Unequal Inequality

WLOG let $A+B \geq C+D$ . Then, note that $\frac{1}{C+D} \geq \frac{1}{A+B}$ Now note that $A+B+C+D \leq (B+C) + (B+C)= 2(B+C) \text{ .....(i)}$ We have $\frac{B}{C+D} + \frac{C}{A+B} = \frac{B+C}{C+D} - C \left ( \frac{1}{C+D} - \frac{1}{A+B} \right )$ From $(i)$ , we obtain: $\frac{B+C}{C+D} - C \times \left ( \frac{1}{C+D} - \frac{1}{A+B} \right ) \geq \frac{1}{2} \times \frac{A+B+C+D}{C+D} - (C+D) \times \left ( \frac{1}{C+D} - \frac{1}{A+B} \right )$ $= \frac{1}{2} \times \frac{A+B}{C+D} + \frac{C+D}{A+B} - \frac{1}{2}$ $= \frac{1}{2} \times \frac{A+B}{C+D} + \frac{1}{2} \times {2} \times \frac{C+D}{A+B} - \frac{1}{2}$ From AM-GM inequality, we obtain: $\frac{1}{2} \times \frac{A+B}{C+D} + \frac{1}{2} \times {2} \times \frac{C+D}{A+B} - \frac{1}{2} \geq \sqrt{2 \times \frac{A+B}{C+D} \times \frac{C+D}{A+B}} - \frac{1}{2}$ $\implies \frac{B}{C+D} + \frac{C}{A+B} \geq \sqrt{2}-\frac{1}{2}$ $\implies M \geq \sqrt{2}-\frac{1}{2}$ Then, note that $\lfloor 200M \rfloor \geq \lfloor 200\sqrt{2} - 100 \rfloor = 182$ It can be seen that equality holds for $(A,B,C,D)= (\sqrt{2}+1, \sqrt{2}-1, 2, 0)$ . These values can be found from the equality conditions $D=0$ , $A+B= C+D \implies A+B= C$ , and $\frac{A+B}{C+D}= \sqrt{2}$ . We thus conclude the minimum value of $\lfloor 200M \rfloor$ is $\boxed{182}$ .

Since the expression for $M$ is homogeneous, we may assume WLOG that
$A+D=0$ , in which case $A=D=0$ so
$M = \frac{B}{C} + \frac{C}{B} \geq 2$ by AM-GM, or
$A+D=1$ . Now we examine the partial derivatives
$\frac{dM}{dB} = \frac{1}{C+D} - \frac{C}{(A+B)^2} \geq \frac{1}{C+D} - \frac{C+D}{(A+B)^2} = \frac{(A+B)^2 - (C+D)^2}{(A+B)^2(C+D)}$
$\frac{dM}{dC} = \frac{1}{A+B} - \frac{B}{(C+D)^2} \geq \frac{1}{A+B} - \frac{A+B}{(C+D)^2} = \frac{(C+D)^2 - (A+B)^2}{(C+D)^2(A+B)}$
Since $A$ and $D$ cannot both be $0$ , at least one of these inequalities must be strict.
Thus, at least one of the partial derivatives must be positive,
so decreasing either $B$ or $C$ would decrease $M$ .
Hence, to minimize $M$ let $B + C = A+D = 1$ .
Let $f(A,B) = M = \frac{B}{2-A-B} + \frac{1-B}{A+B} = -2 + \frac{2-A}{2-A-B} + \frac{A+1}{A+B}$ .
We would like to minimize $f$ over the domain $[0,1]\times(0,1)$ . Note that $(0,1)$ is an open interval, and switching $A$ with $D$ while switching $B$ with $C$ does not change $M$ , so we only need to check one of the cases $A = 1 \implies D = 0$ or $A = 0 \implies D = 1$ . Thus, it suffices to check relative extrema and $A = 0$ . We solve for relative extrema by setting
$0 = \frac{dM}{dB} = \frac{2-A}{(2-A-B)^2} - \frac{A+1}{(A+B)^2}$
$(A+1)(2-A-B)^2 = (2-A)(A+B)^2$ (eq. 1)
$0 = \frac{dM}{dA} = \frac{B}{(2-A-B)^2} - \frac{1-B}{(A+B)^2}$
$(1-B)(2-A-B)^2 = B(A+B)^2$ (eq. 2)
Dividing eq. 1 and eq. 2, we find
$\frac{A+1}{1-B} = \frac{2-A}{B}$
$AB + B = 2 - A - 2B + AB$
$A = 2 - 3B$
Plugging into eq. 1, we have
$(3-3B)(2B)^2 = (3B)(2-2B)^2$
$12B^2-12B^3 = 12B-24B^2+12B^3$
$24B^3 - 36B^2 + 12B = 0$
$12B(2B-1)(B-1) = 0$ and, since $B \in (0,1)$
$B = \frac{1}{2}$ whereby
$M = f(A,B) = 1$ .
Now, to check the endpoints of the domain, we plug A = 0 into eq. 1 to obtain
$(2-B)^2 = 2B^2$
$4 - 4B + B^2 = 2B^2$
$8 = B^2 + 4 B + 4$
$\pm 2\sqrt{2} = B + 2$ and, since $B \in (0,1)$
$B = 2\sqrt{2} - 2$ whereby
$M = f(A,B) = \sqrt{2} - \frac{1}{2} < 1 < 2$ . Since we have checked all the cases, this is minimum value of $M$ given the constraints in the problem, and the minimum value of $\lfloor 200M \rfloor$ is thus
$\boxed{182}$ .

