# Inequality with strange equality conditions

Question: For positive real numbers $a,b,c$ and $d$, find the smallest possible value of the following expression: $$\large{\left \lfloor{\frac{b+c+d}{a}} \right \rfloor+ \left \lfloor{\frac{a+c+d}{b}} \right \rfloor+\left \lfloor{\frac{a+b+d}{c}} \right \rfloor+\left \lfloor{\frac{a+b+c}{d}} \right \rfloor}$$

Note: For this question, the expression $\lfloor x \rfloor$ stands for the greatest integer function of x. For example, $\lfloor 2.5 \rfloor=2$, while $\lfloor -2.5 \rfloor=-3$

Bonus: Generalise the problem to $n$ numbers $a_1,a_2,\ldots,a_n$.

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We use the following two tools:

• Tool 1: Every real number $x$ satisfies $\lfloor x\rfloor>x-1$ (with strict inequality).
• Tool 2: All positive real numbers $y$ and $z$ satisfy $\frac{x}{y}+\frac{y}{x}\ge2$; indeed this inequality is equivalent to $y^2+z^2\ge2yz$ and $(y-z)^2\ge0$.

Now the expression $E$ that is to be minimized can be estimated (by using these two tools) as follows: \begin{eqnarray*} E &=& \left\lfloor\frac{b+c+d}{a} \right\rfloor+ \left\lfloor\frac{a+c+d}{b} \right\rfloor+ \left\lfloor\frac{a+b+d}{c} \right\rfloor+ \left\lfloor\frac{a+b+c}{d} \right\rfloor \\ &>& \left(\frac{b+c+d}{a}-1\right)+ \left(\frac{a+c+d}{b}-1\right) + \left(\frac{a+b+d}{c}-1\right) + \left(\frac{a+b+c}{d}-1\right) \\ &=& \left(\frac{a}{b}+\frac{b}{a}\right)+ \left(\frac{a}{c}+\frac{c}{a}\right)+ \left(\frac{a}{d}+\frac{d}{a}\right)+ \left(\frac{b}{c}+\frac{c}{b}\right)+ \left(\frac{b}{d}+\frac{d}{b}\right)+ \left(\frac{c}{d}+\frac{d}{c}\right) -4 \\ &\ge& ~6\cdot2-4 ~~=~~ 8. \end{eqnarray*} We get $E>8$. Since $E$ is the sum of four integers and hence integer itself, this actually implies the lower bound $E\ge9$. Finally, for $a=4$ and $b=c=d=5$ the expression attains the value $9$.

Hence the smallest possible value of the above expression is

$~~~9~$ (nine)

For the Bonus question, the above argument easily generalizes. The resulting lower bound is $$E ~>~ {n\choose 2}\cdot 2 -n,$$ which implies $E\ge(n-1)^2$. By setting $a_1=n$ and $a_2=a_3=\cdots=a_n=n+1$, we get $$\left\lfloor\frac{a_2+a_3+\cdots+a_n}{a_1} \right\rfloor = \left\lfloor\frac{(n-1)(n+1)}{n} \right\rfloor = \left\lfloor n-\frac{1}{n} \right\rfloor = n-1,$$ and $$\left\lfloor\frac{a_1+a_3+\cdots+a_n}{a_2} \right\rfloor = \left\lfloor\frac{(n-2)(n+1)+n}{n+1} \right\rfloor = \left\lfloor (n-2)+\frac{n}{n+1} \right\rfloor = n-2.$$ The resulting value of the expression is $(n-1)+(n-1)(n-2)=(n-1)^2$, and hence coincides with the lower bound.

Summarizing, the smallest possible value of the expression in the Bonus question

$~~~(n-1)^2$

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That's it.You nailed it. – Roby5 Mar 19 at 17:49