Suppose that $a,b,c,d>0$. Is there a proof that $$ a\sqrt[3]{\frac{1+d}{b^3+abcd}}+b\sqrt[3]{\frac{1+d}{c^3+abcd}}+c\sqrt[3]{\frac{1+d}{a^3+abcd}}\geq 3?$$ I tried for example Jensen, Karamata, Power mean and Minkowski's inequality without success.
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The inequality can be written as $$(1+d)\left(\sum_{\text{cyc}}\frac{a}{\sqrt[3]{b^3+abcd}}\right)^3\geq27$$ By the generalized Hölder's inequality $$\left(\sum_{\text{cyc}}\frac{a}{\sqrt[3]{b(b^2+d\,ca)}}\right)^3\left(\sum_{\text{cyc}}ab\right)\left(\sum_{\text{cyc}}a(b^2+d\,ca)\right)\geq\left(\sum_{\text{cyc}}a\right)^5$$ This reduces the problem to $$(1+d)\left(\sum_{\text{cyc}}a\right)^5\geq27\left(\sum_{\text{cyc}}ab\right)\left(\sum_{\text{cyc}}a(b^2+d\,ca)\right)$$ But $$\sum_{\text{cyc}}a(b^2+d\,ca)=\sum_{\text{cyc}}ab^2+d\sum_{\text{cyc}}ca^2=(1+d)\sum_{\text{cyc}}ab^2$$ So it's just $$\left(\sum_{\text{cyc}}a\right)^5\geq27\left(\sum_{\text{cyc}}ab\right)\left(\sum_{\text{cyc}}ab^2\right)$$
And see Inequality problem $(a+b+c)^5\geq27(ab+bc+ca)(ab^2+bc^2+ca^2)$.
user2345215
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Nice +1. BTW Holder directly gives $$\left(\sum a\sqrt[3]{\frac{1+d}{b^3+a b c d}}\right)^3(\sum \text{ab})\left(\sum a\left(b^2+a c d\right)\right)\geq (1+d)(\sum a)^5$$ – Macavity Mar 13 '14 at 07:39
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@Macavity Nice shortcut! I'll edit it. – user2345215 Mar 13 '14 at 08:30
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Where did you get that first inequality? Is there some magic in the cyc-notation as I don't see where the term $\frac{b}{\sqrt[3]{c^3+abcd}}$ is? – Student Mar 13 '14 at 10:23
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@Student: Yes, there is, $\sum\limits_\text{cyc}E(a,b,c)$ means a cyclic sum so it's $E(a,b,c)+E(b,c,a)+E(c,a,b)$. (d is treated as a constant there) – user2345215 Mar 13 '14 at 10:49