We define $f(A,B,C,D)=\frac{B}{C+D}+\frac{C}{A+B}$ and let the minimum value of $f$ be $m$ within the conditions in the question.

First, note that if $B+C>A+D$ then $B+C-A>D \Rightarrow f(A,B,C,B+C-A)<f(A,B,C,D)$ because only the denominator of the first fraction increases and all the values are non-negative so the value decreases thus $f(A,B,C,D)$ cannot be minimum so it must occur when $B+C=A+D$

Second, note that $f(A,B,C,D)=f(kA,kB,kC,kD)$ for any non-zero constant $k$ by direct substitution so $f(A,B,C,D)=m \Rightarrow f(\frac{A}{A+D},\frac{B}{B+C},\frac{C}{B+C},\frac{D}{A+D})=m$ since we have shown above that for the minimum to occur $B+C=A+D$ thus $\exists(A',C')$ such that $f(A',1-C',C',1-A')=m$ based on the construction we have just found.

We proceed by calculus. Consider the function $f(A',1-C',C',1-A')=\frac{1-C’}{C’+1-A}+\frac{C’}{1-C’+A’}$ where $0 \leq A’ \leq 1$ and $0 < C’ < 1$ $\frac{\partial f(A',1-C',C',1-A')}{\partial C'} = \frac{A'-2}{(C'+1-A')^2} + \frac{A'+1}{(1-C'+A')^2}$ $\frac{\partial^2 f(A',1-C',C',1-A')}{\partial^2 C'} = \frac{2-A’}{(C'+1-A')^3} + \frac{A'+1}{(1-C'+A')^3}$ First note that as $|C’-A’|\rightarrow 1^{-}$ , $f(A',1-C',C',1-A') \rightarrow +\infty$ and in the given range, $|A’-C’|<1$ so $f(A',1-C',C',1-A')>0$ and $\frac{\partial^2 f(A',1-C',C',1-A')}{\partial^2 C'} > 0$

From these we can interpret that there is exactly one local minimum in the range $A’-1 < C’ < A’+1$ which occurs when $\frac{\partial f(A',1-C',C',1-A')}{\partial C'}=0$ $\Rightarrow (A'-2)(1-C'+A')^2 + (A'+1)(C'+1-A')^2 =0$ $\Rightarrow (2A'-1)C'^2-(4A'^2-2A'-6)+(2A'^3-A'^2-4A'-1) = 0$ Solving this quadratic in $C'$ gives us $C' = \frac{2A'^2-A'-3 \pm 2\sqrt{2+A'-A'^2}}{2A'-1}$ and substituting back into $f$ gives us $-\frac{1}{2} \pm \sqrt{2+A’-A’^2}$

We have shown that the minimum must be positive so we can reject the negative solution hence we aim to find the minimum value of $\sqrt{2+A’-A’^2}-\frac{1}{2} = \sqrt{\frac{9}{4}-(A’-\frac{1}{2})^2}-\frac{1}{2}$ which is equivalent to finding the maximum value of $(A’-\frac{1}{2})^2$ which in the range given is $\frac{1}{4}$ when $A’=0,1$

Substituting back we get $m=f(0,2\sqrt{2}-2,3-2\sqrt{2},1)= f(1,3-2\sqrt{2},2\sqrt{2}-2,0)=\sqrt{2}-\frac{1}{2}$ so the minimum value we need is $\lfloor 200\sqrt{2}-100 \rfloor$ and since $282 = \sqrt{79524} < \sqrt{80000} = 200\sqrt{2} < \sqrt{80089} = 283$ the answer we require is $282 - 100 = 182$

Plug in B=5x, C=x, A=6x, D=0 for all positive real x, we will get the required minimum value, that is 182. Since both A and D only appear in the denominator, A+D must be maximum to obtain the minimum value of M. Therefore, A+D=B+C. Denote A+D=B+C=6x for positive real x. M=B/(C+A)+C/(B+D)=B/(6x+A-B)+(6x-B)/(6x-A+B). Using inequalities and argument by contradiction, we can show that when A=6x and B=5x or A=0 and C=x, M achieves its minimum value and [200M] achieves its minimum value.