Proof that one of three terms with 3 variables is always smaller or equal to 1

Question

Let $x, y$ and $z$ positive real numbers with $x + y + z = 3$. Prove that at least one of the three numbers

$x (x + y − z), y (y + z − x)$ or $z (z + x − y)$

is less than or equal to $1$.

Can someone give me a hint where to start with this?

Answer 1

The three quantities are $x(3-2 z), y(3-2 x), z(3-2y).$ One of $x, y, z$ is not greater than $1.$ Suppose, without log of generality, that it is $x.$ Then, if both $y, z$ are $\geq 1,$ then the first sum is $\leq 1.$ So, let us suppose that $x, y \leq 1,$ and so $z\geq 1,$ then the first quantity is less than $1.$ The alternative is that $x, z \leq 1, y\geq 1,$ then the the third quantity is $\leq 1.$

Answer 2

• If any of (cyclic) $x+y - z $ is negative, then we are done.

• Otherwise, show that $0 < xyz \leq 1$ and $ 0 \leq \prod (x+y-z) \leq 1$.

• Hence, the product of those 3 numbers is between 0 and 1, hence at least one of them is $ \leq 1$.

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Note: Try showing that at least one of them is $ \geq 1$.