optimization
I want to find the parameters $a$ $(1/2<a<2/3)$ and $b$ $(0<b<1)$ such that the maximum of functions $f_1$, $f_2$, $f_3$, and $f_4$ becomes minimized, i.e., I am looking for $min(max(f_1, f_2, f_3, f_4))$:
$f_1(a) = 1.5+\frac{4a^2-4a+1}{6-4a}$
$f_2(b) = 1.19+\frac{b}{1+b}$
$f_3(a, b) = 1+\frac{1}{1+b}+\frac{4+9a^2-12a}{3+9a}$
$f_4(b) = \frac{5}{12}+\frac{2}{1+b}$
What's the best way to do so?
Thanks in advance.
I think the answer is no. Consider the functions in the following picture:
The minima of the sum is the entire interval between the minima of each individual function. You can imagine if you shifted either function up or down slightly, the minima would suddenly snap to one side or the other, regardless of the coefficients $a_i$.
These functions are not differentiable but that's not really an issue - just imagine smoothing out the tip of the functions a little bit and the same basic argument applies.
The conclusion that $f_1(x,y)=f_2(x,y)=f_3(x,y)$ in optimum can be argued by showing that if $f_i(x,y)>\max\{f_j(x,y), f_k(x,y)\}$ ($i,j,k$ is a permutation of $1,2,3$), then we can always find some $(x',y')$ in the neighborhood of $(x,y)$ such that $f_i(x,y)>f_i(x',y')>\max\{f_j(x',y'), f_k(x',y')\}$, which then implies that $(x,y)$ does not achieve the minimum in your problem.
The argument should not be difficult, but is just a little bit mass. A straightforward way is to consider the partial derivatives of each of the three functions (notice that the absolute value sign in $f_1$ and $f_2$ can be dropped for free, while $f_3(x,y)$ can be rewritten as $(x-y)^2/4$, by which all three are actually smooth and calculus works).
For example, we argue that it is impossible to have \begin{equation}f_1(x,y)>\max\{f_2(x,y), f_3(x,y)\}\end{equation} in optimum. To this end, let $A=f_1(x,y)$ and $B=\max\{f_2(x,y), f_3(x,y)\}$, where $A>B$, and we denote $\theta=A-B$. It is clear that $A>B\ge 0$, which then implies that $x,y<1$, and thus there exists some $\delta>0$ such that $\max\{x,y\}+\varepsilon<1$ for all $\varepsilon\in (0,\delta)$. Since $$\frac{\partial f_1}{\partial x}=y-1<0,\,\,\,\,\,\frac{\partial f_2}{\partial y}=x-1<0,$$ it is clear that for all $\varepsilon\in (0,\delta)$, $f_1(x+\varepsilon, y+\varepsilon)<f_1(x,y)$. Also, since all three functions are continuous, there exists some $\delta'>0$ such that for all $(x',y')$ that are at most $\delta'$-distant from $(x,y)$, $|f_i(x',y')-f_i(x,y)|<\theta/3$, $i=1,2,3$. Therefore, pick $x''=x+\min\{\delta, \delta'\}/\sqrt 2$ and $y''=y+\min\{\delta, \delta'\}/\sqrt 2$. We thus have $$f_1(x'',y'')>f_1(x,y)-\frac{\theta}{3}>\max\{f_2(x,y), f_3(x,y)\}+\frac\theta 3\ge\max\{f_2(x'', y''), f_3(x'', y'')\}$$ while $$f_1(x'',y'')<f_1(x,y),$$ showing that $(x,y)$ fails to achieve the optimum to the minmax problem.
Best Answer
Notice that $f_2$ is increasing and and $f_4$ is decreasing. At $b_0 = 92/133 \approx 0.691729$ we have $f_2(b_0) = f_4(b_0)$. Let $v = f_2(b_0)$ and observe that $v = 1439/900 \approx 1.59889$.
It follows that $\max(f_1, f_2,f_3, f_4)\geqslant \max(f_2,f_4) \geqslant v$. The best we can hope for our minimum is hence $v$.
Notice that $f_1$ is increasing. When $a$ approaches $2/3$, $f_1(a)$ approaches $1.53 < v$. It follows from our calculations above that regardless of our choice of $a$, $f_1$ will always be less than $\max(f_2,f_4)$. Hence,
$$\max(f_1,f_2,f_3,f_4) = \max(f_2,f_3,f_4).$$
Notice that $f_3$ is decreasing in $b$ and in $a$. When $a$ approaches $2/3$, $\frac{4+9a^2-12a}{3+9a}$ approaches $0$.
We have that $f_3(2/3,b_0) \approx 1.59111 < v$. It follows that indeed $\min(\max(f_1,f_2,f_3,f_4)) =v$, with the minimum being attained at $b=b_0$ and $a$ near $2/3$.
Indeed, we have
$$f_3(a,b_0) = \frac{358}{225} + \frac{4+9a^2-12a}{3+9a}.$$
Checking for $\frac{358}{225} + \frac{4+9a^2-12a}{3+9a} = \frac{1439}{900}$ when $1/2<a<2/3$, we find that any $a\geqslant a_0$ works, where
$$a_0 = \frac{1207 - \sqrt{25249}}{1800} \approx 0.582278$$