$x,y,z \geqslant 0$ and $x^2+y^2+z^2+xyz=4$, prove $x^{\frac85}+y^{\frac85}+z^{\frac85} \geqslant 3$

Question

$x,y,z \geqslant 0$ and $x^2+y^2+z^2+xyz=4$, prove $$x^{\frac85}+y^{\frac85}+z^{\frac85} \geqslant 3$$
1) The equality occurs only at $x,y,z=1$. Let's assume $F=x^n+y^n+z^n$, I noticed that $n=1$ then $F \leqslant 3$ and $n=2$ then $F \geqslant 3$. I believe $n=\frac85=1.6$ is a very sharp inequality.
2) I try trig substitution but cannot remove the radical . $n=\frac85$ is too much for algebraic manipulation.

Answer 1

Let $x=2\cos\alpha$ and $y=2\cos\beta$, where $\{\alpha,\beta\}\subset\left[0,\frac{\pi}{2}\right]$.

Hence, $z=2\cos\gamma$, where $\alpha+\beta+\gamma=\pi$ and $\gamma\in\left[0,\frac{\pi}{2}\right]$.

Let $f(x)=\left(\cos x\right)^{\frac{8}{5}}$.

Since $f''(x)=\frac{-8(1+4\cos2x)}{25\sqrt[5]{\cos^2x}}\geq0$ for all $x\in\left[\frac{\pi}{3},\frac{\pi}{2}\right)$, by Vasc's RCF Theorem

it remains to prove our inequality for $\beta=\alpha$ and $\gamma=\pi-2\alpha$ or

for $y=x$ and $z=2-x^2$, where $0\leq x\leq\sqrt2$.

Id est, it remains to prove that $2x^{\frac{8}{5}}+(2-x^2)^{\frac{8}{5}}\geq3$, which is obvious.

Answer 2

Remark: I give two proofs.

Proof 1

Let $x = \sqrt{a}, y = \sqrt{b}, z = \sqrt{c}$. The condition becomes $a, b, c\ge 0$ with $a + b + c + \sqrt{abc} = 4$. We need to prove that $a^{4/5} + b^{4/5} + c^{4/5} \ge 3$.

We split into two cases:

1) $abc = 0$: WLOG, assume $c=0$. It suffices to prove that $(4-b)^{4/5} + b^{4/5} \ge 3$ for $b\in [0, 4]$. Let $f(b) = (4-b)^{4/5} + b^{4/5}$. Clearly, $f(b)$ is concave on $[0, 4]$. Also, $f(0) = f(4) = 4^{4/5} > 3$. Thus, $f(b) > 3$ for $b\in [0, 4]$.

2) $abc > 0$: From Vasc's Equal Variable Theorem [1, Corollary 1.9, Case 1 (b)], there exists $a_1=b_1\ge 0, c_1\ge 0$ such that $a_1+b_1+c_1 = a+b+c$, $a_1b_1c_1=abc$, and $a_1^{4/5}+b_1^{4/5} + c_1^{4/5} \le a^{4/5} + b^{4/5} + c^{4/5}$. Thus, we only need to prove the case when $a=b$. From $a, c > 0$ and $2a+c + a\sqrt{c} = 4$, we have $a\in (0, 2)$ and $c = (2-a)^2$. We need to prove that $2a^{4/5} + (2-a)^{8/5} \ge 3$ for $a\in (0, 2)$.

Let $x = a^{4/5}$. It suffices to prove that $2x + (2-x^{5/4})^{8/5} \ge 3$ for $x \in (0, 2^{4/5})$. Let $g(x) = 2x + (2-x^{5/4})^{8/5}$. We have $g'(x) = 2 - 2(2-x^{5/4})^{3/5}x^{1/4}$ and $g''(x) = (2x^{5/4}-1)(2-x^{5/4})^{-2/5}x^{-3/4}$. Thus, $g''(x) \le 0$ for $x\in (0, 2^{-4/5}]$ and $g''(x) > 0$ for $x\in (2^{-4/5}, 2^{4/5})$. Thus, $g(x)$ is concave on $(0, 2^{-4/5}]$, and $g(x)$ is convex on $(2^{-4/5}, 2^{4/5})$. Since $g(0) > 3$ and $g(2^{-4/5}) > 3$, we have $g(x) > 3$ on $(0, 2^{-4/5}]$. Since $g'(1) = 0$, we have $g(x) \ge g(1) = 3$ on $(2^{-4/5}, 2^{4/5})$. Thus, $g(x) \ge 3$ on $(0, 2^{4/5})$.

We are done.

Reference

[1] Vasile Cirtoaje, “The Equal Variable Method”, J. Inequal. Pure and Appl. Math., 8(1), 2007. https://www.emis.de/journals/JIPAM/images/059_06_JIPAM/059_06.pdf

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Proof 2

We will use the following bound (the proof is not hard and thus omitted): $$u^{8/5} \ge \frac{3}{20}u^4 - \frac{7}{10}u^3 + \frac{31}{20}u^2, \quad u \in [0, 2].$$

With the bounds above, it suffices to prove that $$\frac{3}{20}(x^4+y^4+z^4) - \frac{7}{10}(x^3+y^3+z^3) + \frac{31}{20}(x^2+y^2+z^2) \ge 3.$$

We may prove it by using e.g. pqr method, though it is ugly. I omitted the proof here. I hope to see nice proofs.