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.


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: $$x^2y^2+x^2+y^2=x^2y^2+\frac{1}{2}(1^2+1^2)(x^2+y^2)\geq$$ $$\geq x^2y^2+\frac{1}{2}(x+y)^2\geq2\sqrt{x^2y^2\cdot\frac{1}{2}(x+y)^2}=\sqrt2(x+y)xy;$$

  1. $y=z$.

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!

  • $\begingroup$ This is not nearly enough detail to be self-sufficient as an answer, at least not one I could follow. What is meant by the substitutions with $u, v$ and $w$? What happened to "in the following cases"? $\endgroup$ – Chris Aug 8 '17 at 4:10
  • $\begingroup$ I see the added detail about the explicit inequalities you have, but I don't see how you made the substitutions and obtained the "linear inequality" in $u, v$ and $w$. (I see you used the elementary symmetric polynomials, so that part isn't as important, though.) I also don't understand the part where you pass to considering $w^3 = 0$, and why that's seemingly the only case that's important to your answer. $\endgroup$ – Chris Aug 8 '17 at 4:33
  • $\begingroup$ I'm sorry, it's difficult for me to parse this, and I believe you have some typos which also make it hard to tell what you're arguing. (For example, I don't believe you meant to write $x^2 + y^2 + x^2 + y^2 \ge \sqrt{2}(x + y)xy$, but rather $x^2y^2 + x^2 + y^2 \ge \sqrt{2}(x+y)xy$.) I'm also wondering how you obtained the form of the inequality with $u, v$ and $w$. Did you use the elementary symmetric polynomials and something like this? $\endgroup$ – Chris Aug 8 '17 at 4:59
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    $\begingroup$ @Chris It was typo. Thank you! $x^2y^2+x^2z^2+y^2z^2=(xy+xz+yz)^2-2(x+y+z)xyz=9v^4-6uw^3$; $x^2+y^2+z^2=(x+y+z)^2-2(xy+xz+yz)=9u^2-6v^2$; $(x+y)(x+z)(y+z)=(x+y+z)(xy+xz+yz)-xyz=9uv^2-w^3$. $\endgroup$ – Michael Rozenberg Aug 8 '17 at 5:04
  • $\begingroup$ Okay - there's also a typo on the exponent $9v^4 - 6uw^2 \cdots$ in the OP, which should be $9v^4 - 6uw^3 \cdots$ there. I think I'm going to put this answer aside for now though. $\endgroup$ – Chris Aug 8 '17 at 5:10

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