We have that \[a+b+c\leq 6\iff (a+b+c)^3\leq 36(a+b+c)\iff \dfrac{1}{27}(a+b+c)^3\leq\dfrac{4}{3}(a+b+c)\]However AM-GM gives us \[abc\leq\dfrac{1}{27}(a+b+c)^3\]proving the inequality.

parmenides51 wrote: Let $a, b$, and $c$ be positive real numbers not greater than $2$. Prove the inequality $\frac{abc}{a + b + c} \le \frac43$ $$\frac{abc}{a + b + c} \leq\dfrac{1}{27}(a+b+c)^2\le \frac43$$

Let $a, b,c\in (0,2].$ Prove that $$\frac{abc}{a + 2b + c} \le 1$$$$\frac{abc}{a + kb + c} \le \frac{4}{k+2}$$Where $k>0.$

sqing wrote: Let $a, b,c\in (0,2].$ Prove that $$\frac{abc}{a + kb + c} \le \frac{4}{k+2}$$Where $k>0.$

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