Inequality $(xy+z)^2+(yz+x)^2+(zx+y)^2\geq\sqrt{2}(x+y)(y+z)(z+x)$

Question

Let $x,y,z\geq 0$. Prove that $$(xy+z)^2+(yz+x)^2+(zx+y)^2\geq\sqrt{2}(x+y)(y+z)(z+x).$$

Expanding gives $$\sum x^2y^2+\sum x^2+6xyz\geq \sqrt{2}\sum x^2y+2\sqrt{2}xyz.$$

If we use AM-GM on the left-hand side, we get $$\frac{1}{2}(x^2y^2+x^2)\geq x^2y$$ so the left-hand side is at least $\sum x^2y+6xyz$, but this is not enough.

Answer 1

Let $x+y+z=3u$, $xy+xz+yz=3v^2$ and $xyz=w^3$.

Hence, our inequality it's $$\sum_{cyc}(x^2y^2+2xyz+x^2)\geq\sqrt2\sum_{cyc}\left(x^2y+x^2z+\frac{2}{3}xyz\right)$$ or $$9v^4-6uw^3+6w^3+9u^2-6v^2\geq\sqrt2(9uv^2-w^3),$$ which is a linear inequality of $w^3$,

which says that it's enough to prove our inequality for an extremal value of $w^3$.

Now,we see that $x$, $y$ and $z$ are non-negative roots of the following equation. $$(X-x)(X-y)(X-z)=0$$ or $$X^3-3uX^2+3v^2X-w^3=0$$ or $$X^3-3uX^2+3v^2X=w^3,$$ which says that the graph of $f(X)=X^3-3uX^2+3v^2X$ and the line $Y=w^3$

have three common points: $(x,f(x))$, $(y,f(y))$ and $(z,f(z))$.

Now, let $u$ and $v^2$ be constants and $w^3$ changes.

We see that $w^3$ gets a maximal value, when the line $Y=w^3$ will touch to the graph of $f$,

which happens for equality case of two variables.

Also, we see that $w^3$ gets a minimal value, when the line $Y=w^3$ will touch to the graph of $f$,

which happens for equality case of two variables, or when $w^3=0$.

Thus, it's enough to prove our inequality in the following cases.

1. $w^3=0$.

Let $z=0$.

Hence, we need to prove here that $$x^2y^2+x^2+y^2\geq\sqrt2(x+y)xy,$$ which is C-S and AM-GM: $$x^2y^2+x^2+y^2=x^2y^2+\frac{1}{2}(1^2+1^2)(x^2+y^2)\geq$$ $$\geq x^2y^2+\frac{1}{2}(x+y)^2\geq2\sqrt{x^2y^2\cdot\frac{1}{2}(x+y)^2}=\sqrt2(x+y)xy;$$

2. $y=z$.

We need to prove $$2(xy+y)^2+(x+y^2)^2\geq2\sqrt2y(x+y)^2,$$ which is AM-GM and C-S: $$2(xy+y)^2+(x+y^2)^2=2y^2(x+1)^2+(x+y^2)^2\geq$$ $$\geq2\sqrt{2}y(x+1)(x+y^2)\geq2\sqrt2y(x+y)^2.$$ Done!