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I need the following estimate for something I am working on, but I don't immediately see how to establish it.

For $x, y, z \in \mathbb{R}_{\ge 0}$, show that $$2xyz + x^2 + y^2 + z^2 + 1 \ge 2(xy + yz + zx),$$ and I suspect the only point of equality is (1,1,1).

It feels like the sort of thing that ought to have a simple Olympiad-style proof using standard inequalities, but I haven't had any luck thus far; of course, I'll take any proof that I can get.

Also, if this inequality is reminiscent of any others, I would be grateful for references!

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    $\begingroup$ The minimum of the parabola $x^2+y^2+z^2+2xyz-2(xy+yz+zx)$ is equal to $-y(y-2)z(z-2)$. I let you go from there. $\endgroup$ – Henri Cohen Oct 27 at 16:43
  • $\begingroup$ Many thanks to Henri, Iosif, Sean And Fedor for their nice solutions! $\endgroup$ – BPN Oct 28 at 13:12
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Denote $x^2=a^3,y^2=b^3,z^2=c^3$. By AM-GM we have $1+2xyz=1+(abc)^{3/2}+(abc)^{3/2}\geqslant 3\sqrt[3]{1\cdot (abc)^{3/2}\cdot (abc)^{3/2}}=3abc$, so LHS is not less then $$a^3+b^3+c^3+3abc\geqslant ab(a+b)+bc(b+c)+ac(a+c)\\ \geqslant 2(ab)^{3/2}+2(bc)^{3/2}+2(ca)^{3/2}=2(xy+yz+zx),$$ the first inequality is Schur, the second is three AM-GM's.

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Write $x = 1-X$, $y=1-Y$, $z=1-Z$. Then the inequality reduces to $$2XYZ \leq X^2 + Y^2 + Z^2$$ for $X, Y, Z \leq 1$. If $X, Y, Z < 0$ then the inequality is trivial, since LHS < 0. Otherwise suppose $X \in [0, 1]$ wlog. Then $$\text{RHS} - \text{LHS} = X^2 + (1-X^2)Y^2 + (Z-XY)^2 \geq 0.$$ Equality holds iff $X = Y = Z = 0$.

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Let $f(x,y,z)$ denote the difference between the left- and right-hand sides of your inequality. We have to show that $f(x,y,z)\ge0$ if $x,y,z\ge0$.

The minimum of $f(x,y,z)$ in $z\ge0$ is attained at $z_*:=\max(0,x+y-xy)$. If $x+y\le xy$ then $z_*=0$, whereas $f(x,y,0)=1+(x-y)^2>0$. So, without loss of generality (wlog) $x+y\ge xy$, and it remains to show that $$g(x,y):=f(x,y,x+y-xy)\ge0$$ if $$x+y\ge xy.$$ We have $g(x,y)=1-xy(x-2)(y-2)$, and so, $g(x,y)\ge1\ge0$ if $x\le2\le y$ or $y\le2\le x$. So, wlog either $x,y\le2$ or $x,y\ge2$.

Note that $$g(x,y)=1+xy[4(x+y)-xy-4]\ge1+xy[4xy-xy-4]=3(xy)^2-4xy+1=(3xy-1)(xy-1)>0$$ if $x,y\ge2$.

So, wlog $x,y\le2$. Then $0\le x(2-x)\le1$, $0\le y(2-y)\le1$, and hence $$g(x,y)=1-[x(2-x)]\,[y(2-y)]\ge0.$$

It follows that the minimum of $f$ is $0$, and it is attained only at $(1,1)$.

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  • $\begingroup$ Many thanks to both you and Henri, I ended up bashing through a bit of calculus to arrive at precisely this eventually as well. However, it would be nice if there were an elementary proof of some sort: this inequality seems to be the first in a family of more complicated inequalities, and my reason for hoping for an "Olympiad-style" solution was that such an approach seems much more likely to generalise. $\endgroup$ – BPN Oct 27 at 17:18
  • $\begingroup$ If no elementary solutions materialise in a day or two, I'll happily accept this answer. $\endgroup$ – BPN Oct 27 at 17:23
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    $\begingroup$ @BPN : The minimization in $z$ can be done calculus-free, since $f(x,y,z)$ is quadratic in $z$. The case $x,y\le2$ is now also done in a calculus-free way. So, the entire proof is now calculus-free. $\endgroup$ – Iosif Pinelis Oct 27 at 17:30
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Another way.

Since $$\prod_{cyc}((x-1)(y-1))=\prod_{cyc}(x-y)^2\geq0,$$ we can assume that $$(x-1)(y-1)\geq0,$$ which gives $$z(x-1)(y-1)\geq0$$ or $$xyz\geq xz+yz-z.$$ Id est, it's enough to prove that: $$2xz+2yz-2z+x^2+y^2+z^2+1\geq2(xy+xz+yz)$$ or $$(x-y)^2+(z-1)^2\geq0$$ and we are done!

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  • $\begingroup$ Dear Michael, I really quite like this proof and wrote to what appears to be a publicly listed email address for you (english.m.tau.ac.il/profile/michaelro) with the statement of a more complicated inequality that seems to be next in line in this family. If you're interested, do take a look (assuming this email is accurate). Thanks! $\endgroup$ – BPN Nov 13 at 13:22
  • $\begingroup$ @BPN You are welcome! $\endgroup$ – Michael Rozenberg Nov 13 at 16:30
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Another way.

The first step is a homogenization as Fedor Petrov.

We need to prove that: $$x^2+y^2+z^2+3\sqrt[3]{x^2y^2z^2}\geq2(xy+xz+yz).$$ Now, for $xyz=0$ our inequality is obvious.

let $xyz>0$, $x=e^{\frac{a}{2}}$,$y=e^{\frac{b}{2}}$ and $z=e^{\frac{c}{2}}.$

Thus, we need to prove that $$\sum_{cyc}e^a+3e^{\frac{a+b+c}{3}}\geq2\sum_{cyc}e^{\frac{a+b}{2}},$$ which is the T.Popoviciu's inequality for the convex function $f(x)=e^x$.

About Popoviciu see here: https://en.wikipedia.org/wiki/Popoviciu%27s_inequality

Also, the inequality $$x^2+y^2+z^2+3\sqrt[3]{x^2y^2z^2}\geq2(xy+xz+yz)$$ we can get for $n=3$ from the following F.Shleifer's inequality.

Let $x_i\geq0$. Prove that: $$(n-1)\sum_{i=1}^nx_i^2+n\sqrt[n]{\prod_{i=1}^nx_i^2}\geq\left(\sum_{i=1}^nx_i\right)^2.$$

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