Prove that $(xy+yz+xz)(x^2+y^2+z^2+x+y+z)≥2(x+y+z)^2$

Question

Let, $x,y,z>0$ such that $xyz=1$, then prove that

$$(xy+yz+xz)(x^2+y^2+z^2+x+y+z)≥2(x+y+z)^2$$

My progress:

Using the Cauchy-Schwars inequality I got

$$(xy+yz+xz)(x^2+y^2+z^2+x+y+z)≥2(xy+yz+xz)(x+y+z)≥2(x+y+z)^2 \implies xy+yz+xz≥x+y+z$$

But, this is not always true.

I also tried

$$x^2+y^2+z^2≥\frac{(x+y+z)^2}{3}$$

I got

$$(xy+yz+xz)(x^2+y^2+z^2+x+y+z)≥(xy+yz+xz)\left(\frac{(x+y+z)^2}{3}+x+y+z\right)≥2(x+y+z)^2 \implies (xy+yz+xz)(x+y+z+3)≥6(x+y+z)$$

But, again I failed.

Answer 1

This is not an answer! But this is what I have so far. One can consider symmetric polynomials $\sigma_1=x+y+z,\ \sigma_2=xy+yz+zz,\ \sigma_3=xyz=1\ $.

We see that $x^2+y^2+z^2=(x+y+z)^2-2(xy+yx+xz)=\sigma_1^2-2\sigma_2.$ Using that we can rewrite your inequality in the following way.

$$\sigma_2(\sigma_1^2-2\sigma_2+\sigma_1)\geq2\sigma_1^2 \Leftrightarrow$$ $$\sigma_1^2\sigma_2-2\sigma_2^2+\sigma_1\sigma_2-2\sigma_1^2\geq0.$$

However, you can show that the following inequalities are true: i) $\sigma_1^2\geq 3\sigma_2;$ 2) $\sigma_2^2\geq 3\sigma_2$; 3) $\sigma_1\sigma_2\geq 9$.

I believe that using these inequalties one can show that your inequality is also true.

Answer 2

Let $S=x+y+z, T=xy+xz+yz$ and note that $S,T \ge 3$ by AMGM and $xyz=1$

The inequality is: $T(S^2+S-2T) \ge 2S^2$

Case 1: $S \ge T$ then $S^2(T-2) \ge S^2 \ge T^2 \ge T(2T-S)$ which rewrites to the required inequality

Case 2: $S \le T$; then we have $T(S^2+S-2T) \ge S(S^2+S-2T)$ so we need to prove only that $S^2+S-2T \ge 2S$ or that $S^2 \ge S+2T$. But (by C-S) $S^2 \le 3(S^2-2T)$ so $S^2 \ge 3T \ge 2T+S$ and we are done

Answer 3

pqr method:

Let $p = x + y + z, q = xy + yz + zx, r = xyz = 1$. Using AM-GM, we have $p \ge 3\sqrt[3]{r} = 3$ and $q \ge 3\sqrt[3]{r^2} = 3$.

The inequality is written as $$q(p^2 - 2q + p) \ge 2p^2$$ or $$(q - 2)p^2 + q(-2q + p) \ge 0.$$ Using $q\ge 3$ and $p^2 \ge 3q$, it suffices to prove that $$(q - 2)\cdot 3q + q(-2q + p) \ge 0$$ or $$q(q + p - 6) \ge 0$$ which is true.

We are done.