Prove that $\sum\limits_{\mathrm{cyc}}\sqrt{\frac{xy}{z}}+6\sqrt{xyz}\ge \sum\limits_{\mathrm{cyc}} \sqrt{3x^3+6xyz}$

Question

Context. I saw a problem on facebook.It's

Let positive real numbers $x,y,z$ such that $x+y+z=x^2+y^2+z^2$. Prove that$$\sqrt{\frac{xy}{z}}+\sqrt{\frac{yz}{x}}+\sqrt{\frac{zx}{y}}+6\sqrt{xyz}\ge \sqrt{3x^3+6xyz}+\sqrt{3y^3+6xyz}+\sqrt{3z^3+6xyz}.$$ Source: Phan Ngoc Chau.

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Here is my attempt.

Since equality occurs at $x=y=z=1,$ I used AM-GM $2\sqrt{3x^3+6xyz}=2\sqrt{3x(x^2+2yz)}\le 3x+x^2+2yz.$

Similarly, we need to prove $$\sqrt{\frac{xy}{z}}+\sqrt{\frac{yz}{x}}+\sqrt{\frac{zx}{y}}+6\sqrt{xyz}\ge\frac{(x+y+z)^2+3(x+y+z)}{2}.$$ Or$$\frac{xy+yz+zx+6xyz}{\sqrt{xyz}}\ge\frac{(x+y+z)^2+3(x+y+z)}{2}.$$ Let $a+b+c=p; ab+bc+ca=q; abc=r.$ By the hypothesis $p^2=2q+p\implies q=\dfrac{p^2-p}{2}$ and we'll prove $$\frac{p^2-p+12r}{\sqrt{r}}\ge p^2+3p.$$I was stuck here.

Can you help me? Thank you.

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Update.

As Calvin Lin point out, my approach is not correct. Hope there is a better idea.

The C-S below doesn't help.$$\sqrt{\frac{xy}{z}}+\sqrt{\frac{yz}{x}}+\sqrt{\frac{zx}{y}}+6\sqrt{xyz}\ge \sqrt{3}\cdot\sqrt{3(x^3+y^3+z^3)+18xyz}.$$

Answer 1

We need to prove that: $$\frac{\sum\limits_{cyc}x^2y^2\sum\limits_{cyc}x^4}{xyz\sum\limits_{cyc}x^2}+6xyz\geq\sqrt3\sum_{cyc}\sqrt{x^2(x^4+2y^2z^2)}$$ Now, by C-S $$\sum_{cyc}\sqrt{x^2(x^4+2y^2z^2)}\leq\sqrt{(x+y+z)\sum_{cyc}(x^5+2xy^2z^2)}$$ and it's enough to prove that: $$\frac{\sum\limits_{cyc}x^2y^2\sum\limits_{cyc}x^4}{xyz\sum\limits_{cyc}x^2}+6xyz\geq\sqrt{3(x+y+z)\sum_{cyc}(x^5+2xy^2z^2)}$$ and the rest is smooth.

Indeed, let $x=\min\{x,y,z\}$, $x=a$, $y=a+u$, $z=a+v$ and $u=tv$.

Thus, BW gives: https://www.wolframalpha.com/input?i=%28%28x%5E2y%5E2%2Bx%5E2z%5E2%2By%5E2z%5E2%29%28x%5E4%2By%5E4%2Bz%5E4%29%2B6x%5E2y%5E2z%5E2%28x%5E2%2By%5E2%2Bz%5E2%29%29%5E2-3x%5E2y%5E2z%5E2%28x%5E2%2By%5E2%2Bz%5E2%29%5E2%28x%2By%2Bz%29%28x%5E5%2By%5E5%2Bz%5E5%2B2xyz%28xy%2Bxz%2Byz%29%29%2Cx%3Da%2Cy%3Da%2Bu%2Cz%3Da%2Bv

Thus, since $$54(u^2-uv+v^2)=27(2u^2-2uv+2v^2)\geq27(u^2+v^2),$$ it's enough to prove that: $$27t^{10}+144t^9+175t^8+82t^7-54t^6-86t^5-28t^4+14t^3+16t^2+16t+1\geq0$$ and $$162t^9+586t^8+815t^7+613t^6+12005t^5-275t^4-175t^3-31t^2+15t+9\geq0$$ and since both last inequalities are true, we are done.

$uvw$ does not help here. At least, I don't see, how we can use $uvw$.

Answer 2

Proof.

We use the so-called isolated fudging.

