let $a,b,c>0$,show that:


maybe this inequality can Use Holder inequality to solve it $$\left(\sum_{cyc}\sqrt{\dfrac{ab}{2a^2+bc+ca}}\right)^2\sum_{cyc}(2a^2+bc+ca)\ge(\sum_{cyc}\sqrt[3]{ab})^3$$

then I can't prove it


We can use also, AM-GM, C-S and uvw.

Indeed, $$\sum_{cyc}\sqrt{\frac{ab}{2a^2+bc+ca}}=\sum_{cyc}\frac{4ab}{2\sqrt{4ab(2a^2+bc+ca)}}\geq$$ $$\geq\sum_{cyc}\frac{4ab}{2a^2+bc+ca+4ab}=\sum_{cyc}\frac{4b^2}{\frac{b}{a}(2a^2+bc+ca+4ab)}\geq$$ $$\geq\frac{4(a+b+c)^2}{\sum\limits_{cyc}\frac{b}{a}(2a^2+bc+ca+4ab)}.$$ Thus, it's enough to prove that $$8(a+b+c)^5\geq81\sum_{cyc}b^2c(2a^2+bc+ca+4ab)$$ or $$8(a+b+c)^5\geq81\sum_{cyc}(a^3b^2+4a^3bc+3a^2b^2c)$$ or $$8(a+b+c)^5-81\sum_{cyc}(a^3b^2+a^3c^2+4a^3bc+3a^2b^2c)+81\sum_{cyc}a^3c^2\geq0.$$ But by Rearrangement $$\sum_{cyc}a^3c^2=a^2b^2c^2\sum_{cyc}\frac{a}{b^2}\geq a^2b^2c^2\sum_{cyc}\left(a\cdot\frac{1}{a^2}\right)=\sum_{cyc}a^2b^2c.$$ Id east, it's enough to prove that $$8(a+b+c)^5-81\sum_{cyc}(a^3b^2+a^3c^2+4a^3bc+2a^2b^2c)\geq0.$$ Now, let $a+b+c=3u$, $ab+ac+bc=3v^2$ and $abc=w^3$.

Thus, we need to prove a linear inequality of $w^3$, which says that it's enough to prove

the last inequality for an extreme value of $w^3,$ which happens in the following cases.

  1. $w^3\rightarrow0^+$.

Let $c\rightarrow0^+$ and $b=1$.

Thus, we need to prove that $$8(a+1)^5\geq81a^2(a+1),$$ which is true by AM-GM: $$8(a+1)^5=8(a+1)^4(a+1)\geq8\left(2\sqrt{a}\right)^4(a+1)=128a^2(a+1)>81a^2(a+1);$$ 2. $b=c=1$.

We need to prove that $$8(a+2)^5\geq81(2a^3+2a^2+2+4a^3+8a+4a^2+2a)$$ or $$(a-1)^2(4a^3+48a^2+9a+47)\geq0.$$ Done!

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By Hölder's inequality we have $$\left(\sum_{cyc}\sqrt{\frac{a}{2bc + ca + abc}}\right)^2 \cdot \left(\sum_{cyc}\frac{2bc + ca + abc}{a}\right) \geq 27$$ Hence it suffices to prove that $$27 \geq \frac{81}{2} \cdot \frac{abc}{(a + b + c)^3} \cdot \left(\sum_{cyc}\frac{2bc + ca + abc}{a}\right)$$ which is obvious after homogenization and expansion

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