If $x^2+y^2+z^2=1,$ prove $\frac{x(z^2-y^2+2y)}{z}+\frac{y(x^2-z^2+2z)}{x}+\frac{z(y^2-x^2+2x)}{y}\ge 2$

Question

Let $x,y,z>0: x^2+y^2+z^2=1.$ Prove that$$\frac{x(z^2-y^2+2y)}{z}+\frac{y(x^2-z^2+2z)}{x}+\frac{z(y^2-x^2+2x)}{y}\ge 2$$ I tried to use AM-GM $$\frac{x(z^2-y^2+2y)}{z}+\frac{y(x^2-z^2+2z)}{x}+\frac{z(y^2-x^2+2x)}{y}\ge 3\sqrt[3]{\frac{x(z^2-y^2+2y)}{z}.\frac{y(x^2-z^2+2z)}{x}.\frac{z(y^2-x^2+2x)}{y}}=3\sqrt[3]{(z^2-y^2+2y)(x^2-z^2+2z)(y^2-x^2+2x)}.$$ Now, how to prove $$(z^2-y^2+2y)(x^2-z^2+2z)(y^2-x^2+2x)\ge \frac{8}{27}. \tag{1}$$Also, by $x=y=z=\dfrac{\sqrt{3}}{3},$ is the following true ? $$(z^2-y^2+2y)(x^2-z^2+2z)(y^2-x^2+2x)\ge\frac{8\sqrt{3}}{9}.\tag{2}$$ I need a proof for $(1),(2).$ Furthermore, is there other ideas? Thank you for sharings.

Answer 1

Proof.

Remark. In general, my proof is similar to Michael Rozenberg's proof. Anyway, I found it independently, so I posted it.

Due to the given condition, the OP can be rewritten as$$\frac{x(z^2-y^2)}{z}+\frac{y(x^2-z^2)}{x}+\frac{z(y^2-x^2)}{y}\ge 2\sqrt{x^2+y^2+z^2}\left(\sqrt{x^2+y^2+z^2}-\frac{xy}{z}-\frac{yz}{x}-\frac{zx}{y}\right). \tag{*}$$ Now, denoting $ab=x^2;bc=y^2;ca=z^2 \implies x,y,z\ge 0; a=\dfrac{zx}{y}>0,b=\dfrac{yx}{z}>0,c=\dfrac{zy}{x}>0.$

It is easily to make the inequality to$$(a-b)\sqrt{bc}+(b-c)\sqrt{ca}+(c-a)\sqrt{ab}\ge 2\sqrt{ab+bc+ca}(\sqrt{ab+bc+ca}-a-b-c). \tag{*}$$ WLOG, assume $a\ge b\ge c\ge 0.$

Apply Michael Rozenberg's CBS using, we obtain$$(a+b+c)\sqrt{ab+bc+ca}\ge \sqrt{ab}(a+b-c)+2c(a+b). $$$$(a+b+c)\sqrt{ab+bc+ca}\ge \sqrt{ac}(a+c-b)+2b(a+c).$$ It implies\begin{align*} &2(a+b+c)\sqrt{ab+bc+ca}\\&\ge 2(ab+bc+ca)+2bc+\left(a\sqrt{ab}+c\sqrt{ca}\right)+\left(b\sqrt{ab}+a\sqrt{ca}\right)-\sqrt{abc}(\sqrt{b}+\sqrt{c}). \end{align*} Id est, it is enough to prove that$$\sqrt{ab}(a+b)+\sqrt{ca}(c+a)-\sqrt{abc}(\sqrt{b}+\sqrt{c})\ge (a-b)\sqrt{bc}+(b-c)\sqrt{ca}+(c-a)\sqrt{ab}.$$This inequality is equivalent to$$ \sqrt{ab}(2a+b-c)+\sqrt{ca}(2c+a-b)+2bc\ge \sqrt{bc}(a-b)+\sqrt{abc}(\sqrt{b}+\sqrt{c}),$$$$\iff 2bc+2c\sqrt{ca}+(a-b)(\sqrt{ca}-\sqrt{bc})+\sqrt{ab}(2a+b-2c-\sqrt{bc})\ge0,$$which is obvious by $a\ge b\ge c\ge 0.$

Hence, $(*)$ is proven. The proof is done.

