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.
 A: 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
$$f\ge\frac{a}{a+b}2\sqrt a+\frac{b}{a+b}\frac{3\cdot2^{1/3}}{2}b^{1/3}.$$
A: As Nishant Chandgotia sugessted: simply write
$f(x) = \left(p\cdot \frac{1}{x} + ax\right) + \left((1-p)\frac{1}{x}+bx^2 \right)$
for some parametr $p\in[0,1]$.
For the first term, minimizer is equal to $ p^{\frac{1}{2}}a^{-\frac{1}{2}}$ and the minimal value is $p^{\frac{1}{2}} 2a^{\frac{1}{2}}$.
For the second therm, minimizer is equal to $(1-p)^{\frac{1}{3}}(2b)^{-\frac{1}{3}} $ and the minimal value is $(1-p)^{\frac{2}{3}} \cdot\frac{3\cdot 2^{\frac{1}{3}} }{2} b^{\frac{1}{3}}$
The best estimate is achieved when both minimizers are equal, which means, in therms of $p$, that 
\begin{equation}
\left(\frac{p}{a}\right)^{3} = \left( \frac{1-p}{2b}\right)^{2} 
\end{equation}
Note, that this equation has a solution in interval $0 < p < 1$  by Mean-value theorem, unfortunately not expressible in nice way. 
Any way, we get for any $p\in[0,1]$ the following estimate:
\begin{equation}
f(x) \geqslant p^{\frac{1}{2}} \cdot 2a^{\frac{1}{2}} + (1-p)^{\frac{2}{3}} \cdot\frac{3\cdot 2^{\frac{1}{3}} }{2} b^{\frac{1}{3}}
\end{equation}
Finally, we check that $p^{\frac{1}{2}} + (1-p)^{\frac{2}{3}} \geqslant 1$. This inequality implies, that estimate reamins valid after using any convex combination instead of weights $p^{\frac{1}{2}}, (1-p)^{\frac{2}{3}}$, i.e.
\begin{equation}
f(x) \geqslant \alpha \cdot 2a^{\frac{1}{2}} + (1-\alpha) \cdot\frac{3\cdot 2^{\frac{1}{3}} }{2} b^{\frac{1}{3}}
\end{equation}
This can be seen immediately, however, by the inequality
$f \geqslant \alpha f_0 + (1-\alpha) f_1$.
My goal was to complete presented ideas and to show when exact optimum is attained.
