I have this inequality: for every $a,b,c > 0$ prove that $$ \frac{1}{a+2b}+\frac{1}{b+2c}+\frac{1}{c+2a}\leq \frac{1}{\sqrt[3]{abc}}$$ I have tried AM-GM, C-B-S and other ones. Please give me a hint. This is what I have tried: $$\sum_{cyc}\frac{1}{a+2b}\leq\sum_{cyc}\frac{1}{9}(\frac{1}{a}+\frac{1}{b}+\frac{1}{b}) $$ $$\sum_{cyc}\frac{1}{a+2b}\leq \frac{1}{3}(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})\leq \frac{1}{\sqrt[3]{abc}}$$ but $\frac{1}{3}(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})\leq \frac{1}{\sqrt[3]{abc}}\Leftrightarrow \sqrt[3]{abc}\leq \frac{3}{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}} $ and it's not true.
2026-03-26 01:28:31.1774488511
Inequality with a,b,c >0 with GM
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it's wrong!
Try $$(a,b,c)=(100,0.1,0.1).$$