Kvant Math Problem 51
Let the numbers be $a,b,c>0$ with $abc=1$.
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Problem
Prove that if the product of three positive numbers is equal to 1, and the sum of these numbers is strictly greater than the sum of their reciprocals, then exactly one of these numbers is greater than 1.
All-Union Mathematical Olympiad (1970, Grade 8)
Exploration
Let the numbers be $a,b,c>0$ with $abc=1$. Since
$$\frac1a=bc,\qquad \frac1b=ca,\qquad \frac1c=ab,$$
the given inequality becomes
$$a+b+c>ab+bc+ca.$$
The conclusion concerns how many of $a,b,c$ exceed $1$. Because their product is $1$, all three cannot be greater than $1$, and exactly two greater than $1$ forces the third to be less than $1$. The possible distributions are therefore:
$$(0\text{ numbers}>1),\qquad (1\text{ number}>1),\qquad (2\text{ numbers}>1).$$
The first possibility means $a,b,c\le 1$. Since $abc=1$, this can happen only when $a=b=c=1$, because if one of them were strictly less than $1$, the product would be strictly less than $1$. In that case
$$a+b+c=ab+bc+ca=3,$$
which contradicts the strict inequality.
The remaining issue is to exclude the case of exactly two numbers greater than $1$. Trying an example,
$$a=2,\quad b=2,\quad c=\frac14,$$
gives
$$a+b+c=4.25,\qquad ab+bc+ca=5,$$
so the required inequality fails. This suggests that whenever two numbers exceed $1$,
$$ab+bc+ca\ge a+b+c.$$
The crucial step is proving this rigorously. If $a>1$, $b>1$, and $c=\frac1{ab}$, then
$$ab+bc+ca-(a+b+c) =(ab-a-b+1)+\left(\frac1a+\frac1b-\frac1{ab}-1\right).$$
Each bracket factors:
$$ab-a-b+1=(a-1)(b-1),$$
$$\frac1a+\frac1b-\frac1{ab}-1 =-\frac{(a-1)(b-1)}{ab}.$$
Hence
$$ab+bc+ca-(a+b+c) =(a-1)(b-1)\left(1-\frac1{ab}\right).$$
When $a,b>1$, both factors are positive, so the difference is positive. Thus the given inequality cannot hold when two numbers exceed $1$.
This leaves exactly one number greater than $1$.
Problem Understanding
We are given positive numbers $a,b,c$ satisfying $abc=1$ and
$$a+b+c>\frac1a+\frac1b+\frac1c.$$
We must prove that exactly one of $a,b,c$ is greater than $1$.
This is a Type B problem, a pure proof.
Using $abc=1$, the inequality can be rewritten as
$$a+b+c>ab+bc+ca.$$
The core difficulty is showing that the inequality is incompatible with the possibility that two of the numbers exceed $1$.
Proof Architecture
First, rewrite the inequality as $a+b+c>ab+bc+ca$ by using $abc=1$.
Next, show that the case in which none of the numbers exceeds $1$ is impossible. Since $abc=1$ and all three numbers are at most $1$, one must have $a=b=c=1$, which contradicts the strict inequality.
Then assume exactly two numbers exceed $1$, say $a>1$ and $b>1$. Express $c$ as $1/(ab)$ and compute
$$ab+bc+ca-(a+b+c).$$
Factor the result as
$$(a-1)(b-1)\left(1-\frac1{ab}\right),$$
which is positive. Hence $ab+bc+ca>a+b+c$, contradicting the hypothesis.
Since the only possible counts of numbers exceeding $1$ are $0$, $1$, and $2$, and the first and third have been excluded, exactly one number exceeds $1$.
The hardest part is the exclusion of the case of two numbers greater than $1$, because it requires a correct algebraic factorization and sign analysis.
Solution
Let
$$a,b,c>0,\qquad abc=1,$$
and suppose that
$$a+b+c>\frac1a+\frac1b+\frac1c.$$
Since $abc=1$,
$$\frac1a=bc,\qquad \frac1b=ca,\qquad \frac1c=ab.$$
Therefore the given inequality is equivalent to
$$a+b+c>ab+bc+ca.$$
We examine the possible numbers of elements among $a,b,c$ that are greater than $1$.
