$$3 = 3 \frac{x^5+y^5+z^5}{xy+yz+xz} \geq x^3+y^3+z^3 \geq x^2y + y^2z +z^2x$$

$$\sum_{cyc} x^5+y^5+1+1+1 \ge \sum_{cyc} 5xy$$$$3 \ge x^5+y^5+z^5 (1)$$$$\sum_{cyc} x^5+x^5+y^5+1+1 \ge \sum_{cyc} 5x^2y$$$$ 15 \ge \sum_{cyc} 3x^5+2 \ge \sum_{cyc} 5x^2y$$Solved by Mehmet Burak Gonul

Alternative: Set $S_n = x^n+y^n+z^n$. By power mean inequality, $(S_5/3)^{1/5}\geqslant S_1/3$, that is, $S_5\geqslant S_1^5/3^4$. Now we also have $S_5=xy+yz+zx\leqslant S_1^2/3$. These two yield $S_1^2/3\geqslant S_1^5/3^4$, that is, $S_1\leqslant 3$, and $S_5\leqslant 3$. Now, using Power mean once more, we have $S_4\leqslant 3$ and $S_2\leqslant 3$. Finally, by Cauchy-Schwarz, $9\geqslant S_4S_2 =(x^4+y^4+z^4)(y^2+z^2+x^2)\geqslant (x^2y+y^2z+z^2x)^2$, yielding the desired result.

electrovector wrote: $x,y,z \in \mathbb{R}^+$ and $x^5+y^5+z^5=xy+yz+zx$. Prove that $$3 \ge x^2y+y^2z+z^2x$$

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S.Lenny wrote: $$3 = 3 \frac{x^5+y^5+z^5}{xy+yz+xz} \geq x^3+y^3+z^3 \geq x^2y + y^2z +z^2x$$ why is that $ 3 \frac{x^5+y^5+z^5}{xy+yz+xz} \geq x^3+y^3+z^3$ ?

Multiply both sides with $xy+yz+zx$. Rest is rearrangement.

Homogenizing, it suffices to show that: $$x^2y+y^2z+z^2x\le3\frac{x^5+y^5+z^5}{xy+yz+zx}.$$This is equivalent to: $$3(x^5+y^5+z^5)\ge x^3y^2+y^3z^2+z^3x^2+x^3yz+xy^3z+xyz^3+x^2y^2z+xy^2z^2+x^2yz^2,$$which is true since Muirhead $(5,0,0)\succ(3,1,1)$, $(5,0,0)\succ(2,2,1)$, and $x^5+y^5+z^5\ge x^3y^2+y^3z^2+z^3x^2$ by Rearrangement.

By Rearrangement or simply by AM-GM, we have $$3(x^5+y^5+z^5)\ge (x^3+y^3+z^3)(x^2+y^2+z^2)$$On the other hand, we have $$3(x^2+y^2+z^2)\ge 3(xy+yz+zx)=3(x^5+y^5+z^5)$$Therefore, $$3\ge (x^3+y^3+z^3)$$Again, by Rearrangement or simply by AM-GM $$(x^3+y^3+z^3)\ge x^2y+y^2z+z^2x$$Hence, proved.

Generalization 1 Let $x,y,z$ be positive reals such that $x^{p}y^{r}+y^pz^r+z^px^r=x^{k}+y^k+z^k$ equality holds. Then prove that $$x^{\alpha}y^{k-p-r-\alpha}+y^{\alpha}z^{k-p-r-\alpha}+z^{\alpha}x^{k-p-r-\alpha}\leq 3$$

Generalization 2 Let $x_{1},x_{2},x_{3}$ be positive reals such that $\sum_{cyc}{x_{1}^{p}x_{2}^{r}}=x_{1}^{k}+x_{2}^k+x_{3}^k$ equality holds. Then prove that $$\sum_{cyc}{x_{1}^{\alpha}x_{2}^{k-p-r-\alpha}}\leq n$$

Generalization 3 Let $x_{1},x_{2},\cdots,x_{n}\in \mathbf{R^+}$ and $p_{1},p_{2},\cdots,p_{n},r_{1},r_{2},\cdots,r_{n}$ be non negative reals such that $$\sum_{cyc}{x_{1}^{p_{1}}x_{2}^{p_{2}}\cdots x_{n}^{p_{n}}}=\sum_{cyc}{x_{1}^{k}}$$$$k=\sum_{cyc}{r_{1}}+\sum_{cyc}{p_{1}}$$equalities take place. Then prove that $$\sum_{cyc}{x_{1}^{r_{1}}x_{2}^{r_{2}}\cdots x_{n}^{r_{n}}}\leq n$$