4 Changed the convex combination

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.

3 Removed the bit about a 1.5 approximation

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 valueand that the two differ by no more than a factor of $1.5$ (provided that $a$ and $b$ aren't really small). 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!

2 Added a caveat on a and b

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 and that the two differ by no more than a factor of $1.5$. 1.5$(provided that$a$and$b$aren't really small). Does this follow from some nice result about parameterized convex functions? It seems like it shouldn't be hard to prove. I guess in a nutshell I just want to prove that for all$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!

1