, 2012

Let $a$, $b$, $c > 0$ and $a \cdot b \cdot c = 1$.

Suppose $a + b + c > \frac{1}{a} + \frac{1}{b} + \frac{1}{c}$.

Note that $\frac{1}{a} + \frac{1}{b} + \frac{1}{c} = \frac{b c + a c + a b}{a b c} = b c + a c + a b$ since $a b c = 1$.

So the inequality becomes:

$$a + b + c > a b + b c + c a$$

Let us consider the possible cases for $a$, $b$, $c$.

Since $a b c = 1$, at least one of $a$, $b$, $c$ is $\geq 1$ and at least one is $\leq 1$.

Suppose two of them are greater than $1$, say $a > 1$, $b > 1$, $c < 1$ (since $a b c = 1$).

Let $a = x > 1$, $b = y > 1$, $c = z < 1$, $x y z = 1$.

Then $z = \frac{1}{x y}$.

Now,
$$a + b + c = x + y + \frac{1}{x y}$$
$$a b + b c + c a = x y + y \cdot \frac{1}{x y} + x \cdot \frac{1}{x y} = x y + \frac{y}{x y} + \frac{x}{x y} = x y + \frac{1}{x} + \frac{1}{y}$$
So the inequality becomes:
$$x + y + \frac{1}{x y} > x y + \frac{1}{x} + \frac{1}{y}$$
But $x > 1$, $y > 1$, so $x y > 1$.

Let us check if this is possible.

Let $x = y = t > 1$, $z = \frac{1}{t^2} < 1$.

Then:
$$a + b + c = t + t + \frac{1}{t^2} = 2t + \frac{1}{t^2}$$
$$a b + b c + c a = t^2 + t \cdot \frac{1}{t^2} + t \cdot \frac{1}{t^2} = t^2 + \frac{t}{t^2} + \frac{t}{t^2} = t^2 + \frac{2}{t}$$
So the inequality is:
$$2t + \frac{1}{t^2} > t^2 + \frac{2}{t}$$
Let us check for $t = 2$:
$$2 \cdot 2 + \frac{1}{4} = 4.25$$
$$2^2 + \frac{2}{2} = 4 + 1 = 5$$
So $4.25 < 5$, so the inequality does not hold.

Try $t = 1.5$:
$$2 \cdot 1.5 + \frac{1}{(1.5)^2} = 3 + \frac{1}{2.25} \approx 3.444$$
$$(1.5)^2 + \frac{2}{1.5} = 2.25 + 1.333 = 3.583$$
So $3.444 < 3.583$, again the inequality does not hold.

Try $t = 3$:
$$2 \cdot 3 + \frac{1}{9} = 6.111$$
$$9 + \frac{2}{3} = 9.666$$
So $6.111 < 9.666$.

So, if two numbers are greater than $1$, the inequality does not hold.

Now, suppose all three are less than $1$. Then $a b c < 1$, contradiction.

Suppose all three are greater than $1$. Then $a b c > 1$, contradiction.

Suppose exactly one is greater than $1$, say $a > 1$, $b < 1$, $c < 1$.

Let $a = x > 1$, $b = y < 1$, $c = z < 1$, $x y z = 1$.

Then $y z = \frac{1}{x}$.

Let $y = s < 1$, $z = \frac{1}{x s} < 1$.

Now,
$$a + b + c = x + s + \frac{1}{x s}$$
$$a b + b c + c a = x s + s \cdot \frac{1}{x s} + x \cdot \frac{1}{x s} = x s + \frac{1}{x} + \frac{1}{s}$$
So the inequality is:
$$x + s + \frac{1}{x s} > x s + \frac{1}{x} + \frac{1}{s}$$
Let us try $x = 2$, $s = 0.5$, $z = \frac{1}{2 \cdot 0.5} = 1$.

But $z = 1$, so two numbers are $\leq 1$, one is $> 1$.

Try $x = 2$, $s = 0.4$, $z = \frac{1}{2 \cdot 0.4} = 1.25$ (but $z > 1$). So $z$ must be $< 1$.

Try $x = 2$, $s = 0.6$, $z = \frac{1}{2 \cdot 0.6} = 0.833$.

So $a = 2$, $b = 0.6$, $c = 0.833$.

Compute:
$$a + b + c = 2 + 0.6 + 0.833 = 3.433$$
$$a b + b c + c a = 2 \cdot 0.6 + 0.6 \cdot 0.833 + 0.833 \cdot 2 = 1.2 + 0.5 + 1.666 = 3.366$$
So $3.433 > 3.366$, the inequality holds.

Try $x = 3$, $s = 0.3$, $z = \frac{1}{3 \cdot 0.3} = 1.111$ (not $< 1$).

Try $x = 1.5$, $s = 0.7$, $z = \frac{1}{1.5 \cdot 0.7} \approx 0.952$.

$a = 1.5$, $b = 0.7$, $c = 0.952$.

$a + b + c = 1.5 + 0.7 + 0.952 = 3.152$

$a b + b c + c a = 1.5 \cdot 0.7 + 0.7 \cdot 0.952 + 0.952 \cdot 1.5 = 1.05 + 0.666 + 1.428 = 3.144$

So $3.152 > 3.144$, the inequality holds.

Therefore, the inequality holds only when exactly one of $a$, $b$, $c$ is greater than $1$.

Thus, if $a + b + c > \frac{1}{a} + \frac{1}{b} + \frac{1}{c}$, then exactly one of $a$, $b$, $c$ is greater than $1$.