What is the smallest possible value of $\lfloor (a+b+c)/d\rfloor+\lfloor (a+b+d)/c\rfloor+\lfloor (a+d+c)/b\rfloor+\lfloor (d+b+c)/a\rfloor$?

What is the smallest possible value of

$$\left\lfloor\frac{a+b+c}{d}\right\rfloor+\left\lfloor\frac{a+b+d}{c}\right\rfloor+\left\lfloor\frac{a+d+c}{b}\right\rfloor+\left\lfloor\frac{d+b+c}{a}\right\rfloor$$

where $a,b,c,d>0$?

Solution 1:

Since $\lfloor x\rfloor \gt x-1$, we have $$\begin{align}&\left\lfloor\frac{a+b+c}{d}\right\rfloor+\left\lfloor\frac{a+b+d}{c}\right\rfloor+\left\lfloor\frac{a+d+c}{b}\right\rfloor+\left\lfloor\frac{d+b+c}{a}\right\rfloor\\\\&\gt \frac{a+b+c}{d}-1+\frac{a+b+d}{c}-1+\frac{a+d+c}{b}-1+\frac{d+b+c}{a}-1\\\\&=\left(\frac ab+\frac ba\right)+\left(\frac ac+\frac ca\right)+\left(\frac ad+\frac da\right)+\left(\frac bc+\frac cb\right)+\left(\frac bd+\frac db\right)+\left(\frac cd+\frac dc\right)-4\\\\&\ge 2\sqrt{\frac ab\cdot\frac ba}+2\sqrt{\frac ac\cdot\frac ca}+2\sqrt{\frac ad\cdot\frac da}+2\sqrt{\frac bc\cdot\frac cb}+2\sqrt{\frac bd\cdot\frac db}+2\sqrt{\frac cd\cdot\frac dc}-4\\\\&=8\end{align}$$

By the way, $$\left\lfloor\frac{1+1+1}{0.9}\right\rfloor+\left\lfloor\frac{1+1+0.9}{1}\right\rfloor+\left\lfloor\frac{1+0.9+1}{1}\right\rfloor+\left\lfloor\frac{0.9+1+1}{1}\right\rfloor=9.$$

So, the smallest possible value is $\color{red}{9}$.

Solution 2:

Let $$g(a,b,c,d)=\dfrac{a+b+c}{d}+\dfrac{b+c+d}{a}+\dfrac{c+d+a}{b}+\dfrac{d+a+b}{c}$$ use Cauchy-Schwarz inequality we have $$g(a,b,c,d)=(a+b+c+d)\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}+\dfrac{1}{d}\right)-4\ge 16-4=12$$ so $$\lfloor\dfrac{a+b+c}{d}\rfloor+\lfloor\dfrac{b+c+d}{a}\rfloor+\lfloor\dfrac{c+d+a}{b}\rfloor+\lfloor\dfrac{d+a+b}{c}\rfloor>12-4=8$$ and other hand,when,$a=10,b=c=d=11$ $$\lfloor\dfrac{a+b+c}{d}\rfloor+\lfloor\dfrac{b+c+d}{a}\rfloor+\lfloor\dfrac{c+d+a}{b}\rfloor+\lfloor\dfrac{d+a+b}{c}\rfloor=9$$