Simple inequality over positive reals

Problem

Let $x,y,z$ be real positive numbers with $xyz=1$. Prove: $$2(x+y+z) \geq 3xyz + xy+yz+zx$$

Note : I don't know whether the inequality is true or not. I couldn't find a prove in the place found it nor a solution to it.

My try

I firstly took $a = \frac{x}{y}$, $b = \frac{y}{z}$ and $c = \frac{z}{x}$ so that the expression becomes $2\sum_{cyc} a^2c \geq 3abc + \sum_{cyc}a^2b$.

Then I tried to think of it as a function $f(a,b,c)$ and say that if we let two variable swap, we would come from one to another, letting us see that the order of the variables has nothing to do with the result. But I couldn't get anything here.

Any hint or solution would be really apreciated.

PD: I tried using the search function, but I couldn't find anything. so, sorry if this is a duplicate

• It seems strange to say "$xyz=1$" and then write "$3xyz$" instead of $3$. – David Richerby Dec 2 '17 at 21:18

If $x=y=4$ and $z=1/16$, then $xyz=1$, and $$2(x+y+z)-(3xyz+xy+yz+zx)=-27/8<0.$$

with your Substitution we get $$a(a-b)(b-c)+b(a-c)(b-c)+c(a-c)(a-b)\geq 0$$ this is true for $$a\geq b\geq c$$

It is false. Consider $x=\frac1{20},$ $y=4,$ and $z=5.$

HINT

Let's observe that:

$2(x+y+z) \geq 3xyz + xy+yz+zx \geq xyz \left( 3 + \frac1x+\frac1y+\frac1z\right)\geq \left( 3 + \frac1x+\frac1y+\frac1z\right)$

and use AM-HM inequality.

It strikes me that the inequality contains all three basic symmetric polynomials in three variables. Watch what happens when we multiply out a degree-three polynomial with roots $x,y,z$:

$$(t-x)(t-y)(t-z) = t^3 - (x+y+z)t^2 + (xy+yz+zx)t - xyz$$

So since apparently $xyz=1$, the question is: is there a polynomial $t^3 - at^2 + bt - 1$ with three positive real roots such that $2a < 3+b$? If so, that would give a counterexample to the given inequality.

And indeed there is, for example: $t^3 - 8t^2 + 16t - 1$. WolframAlpha gives roots, and they aren't pretty. (WA even gets somewhat confused, thinking that one of the roots has a small imaginary part... This is impossible, since in a completely real polynomial, complex roots always occur in pairs; and in any case, the plot tells otherwise of course.) But taking these roots as your values of $x,y,z$ (in any order -- the polynomials in your inequality are symmetric!) will give a counterexample to the inequality, showing that it cannot be always true, like others have already shown.

I thought it would be interesting to show an alternate approach to the more "direct" methods of the other answerers.