The product condition $abc=1$ excludes the possibility that all three are greater than $1$.
Suppose that none of them is greater than $1$. Then
$$a\le1,\qquad b\le1,\qquad c\le1.$$
If one of these inequalities were strict, the product would satisfy $abc<1$, contrary to $abc=1$. Hence
$$a=b=c=1.$$
Then
$$a+b+c=ab+bc+ca=3,$$
which contradicts
$$a+b+c>ab+bc+ca.$$
Thus the case of no numbers greater than $1$ is impossible.
Now suppose that exactly two numbers are greater than $1$. After relabeling, let
$$a>1,\qquad b>1.$$
Since $abc=1$,
$$c=\frac1{ab}.$$
Consider
$$ab+bc+ca-(a+b+c).$$
Substituting $c=\frac1{ab}$ gives
$$ab+\frac1a+\frac1b-a-b-\frac1{ab}.$$
Rewrite this as
$$(ab-a-b+1)+\left(\frac1a+\frac1b-\frac1{ab}-1\right).$$
The first bracket equals
$$(a-1)(b-1).$$
For the second bracket,
$$\frac1a+\frac1b-\frac1{ab}-1 =\frac{a+b-1-ab}{ab} =-\frac{(a-1)(b-1)}{ab}.$$
Hence
$$ab+bc+ca-(a+b+c) =(a-1)(b-1)\left(1-\frac1{ab}\right).$$
Because $a>1$ and $b>1$, we have
$$(a-1)(b-1)>0$$
and
$$ab>1,$$
so
$$1-\frac1{ab}>0.$$
Therefore
$$ab+bc+ca-(a+b+c)>0.$$
Thus
$$ab+bc+ca>a+b+c,$$
contradicting the hypothesis.
The cases of zero numbers greater than $1$ and of two numbers greater than $1$ are both impossible. Since all three numbers cannot be greater than $1$ because their product is $1$, the only remaining possibility is that exactly one of $a,b,c$ is greater than $1$.
This completes the proof.
∎
Verification of Key Steps
The first delicate step is the reduction
$$\frac1a+\frac1b+\frac1c=ab+bc+ca.$$
This uses only $abc=1$:
$$\frac1a=\frac{bc}{abc}=bc,$$
and similarly for the other two reciprocals. Without this substitution, the later factorization does not appear naturally.
The second delicate step is the factorization in the case $a,b>1$. Starting from
$$ab+\frac1a+\frac1b-a-b-\frac1{ab},$$
grouping terms incorrectly can obscure the sign. Computing directly,
$$ab-a-b+1=(a-1)(b-1),$$
while
$$\frac1a+\frac1b-\frac1{ab}-1 =\frac{a+b-1-ab}{ab} =-\frac{(a-1)(b-1)}{ab}.$$
Adding these expressions yields
$$(a-1)(b-1)\left(1-\frac1{ab}\right).$$
Every factor is positive when $a,b>1$.
The third delicate step is excluding the possibility that none of the numbers exceeds $1$. The argument relies on positivity. If
$$0<a,b,c\le1$$
and one of them is strictly less than $1$, then multiplying by the other two numbers, each at most $1$, gives $abc<1$. Hence $abc=1$ forces $a=b=c=1$.
Alternative Approaches
After rewriting the condition as
$$a+b+c>ab+bc+ca,$$
one may use the identity
$$(a-1)(b-1)(c-1) =abc-ab-bc-ca+a+b+c-1.$$
Since $abc=1$, this becomes
$$(a-1)(b-1)(c-1) =a+b+c-ab-bc-ca.$$
The hypothesis implies that the right-hand side is positive, so
$$(a-1)(b-1)(c-1)>0.$$
A positive product of the three factors means either all three factors are positive or exactly two are negative. The first possibility would give $a,b,c>1$, contradicting $abc=1$. Hence exactly two factors are negative and one is positive, which means exactly one of $a,b,c$ exceeds $1$.
This approach is shorter. The main solution was chosen because it proceeds directly from the possible distributions relative to $1$ and isolates the only nontrivial case through an explicit factorization.