Answer 2

We need to prove that: $$\sum_{cyc}\frac{x(z^2-y^2)}{z}+2\sqrt{x^2+y^2+z^2}\sum_{cyc}\frac{xy}{z}\geq2(x^2+y^2+z^2)$$ or $$2\sqrt{x^2+y^2+z^2}\sum_{cyc}x^2y^2\geq\sum_{cyc}(x^3z^2+2x^3yz-x^2y^2z)$$ and since by AM-G (or by Rearrangement) $$\sum_{cyc}(x^3y^2-x^2y^2z)\geq0,$$ it's enough to prove that: $$2\sqrt{x^2+y^2+z^2}\sum_{cyc}x^2y^2\geq\sum_{cyc}(x^3y^2+x^3z^2+2x^3yz-2x^2y^2z).$$ Now, let $a+b+c=3u$, $ab+ac+bc=3v^2$ and $abc=w^3$.

Thus, we need to prove a linear inequality of $w^3$.

But a linear function gets a minimal value for an extremal value of $w^3$, which by $uvw$ happens in the following cases.

1. $w^3\rightarrow0^+$.

Let $z\rightarrow0^+$ and $y=1$.

Thus, we obtain: $$2x^2\sqrt{x^2+1}\geq x^3+x^2$$ or $$4(x^2+1)\geq(x+1)^2,$$ which is true by C-S: $$4(x^2+1)>(1+1)(x^2+1)\geq(x+1)^2.$$ 2. Two variables are equal.

Let $y=z=1$,

Thus, we need to prove that: $$2(2x^2+1)\sqrt{x^2+2}\geq2(x^3+x^2+1)+2x(x^2+2)-2x(x+2)$$ or $$(2x^2+1)\sqrt{x^2+2}\geq2x^3+1$$ or $$12x^4-4x^3+9x^2+1\geq0,$$ which is obvious.

About $uvw$ see here: https://artofproblemsolving.com/community/c6h278791

Answer 3

Another way.

By my previous post we need to prove that:

$$2\sqrt{x^2+y^2+z^2}\sum_{cyc}x^2y^2\geq\sum_{cyc}(x^3z^2+2x^3yz-x^2y^2z).$$ Now, let $x=\max\{x,y,z\}.$

Thus, by C-S $$2\sqrt{x^2+y^2+z^2}\sum_{cyc}x^2y^2=2\sqrt{(x^2+y^2+z^2)(x^2y^2+x^2z^2+y^2z^2)^2}=$$ $$=2\sqrt{(x^2+y^2+z^2)((x^2y^2+x^2z^2-y^2z^2)^2+4x^2y^2z^2(y^2+z^2))}\geq$$ $$\geq2\left(x(x^2y^2+x^2z^2-y^2z^2)+2xyz(y^2+z^2)\right)$$ and it's enough to prove that: $$2x^3y^2+2x^3z^2-2y^2z^2x+4y^3xz+4z^3xy\geq\sum_{cyc}(x^3z^2+2x^3yz-x^2y^2z)$$ or $$2x^3y^2+x^3z^2-y^2z^2x+x^2y^2z+x^2z^2y+2y^3xz+2z^3xy-2x^3yz-y^3x^2-z^3y^2\geq0,$$ which is true because $$x^3y^2+x^3z^2-2x^3yz=x^3(y-z)^2\geq0,$$ $$x^3y^2\geq y^3x^2,$$ $$z^3xy\geq z^3y^2$$ and $$x^2z^2y\geq y^2z^2x.$$

Answer 4

Using $x \ge x^2$, $y \ge y^2$ and $z \ge z^2$, it suffices to prove that $$\frac{x(z^2 + y)}{z} + \frac{y(x^2 + z)}{x} + \frac{z(y^2 + x)}{y} \ge 2$$ or $$(xy + yz + zx)xyz + x^2y^2 + y^2z^2 + z^2x^2 \ge 2xyz$$ or (using $x^2 + y^2 + z^2 = 1$) $$(xy + yz + zx)xyz + x^2y^2 + y^2z^2 + z^2x^2 \ge 2xyz(x^2 + y^2 + z^2). \tag{1}$$

WLOG, assume that $x \ge y \ge z$. Using $x \le 1$, we have \begin{align*} &(xy + yz + zx)xyz + x^2y^2 + y^2z^2 + z^2x^2 - 2xyz(x^2 + y^2 + z^2)\\ \ge{}& (xy + yz + zx)xyz + (x^2y^2 + y^2z^2 + z^2x^2)x - 2xyz(x^2 + y^2 + z^2)\\ ={}& x\Big((y - z)^2x^2 + (y + z)yz \cdot x - 2yz(y^2 - yz + z^2)\Big)\\ \ge{}& x\Big((y - z)^2\cdot y^2 + (y + z)yz \cdot y - 2yz(y^2 - yz + z^2)\Big)\\ ={}& xy (y - z)(y^2 - 2yz + 2z^2)\\ \ge{}& 0. \end{align*} (Note: Acutally, (1) is true for all $x, y, z\in [0, 1]$.)

We are done.