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Please help to prove this.

Assume $x\geq y\geq z>0$, then $${x^2y\over z}+{y^2z\over { x}}+{z^2x\over y}\geq{xy^2\over z}+{yz^2\over { x}}+{zx^2\over y}.$$

Thanks.

user147742
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1 Answers1

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First note that if we set any two of the variables equal then we get equality.

Multiply through by $xyz$, this is then equivalent to showing $x^3y^2+y^3z^2+z^3x^2-x^2y^3-y^2z^3-z^2x^3 \ge 0$

By our initial observation we know that $(x-y)(y-z)(x-z)$ must divide this polynomial, so just carry out the calculation (or use the fact that it's symmetric and you know the degrees) to get that:

$x^3y^2+y^3z^2+z^3x^2-x^2y^3-y^2z^3-z^2x^3 = (x-y)(y-z)(x-z)(xy+yz+xz)$ which is obviously positive for the range of values we care about.

Nate
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  • Thanks for the quick solution. Can I know how you get to know that $x-y$ can divide the polynomial? – user147742 May 10 '14 at 17:10
  • @user147742 Consider it a polynomial in $x$, and check what $P(y)$ is. – Macavity May 10 '14 at 17:18
  • If you have a polynomial in any number of variables for which is $0$ whenever $x=y$ then $(x-y)$ divides the polynomial. Think of this as a generalization of the fact for one variable polynomials that if $r$ is a root of $p(x)$ then $(x-r)$ divides $p(x)$, and in fact you can prove it in essentially the same way. These are both just special instances of a basic fact from algebraic geometry called the Nullstellensatz, which relates vanishing sets of polynomials to ideals in the polynomial ring. – Nate May 10 '14 at 17:18