Matematika wrote: Prove that the following inequality holds for all positive real numbers $a$, $b$, $c$, $d$, $e$ and $f$ $\sqrt[3]{\frac{abc}{a+b+d}}+\sqrt[3]{\frac{def}{c+e+f}} < \sqrt[3]{(a+b+d)(c+e+f)} \text{.}$ Author: Dimitar Trenevski but by $AMGM$ $\sqrt[3]{\frac{abc}{a+b+d}}+\sqrt[3]{\frac{def}{c+e+f}} \le \frac 2 3 \sqrt[3]{(a+b+d)(c+e+f)}$

The inequality is equivalent with $ \sqrt[3]{\frac{abc}{(a+b+d)^2(c+e+f)}}+\sqrt[3]{\frac{def}{(c+e+f)^2(a+b+d)}}<1 $. By Am-Gm $ 3LHS \le\ \frac{a+b}{a+b+d}+\frac{c}{c+e+f}+\frac{e+f}{c+e+f}+\frac{d}{a+b+d}=2<3 $

Another solution Using $ (a+b)^3 \le\ 4(a^3+b^3) $ we need to prove that $ (a+b+d)^2(c+e+f)^2 \ge\ 4abc(c+e+f)+4def(a+b+d) $ which is true after expanding and using simple Am-Gm.

I think it is easy to know and show the proof by using AM-GM ! But, after I get the solution looked like dear andrejilievski's, I still wonder the stronger one

Matematika wrote: Prove that the following inequality holds for all positive real numbers $a$, $b$, $c$, $d$, $e$ and $f$ \[\sqrt[3]{\frac{abc}{a+b+d}}+\sqrt[3]{\frac{def}{c+e+f}} < \sqrt[3]{(a+b+d)(c+e+f)} \text{.}\] Proposed by Dimitar Trenevski.

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Matematika wrote: Prove that the following inequality holds for all positive real numbers $a$, $b$, $c$, $d$, $e$ and $f$ \[\sqrt[3]{\frac{abc}{a+b+d}}+\sqrt[3]{\frac{def}{c+e+f}} < \sqrt[3]{(a+b+d)(c+e+f)} \text{.}\] Proposed by Dimitar Trenevski. Note that by Holder, \[ \sqrt[3]{\frac{abc}{a + b + d}} + \sqrt[3]{\frac{def}{c + e + f}} \le \sqrt[3]{((a + b) + d)(c + (e + f)) \left( \frac{ab}{(a + b)(a + b + d)} + \frac{ef}{(e + f)(c + e + f)} \right) }\]It suffices to prove that \[ \frac{ab}{(a + b)(a + b + d)} + \frac{ef}{(e + f)(c + e + f)} < 1 \]However, this is obvious since \[ \frac{ab}{(a + b)(a + b + d)} + \frac{ef}{(e + f)(c + e + f)} < \frac{ab}{(a + b)^2} + \frac{ef}{(e + f)^2} \le \frac{1}{2} < 1 \] Remark. The only hard part in this problem is how to handle the weird expression above. After figuring that could be done by "manipulatively use of Holder", the rest is straightforward.