three variable inequality $x+y+z\le xyz+2$ with constraint $x^2+y^2+z^2=2$

Question

Let $x$, $y$ and $z$ be three real numbers such that $x^2+y^2+z^2=2$.

it is asked to prove that

$$x+y+z \le xyz+2$$

I tried using Lagrange multipliers but I'm stuck with the following system

$$\begin{cases} x^2 + y^2 + z^2 = 2 \\ 1-yz = 2\lambda x \\1-xz = 2\lambda y \\ 1-xy = 2\lambda z \end{cases} $$

thanks for any advice, hint.

Answer 1

By C-S $$x+y+z-xyz=x\cdot(1-yz)+(y+z)\cdot1\leq$$ $$\leq\sqrt{(x^2+(y+z)^2)((1-yz)^2+1^2)}=\sqrt{2(1+yz)(2-2yz+y^2z^2)}\leq2$$ because the last inequality it's just $$y^2z^2(1-yz)\geq0,$$ which is true because $$yz\leq\frac{y^2+z^2}{2}\leq1.$$

Answer 2

The best proof is using Cauchy-Schwarz. I have another proof

We need to prove $$4-(x+y+z-xyz)^2 \geqslant 0.$$ equivalent to $$4-(x+y+z)^2+2xyz(x+y+z)-4x^2y^2z^2 \geqslant 0.$$ Setting $A=x^2+y^2+z^2=2,$ the inequality equivalent to $$4[4-(x+y+z)^2+2xyz(x+y+z)-x^2y^2z^2] \ge 0$$ or $$2^3-2^2 \cdot [(x+y+z)^2-2]+2 \cdot 4xyz \cdot (x+y+z)-4x^2y^2z^2\geqslant 0,$$ $$A^3-A^2\cdot\left[(x+y+z)^2-A\right]+4 \cdot A \cdot xyz \cdot (x+y+z)-4x^2y^2z^2 \geqslant 0,$$ or $$(2-2xy)(2-2yz)(2-2zx)+4x^2y^2z^2 \geqslant 0.$$ Which is true because $$2-2xy=(x^2-2xy+y^2)+z^2=(x-y)^2+z^2 \geqslant 0.$$ The proof is completed.