1
$\begingroup$

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.

$\endgroup$
1
$\begingroup$

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!

$\endgroup$
  • $\begingroup$ 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
  • 1
    $\begingroup$ @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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.