hint

Let $x=y=z=\sqrt[3]{c+1}$ then $\frac{x^3y+y^3z+z^3x}{x+y+z}+\frac{4c}{xyz} \geq 2c+2 \to c+1+\frac{4c}{c+1} \geq 2c+2 \to (c-1)^2 \leq 0 \to c=1$ For $c=1$: $\frac{x^3y+y^3z+z^3x}{x+y+z}+\frac{4}{xyz} \geq \frac{x^2yz+xy^2z+xyz^2}{x+y+z}+\frac{4}{xyz} =xyz+\frac{4}{xyz} \geq 4=2c+2$

How can we get from AM-GM to $x=y=z$

electrovector wrote: For all $x,y,z$ positive real numbers, find the all $c$ positive real numbers that providing $$\frac{x^3y+y^3z+z^3x}{x+y+z}+\frac{4c}{xyz}\ge2c+2$$ Solution of Zhangyanzong:

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RagvaloD wrote: Let $x=y=z=\sqrt[3]{c+1}$ then $\frac{x^3y+y^3z+z^3x}{x+y+z}+\frac{4c}{xyz} \geq 2c+2 \to c+1+\frac{4c}{c+1} \geq 2c+2 \to (c-1)^2 \leq 0 \to c=1$ For $c=1$: $\frac{x^3y+y^3z+z^3x}{x+y+z}+\frac{4}{xyz} \geq \frac{x^2yz+xy^2z+xyz^2}{x+y+z}+\frac{4}{xyz} =xyz+\frac{4}{xyz} \geq 4=2c+2$ this is original !,very nice

sqing wrote: electrovector wrote: For all $x,y,z$ positive real numbers, find the all $c$ positive real numbers that providing $$\frac{x^3y+y^3z+z^3x}{x+y+z}+\frac{4c}{xyz}\ge2c+2$$ Solution of Zhangyanzong: $$x^3y\geq 2x^2yz-xyz^2, ...$$This is original . Very very nice.

RagvaloD wrote: Let $x=y=z=\sqrt[3]{c+1}$ then $\frac{x^3y+y^3z+z^3x}{x+y+z}+\frac{4c}{xyz} \geq 2c+2 \to c+1+\frac{4c}{c+1} \geq 2c+2 \to (c-1)^2 \leq 0 \to c=1$ For $c=1$: $\frac{x^3y+y^3z+z^3x}{x+y+z}+\frac{4}{xyz} \geq \frac{x^2yz+xy^2z+xyz^2}{x+y+z}+\frac{4}{xyz} =xyz+\frac{4}{xyz} \geq 4=2c+2$ how to prove that it is the The only solution?

We claim that the only such number is $c=1$. For $x=y=z$ the inequality is equivalent to $$(x^3-2)(x^3-2c)\geq 0$$If $x=\sqrt[3]{2+2\varepsilon}$ for $\varepsilon>0$, then $1+\varepsilon \geq c$, if $\sqrt[3]{2-2\Delta}$ for $\Delta>0$, then $1-\Delta \leq c$, whence $1+\varepsilon \geq c \geq 1-\Delta$ for all positive $\Delta, \varepsilon$. Since we can take $\varepsilon$ and $\Delta$ arbitarily small, $c=1$. Now we want to prove the inequality for all positive $x$, $y$, $z$ $$\frac{x^3y+y^3z+z^3x}{x+y+z}+\frac{4}{xyz}\geq 4$$We claim that for all positive $x$, $y$, $z$ $$x^3y+y^3z+z^3x\geq xyz(x+y+z)$$which is true if we cyclically sum the weighted AM-GM $$\frac{4}{7}x^3y+\frac{1}{7}y^3z+\frac{2}{7}z^3x\geq x^2yz$$Then $$\frac{x^3y+y^3z+z^3x}{x+y+z}+\frac{4}{xyz} \stackrel{\text{Claim}}{\geq} xyz+\frac{4}{xyz} \stackrel{\text{AM-GM}}{\geq} 4$$This finishes the proof.