After homogenization, it suffices to prove that, for all $x, y, z > 0$, $$\frac{x^2 + y^2 + z^2}{x + y + z}\left(\sqrt{\frac{xy}{z}}+\sqrt{\frac{yz}{x}}+\sqrt{\frac{zx}{y}}\right) + 6\sqrt{xyz} \ge \sum_{\mathrm{cyc}} \sqrt{3y^3 + 6xyz}. \tag{1}$$

(1) is written as $$\sum_{\mathrm{cyc}} \left(\frac{2y^2 + z^2 }{2(x+y+z)}\sqrt{\frac{xy}{z}} + \frac{x^2 + 2y^2}{2(x+y+z)}\sqrt{\frac{yz}{x}} + 2\sqrt{xyz} - \sqrt{3y^3 + 6xyz}\right) \ge 0. \tag{2}$$

It suffices to prove that, for all $x, y, z > 0$, $$\frac{2y^2 + z^2 }{2x + 2y + 2z}\sqrt{\frac{xy}{z}} + \frac{x^2 + 2y^2}{2x + 2y + 2z}\sqrt{\frac{yz}{x}} + 2\sqrt{xyz} - \sqrt{3y^3 + 6xyz} \ge 0. \tag{3}$$

Let $$a := \sqrt{\frac{xy}{z}}, \quad b := \sqrt{\frac{yz}{x}}, \quad c := \sqrt{\frac{zx}{y}}.$$ Then $x = ca, y = ab, z = bc$. It suffices to prove that, for all $a, b, c > 0$, $$\frac{(2a^2b^2 + b^2c^2)a}{2ab + 2bc + 2ca} + \frac{(c^2a^2 + 2a^2b^2)b}{2ab + 2bc + 2ca} + 2abc - ab\sqrt{3ab + 6c^2} \ge 0$$ or $$\frac{ab(a + b)(2ab + c^2)}{2ab + 2bc + 2ca} + 2abc - ab\sqrt{3ab + 6c^2} \ge 0$$ or $$\frac{2ab + c^2}{\frac{2ab}{a+b} + 2c} + 2c - \sqrt{3ab + 6c^2} \ge 0$$ or (using $\frac{2ab}{a+b} \le \sqrt{ab}$) $$\frac{2ab + c^2}{\sqrt{ab} + 2c} + 2c - \sqrt{3ab + 6c^2} \ge 0$$ or $$\left(\frac{2ab + c^2}{\sqrt{ab} + 2c} + 2c\right)^2 - (3ab + 6c^2) \ge 0$$ or $$\frac{(c - \sqrt{ab})^4}{(\sqrt{ab} + 2c)^2}\ge 0$$ which is true.

We are done.

Answer 3

Proof.

By squaring both side, let $p=x+y+z, q=xy+yz+zx, r=xyz$ $$(RHS)^2=3\sum\left(x^3 + 2xyz + 2 \sqrt{xy(x^2+2yz)(y^2+2zx)}\right).$$ Notice that$$(x^2+2yz)(y^2+2zx)+(y^2+2zx)(z^2+2xy)+(z^2+2xy)(x^2+2yz)$$$$=(xy+yz+xz)\left[2(x+y+z)^2-3(xy+yz+zx)\right].$$ By using Cauchy-Schwarz \begin{align*} &2\sum\sqrt{xy(x^2+2yz)(y^2+2zx)}\\ &\le 2\sqrt{xy+yz+zx}.\sqrt{(x^2+2yz)(y^2+2zx)+(y^2+2zx)(z^2+2xy)+(z^2+2xy)(x^2+2yz)} \\ &= 2 (xy+yz+zx). \sqrt{2(x+y+z)^2-3(xy+yz+zx)}. \end{align*} Also by AM-GM \begin{align*} &2(xy+yz+zx) \sqrt{2(x+y+z)^2-3(xy+yz+zx)}\\& = 2q\sqrt{2p^2-3q}\\&\le \frac{3q}{p}(p^2-q). \end{align*} Thus,$$(RHS)^2 \le 3\left(x^3+y^3+z^3+6xyz+\frac{3q}{p}(p^2-q)\right)=3\left(p^3+9r - \frac{3q^2}{p}\right).$$The condition gives $p^2=p+2q,$ $$(LHS)^2 = r\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}+6\right)^2= \frac{(p^2q-2q^2+6pr)^2}{p^2r}.$$And we'll prove$$\frac{\left(p^2q - 2(q^2-3pr)\right)^2}{rp^2}\ge 3\left(p^3+9r - \frac{3q^2}{p}\right).$$ Or$$\left(p^2q - 2(q^2-3pr)\right)^2 \ge 3rp(p^4+9pr-3q^2).$$It's,\begin{align*} &\left(p^2q - 2(q^2-3pr)\right)^2 -3rp(p^4+9pr-3q^2)\\ &=\left[\sum_{cyc}(ab-ac)^2\right]\left[\frac{a+b+c}{2}\sum_{cyc}a(a-b)(a-c)+\sum_{cyc}(ab-ac)^2\right]\ge 0. \end{align*} The last inequality is true by Schur of third degree.

Hence, the proof is done. Equality holds at $x=y=z=1.$