$(a+1)x^{2} +(b+1)y^{2} +(c+1)z^{2} \geq 2(xy+yz+zx)$ Prove $(a+1)x^{2} +(b+1)y^{2} +(c+1)z^{2} \geq 2(xy+yz+zx)$ for $a,b,c>0$ and $abc\geq1$.
My try: $(a+1)x^2-4xy+(b+1)y^2+(a+1)x^2-4xz+(c+1)z^2+(b+1)y^2-4yz+(c+1)z^2\geq0$
Then I had to prove that $(a+1)x^2-4xy+(b+1)y^2\geq2x^2-4xy+2y^2$, but I realized that there's no way of using $abc\geq1$.
Any help is appreciated!
 A: *

*Clearly we just need to focus on $ abc = 1$. Suppose $x, y, z$ are fixed. How can we minimize the $LHS$ subject to $ abc = 1$?

*If $c$ is fixed, then we have $(a+1)x^2 + (\frac{1}{ac} + 1) y^2 + (c+1) z^2$, and differentiating with respect to $a$ gives $ x^2  - \frac{1}{c a^2} y^2 = 0 $, or that $ \frac{x^2}{y^2} = \frac{b}{a}$. (Check that this yields a minimum.)

*Hence, the minimum occurs when $ x^2 : y^2 : z^2 = \frac{1}{a} : \frac{1}{b} : \frac{1}{c}$ and $ abc = 1$, which yields the solution $ a = \frac{ y^{2/3} z^{2/3} } { x^{4/3}}$ etc. Hence

$$LHS = (a+1)x^2 + (b+1) y^2 + (c+1)z^2 \geq \sum ( \frac{ y^{2/3} z^{2/3} } { x^{4/3}} + 1 ) x^2 . $$

*

*Let $ x^{1/3} = p, y^{1/3} = q, z^{1/3} = r$, then it remains to show that

$$ \sum (p^2q^2r^2 + p^6) \geq 2\sum p ^3 r^3  = RHS$$

*

*This is true by Shurs on $ p^2, q^2 r^2$, which  gives us the first inequality, and the second is just AM-GM:

$$ p^6 + q^6 + r^6 + 3 p^2 q^2 r^2 +  \geq \sum p^2q^2(p^2+q^2) \geq 2 \sum p^3 q^3.$$

*

*Equality holds when

*

*$ p = q = r \Rightarrow x = y = z, a = b = c = 1$

*$ p = 0 , q = r \Rightarrow x = 0, y = z$, but then $a \rightarrow \infty, b = c \rightarrow 0 $ which isn't allowed.



Notes

*

*Yes, we could get to $ x^2 : y^2 : z^2 = \frac{1}{a} : \frac{1}{b} : \frac{1}{c}$ separately, but differentiation was the fastest / easiest.

*It is not true that $ax^2 + by^2 + cz^2 \geq xy+yz+zx$. Likewise, the inequality of $ \sum (1.01 a+1) x^2 \geq 2.01\sum xy$ is also not true.

A: We can assume at least two of ${x}$, ${y}$, ${z}$ non negative, otherwise we take their additive inverse. Let ${y}$, $z\ge0$.
Consider $x{}$ the only variable. The inequality can be regarded as a quadratic function of $x$. Since the coefficient of $x^2$is positive, we only need $\frac\Delta4\le0$, which requires
\[(a+1)\left((b+1)y^2+(c+1)z^2\right)\ge(y+z)^2+2(a+1)yz.\]Or $(ab+a+b)y^2+(ac+a+c)z^2\ge2(a+2)yz$. Using AM-GM, it suffices to show
\[(ab+a+b)(ac+a+c)\ge a+2.\]Expand the left side, since we have $ab\cdot ac\ge a$, $ab\cdot c$, $ac\cdot b\ge1$ and the rest $\ge0$, we're done.
