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Problem. Let $x,y,z\ge 0: x^2+y^2+z^2=1.$ Find minimum $$P=\sqrt[3]{\frac{x}{x^3+2yz}}+\sqrt[3]{\frac{y}{y^3+2zx}}+\sqrt[3]{\frac{z}{z^3+2xy}}.$$

After check some special values of $x,y,z$, I think Min $P=\sqrt[3]{16}$ achieved at $(x,y,z)=\left(0,\dfrac{\sqrt{2}}{2},\dfrac{\sqrt{2}}{2}\right).$

I tried the following Holder (by a hint of Micheal Rozenberg): $$\sum_{cyc}\sqrt[3]{\frac{x}{x^3+2yz}}=$$ $$=\sqrt{\frac{\left(\sum\limits_{cyc}\sqrt[3]{\frac{x}{x^3+2yz}}\right)^3\sum\limits_{cyc}x^3(x^3+2yz)\left(y-\sqrt{\frac{yz}{2}}+z\right)^4}{\sum\limits_{cyc}x^3(x^3+2yz)\left(y-\sqrt{\frac{yz}{2}}+z\right)^4}}\geq$$ $$\geq\sqrt{\frac{\left(\sum\limits_{cyc}\left(2xy-x\sqrt{\frac{yz}{2}}\right)\right)^4}{\sum\limits_{cyc}x^3(x^3+2yz)\left(y-\sqrt{\frac{yz}{2}}+z\right)^4}}\geq\sqrt[3]{16},$$ where the last inequality helps to get a polynomial inequality. I checked and it seems true.

To end proof, we can used BW to verified but it is very ugly. I hope to see a simple idea. Thank you.

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2 Answers2

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Let's use Lagrange multiplier. Let $L(x,y,z,\lambda)=f(x,y,z)-\lambda(x^2+y^2+z^2-1)=\sqrt[3]{\frac{x}{x^3+2yz}}+\sqrt[3]{\frac{y}{y^3+2xz}}+\sqrt[3]{\frac{z}{z^3+2xy}}-\lambda(x^2+y^2+z^2-1).$
Now let's derive:
$L^{'}(x,y,z,\lambda)_x=0=\frac{-2x^3+2yz}{3x^\frac{2}{3}(x^3+2yz)^\frac{4}{3}}-\frac{2zy^\frac{1}{3}}{3(y^2+2xz)^\frac{4}{3}}-\frac{2yz^\frac{1}{3}}{3(z^2+2xy)^\frac{4}{3}}-2\lambda x\\ L^{'}(x,y,z,\lambda)_y=0=-\frac{2zx^\frac{1}{3}}{3(x^3+2yz)^\frac{4}{3}}+\frac{-2y^3+2xz}{3y^\frac{2}{3}(y^3+2xz)^\frac{4}{3}}-\frac{2xz^\frac{1}{3}}{3(z^3+2xy)^\frac{4}{3}}-2\lambda y\\ L^{'}(x,y,z,\lambda)_z=0=-\frac{2yx^\frac{1}{3}}{3(x^3+2yz)^\frac{4}{3}}-\frac{2xy^\frac{1}{3}}{3(y^3+2xz)^\frac{4}{3}}+\frac{-2z^3+2xy}{3z^\frac{2}{3}(z^3+2xy)^\frac{4}{3}}-2\lambda z\\ L^{'}(x,y,z,\lambda)_\lambda=0=0-(x^2+y^2+z^2-1)$

From here you need to solve 4 equaitons with 4 different variables. Couldn't post it as a comment because it was too long. Good luck.

Ariel Fishbein
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Some thoughts.

It suffices to prove that, for all $x, y, z \ge 0$ with $x^2 + y^2 + z^2 = 1$, $$\sum_{\mathrm{cyc}} \sqrt[3]{\frac{x}{x^3+2yz}\cdot \frac{1}{2}} \ge 2.$$

Fact 1. Let $a, b, c \ge 0$ with $a^3b^3 + b^3c^3 + c^3a^3 + 24a^3b^3c^3 \ge 1$. Then $a + b + c \ge 2$.
(The proof is given at the end.)

Let $$a := \sqrt[3]{\frac{x}{x^3+2yz}\cdot \frac{1}{2}}, \quad b := \sqrt[3]{\frac{y}{y^3+2zx}\cdot \frac{1}{2}}, \quad c := \sqrt[3]{\frac{z}{z^3+2xy}\cdot \frac{1}{2}}.$$

We have $$a^3b^3 + b^3c^3 + c^3a^3 + 24a^3b^3c^3 \ge 1. \tag{1}$$ Note: (1) is verified by Mathematica.

By Fact 1, we have $a + b + c \ge 2$.

$\phantom{2}$

Proof of Fact 1.

Equivalently, we need to prove that, for all $a, b, c\ge 0$ with $a + b + c < 2$, $$a^3b^3 + b^3c^3 + c^3a^3 + 24a^3b^3c^3 < 1. \tag{A1}$$

If $a + b = b + c = c + a = 1$ (that is $a = b = c = 1/2$), clearly (A1) is true. In the following, WLOG, assume that $a + b \ne 1$.It suffices to prove that, for all $a, b \ge 0$ with $1 \ne a + b < 2$, $$a^3b^3 + b^3(2 - a - b)^3 + (2 - a - b)^3a^3 + 24a^3b^3(2 - a - b)^3 < 1. \tag{A2}$$

Let $p = a + b$ and $q = ab$. We have $0\le p < 2$ and $p \ne 1$ and $p^2 \ge 4q$.

Using $a^3 + b^3 = (a + b)(a^2 - ab + b^2) = p(p^2 - 3q)$, (A2) is written as $$q^3 + p(p^2 - 3q)\cdot (2 - p)^3 + 24q^3(2 - p)^3 < 1, $$ or \begin{align*} f(q) &:= \left( 24\,{p}^{3}-144\,{p}^{2}+288\,p-193 \right) {q}^{3} \\ &\qquad + \left( -3 \,{p}^{4}+18\,{p}^{3}-36\,{p}^{2}+24\,p \right) q+{p}^{6}-6\,{p}^{5}+ 12\,{p}^{4}-8\,{p}^{3}+1\\ &\ge 0. \end{align*}

Note that $f''(q) = 24\,{p}^{3}-144\,{p}^{2}+288\,p-193 = -1 - 24(2 - p)^3 < 0$. Thus, $f(q)$ is concave. Also, $$f(0) = \left( {p}^{4}-4\,{p}^{3}+3\,{p}^{2}+2\,p+1 \right) \left( p-1 \right) ^{2} > 0, $$ and \begin{align*} &f(p^2/4) \\ ={}& \frac{p-2}{64}\cdot ( 24{p}^{8}-96{p}^{7}+96{p}^{6}+15{p }^{5}-66{p}^{4}+60{p}^{3}-8{p}^{2}-16p-32)\\ >{}& 0. \end{align*} Thus, $f(q) > 0$ on $[0, p^2/4]$. We are done.

River Li
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