# Prove that $0 \le xy+yz+xz - 2xyz \le 7/27, \:\:\:x,y,z \ge 0$ with $x+y+z=1$. [duplicate]

Prove that $$0 \le xy+yz+xz - 2xyz \le 7/27, \:\:\:x,y,z \ge 0$$ with $$x+y+z=1$$.

Attempt:

We must prove $$xy + yz + xz \le 2xyz + 7/27$$

Assume $$x,y,z>0$$, since the case when all equal $$0$$ is clear.

Next, I have that $$xy = \sqrt{x^{2}} \sqrt{y^{2}} \le (x^{2} + y^{2})/2$$ by AM-GM. Same also for the $$yz, xz$$. Summing the inequalities we get $$xy+ yz + xz \le x^{2} +y^{2} +z^{2}$$

Ley $$\alpha =xy+yz+xz$$, now

$$x^{2} +y^{2} +z^{2} =(x+y+z)^{2} - 2 \alpha$$ so we also have $$3 \alpha \le 1 \implies \alpha \le 1/3= \frac{9}{27}.$$

Also $$xyz \le (x+y+z)^{3}/27 = 1/27$$ (by AM-GM).

So $$\alpha =xy + yz + xz \le 7/27 + 2/27$$

In order to finish the proof we must have

$$1/27 \le xyz$$ but instead we have $$xyz \le 1/27$$. This is the tricky part..

This problem is from IMO 1984:there is a solution in this link https://mks.mff.cuni.cz/kalva/imo/isoln/isoln841.html that manipulates $$xy+yz+xz-2xy$$ into something with $$(1-2x)(1-2z)(1-2y)$$ and then use AM-GM. This only proves for the case $$x,y,z < 1/2$$.

How to finish the proof?

• Jun 20, 2019 at 9:41

As mentioned in the solution you have linked at the end,

$$yz+zx+xy-2xyz=\frac14(1 - 2x)(1 - 2y)(1 - 2z) + \frac14$$

Part 1:

If $$x,y$$ and $$z$$ are all less than $$\frac12$$, we get by AM-GM that $$(1-2x)(1-2y)(1-2z)\leq \left(\frac{3-2(x+y+z)}{3}\right)^3=\frac1{27}$$ and thus

$$yz+zx+xy-2xyz\leq \frac{7}{27}$$

In the case where one of $$x,y$$ or $$z$$ is greater than $$\frac12$$ (note that either $$0$$ or exactly $$1$$ of $$x,y$$ and $$z$$ is greater than $$\frac12$$), the term $$(1-2x)(1-2y)(1-2z)<0$$ so $$yz+zx+xy-2xyz=\frac14(1 - 2x)(1 - 2y)(1 - 2z) + \frac14<\frac14<\frac{7}{27}$$ and the inequality holds true.

Part 2:

We must prove that $$yz+zx+xy-2yz\geq0$$

Again, we have that $$yz+zx+xy-2xyz=\frac14(1 - 2x)(1 - 2y)(1 - 2z) + \frac14$$

As $$x,y,z<1$$, $$|1-2x|\leq1$$ and $$|(1-2x)(1-2y)(1-2z)|\leq1$$.

This gives that $$-1\leq (1-2x)(1-2y)(1-2z)\leq 1$$.

Thus, $$yz+zx+xy-2xyz=\frac14(1 - 2x)(1 - 2y)(1 - 2z) + \frac14 \geq \frac{-1}{4}+\frac14=0$$

and the proof is done.

The left inequality is satisfied for $$(x,y,z)=(1,0,0)$$ and the right inequality is satisfied for $$(x,y,z)=(\frac13,\frac13,\frac13)$$