{2}b^{1/3}$. This implies
$$f\ge\frac{a}{a+b}2\sqrt a+\frac{b}{a+b}\frac{3\cdot2^{1/3}}{2}b^{1/3}.$$
