$x+y+z=a+b+c$ and $xyz=abc$. Suppose that $a≤x<y<z≤c$ and $a<b<c$. Prove that $a=x$, $b=y$ and $c=z$.

algebra-precalculussystems of equations

Let $a,b,c,x,y,z$ be six positive real numbers satisfying
$$x+y+z=a+b+c$$ and $$xyz=abc$$ Further, suppose that $a≤x<y<z≤c$ and $a<b<c$. Prove that $a=x$, $b=y$ and $c=z$.

I have tried to convert this problem into two cubic polynomials using Vieta's Theorem.
But I don't have idea to proceed.
I am adding the two graphs for a visualization of the idea.

Best Answer

Let $f(X) = (X-a)(X-b)(X-c), g(X) = (X-x)(X-y)(X-z)$.
By the equality condition $ f(X) - g(X) = k X$, where $ k = ab+bc+ca - (xy+yz+za)$.

Hint: Now, prove that $k = 0$, from which we can conclude that $ f(X) = g(X)$ so $ a=x, b =y, c = z$.

Hint: Which condition have we not used as yet?

By the condition of the roots, $f(a) = 0, g(a) \leq 0$ so $f(a) - g(a) = k a \geq 0 \Rightarrow k \geq 0$.
By the condition of the roots, $ f(c) = 0, g(c) \geq 0 $ so $ f(c) - g(c) = k c \leq 0 \Rightarrow k \leq 0$.

Note: Technically, with $ f(X) = g(X)$, all that we can conclude is$ \{ a, b, c \} = \{ x, y, z \}$. Adding in $ a < b < c$ and $ x < y < z$, we can then conclude that $ a=x, b =y , c = z$.

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