# the least possible value for :$\lfloor \frac{a+b}{c}\rfloor +\lfloor \frac{b+c}{a} \rfloor+\lfloor \frac{c+a}{b} \rfloor$

If we know that for every $a,b,c>0$ ,how we can find the least possible value for :

$$\lfloor \frac{a+b}{c}\rfloor +\lfloor \frac{b+c}{a} \rfloor+\lfloor \frac{c+a}{b} \rfloor$$

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Nice question (+1) – user 1618033 Sep 26 '12 at 13:37

The expression $$E := \frac {a + b} c + \frac {b + c} a + \frac {c + a} b$$ is homogeneous, so we can assume $a + b + c = 1$. Then it becomes $$E = \frac 1 a + \frac 1 b + \frac 1 c - 3$$ By CS inequality we get $$\frac 1 a + \frac 1 b + \frac 1 c \geq \frac {(1 + 1 + 1)^2} {a + b + c} = 9$$ So $E \geq 6$. By the following relation $$\left\lfloor \frac {a + b} c \right\rfloor > \frac {a + b} c - 1$$ (and the other ones got by cyclic permutations of variables), we have $$E' := \left\lfloor \frac {a + b} c \right\rfloor + \left\lfloor \frac {b + c} a \right\rfloor + \left\lfloor \frac {c + a} b \right\rfloor > E - 3 \geq 6 - 3 = 3$$ Being $E'$ an integer, the above inequality is equivalent to $$E' \geq 4$$ To conclude, let's note that for $a = b = 4$ and $c = 3$, $E' = 4$.

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Very nice argument. – Beni Bogosel Sep 21 '12 at 14:56

Put $(a,b,c)=(3,4,4)$ to get 4. I will show this is optimal.

Assume without loss of generality that $a \leq b \leq c$. Two cases:

If $c \geq a+b$ then $\lfloor\frac{c+a}{b}\rfloor \geq 1$ and $\lfloor\frac{c+b}{a} \rfloor \geq \lfloor\frac{a+b+b}{a} \rfloor \geq 3$ so the sum is at least 4.

If $c \leq a+b$ then $\lfloor\frac{a+b}{c}\rfloor \geq 1$, $\lfloor\frac{b+c}{a}\rfloor \geq 2$, $\lfloor\frac{c+a}{b}\rfloor \geq 1$ so the sum is at least 4.

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I would come up with one line proof starting from the trivial fact that $\left\lfloor x \right\rfloor>x-1$. Then

$$\lfloor \frac{a+b}{c}\rfloor +\lfloor \frac{b+c}{a} \rfloor+\lfloor \frac{c+a}{b} \rfloor>\frac{a}{c}+\frac{c}{a}+\frac{b}{a}+\frac{a}{b}+\frac{c}{b}+\frac{b}{c}-3\ge3.$$

Since the left side is an integer, then the smallest value is $4$.

Q.E.D.

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