Problem 1. Let $a,b,c\ge 0: a+b+c=5$. Prove that:
$$ab+bc+ca+\sqrt[3]{abc}+20 \ge 2\left(\sqrt{a(4a+ab+bc+ca)}+\sqrt{b(4b+ab+bc+ca)}+\sqrt{c(4c+ab+bc+ca)}\right)$$
Equality holds iff $(a,b,c)=\left(\dfrac{5}{3},\dfrac{5}{3},\dfrac{5}{3}\right);\left(5,0,0\right)$ and per cyclic.
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Here is my attempt:
Case 1: $abc=0$:
WLOG, assume $a=0:b+c=5$, we just need to prove:$$bc+20\ge 2\left(b\sqrt{(4+c)}+c\sqrt{(4+b)}\right)$$
Put $b=5-c$, it becomes: $$-c^2+5c+20\ge 2\left((5-c)\sqrt{(4+c)}+c\sqrt{(9-c)}\right)$$
which is true $\forall 0\le c\le 5$.
See also here
Case 2: $abc>0$: Notice that $a=b=c=\dfrac{5}{3} \implies 4a+ab+bc+ca=9a$ .Using AM-GM inequality:
$$2\sum_{cyc}\sqrt{a(4a+ab+bc+ca)}\le \frac{13(a+b+c)+3(ab+bc+ca)}{3}$$
Hence, we will prove:
$$ab+bc+ca+\sqrt[3]{abc}+20\ge\frac{13(a+b+c)+3(ab+bc+ca)}{3}$$
Or: $\sqrt[3]{abc}\ge 1$ which is not true.
I also try to prove stronger: $$\color{red}{\sqrt[3]{abc}+\dfrac{5(ab+bc+ca)}{a+b+c}+4(a+b+c)\ge 2\sum_{cyc}\sqrt{a\left(4a+\frac{5(a+b+c)}{3}\right)}}$$
But it is not true for $a=b=x\rightarrow \frac{5}{3}$
I am very appreciate if someone can give me an advice or a hint to continue solve the tough problem.
P/S: Thank you @arqady for inviting me join this website. Hope to see your help!
Edit: arqady's idea inspired me.
Problem 2. Given $a,b,c\ge 0$. Prove that: $$\sqrt[3]{abc}+2+2(a+b+c)\ge$$ $$ \sqrt{a\left(4a+\sqrt[3]{abc}+4\right)}+\sqrt{b\left(4b+\sqrt[3]{abc}+4\right)}+\sqrt{c\left(4c+\sqrt[3]{abc}+4\right)}$$
Equality holds iff $a=b=c=1; a=b\Rightarrow +\infty;c=0$ and pers.
Best Answer
Let $a+b+c=3u$, $ab+ac+bc=3v^2$ and $abc=w^3$.
Thus, we need to prove that: $$3v^2+\frac{3uw}{5}+\frac{36u^2}{5}\geq2\sum_{cyc}\sqrt{\frac{3ua}{5}\left(\frac{12au}{5}+3v^2\right)}$$ or $$12u^2+5v^2+uw\geq2\sum_{cyc}\sqrt{ua(4ua+5v^2)}$$ or $$12u^2+5v^2+uw+\sum_{cyc}\left(u(2a+w)-2\sqrt{ua(4ua+5v^2)}+\frac{a(4ua+5v^2)}{2a+w}\right)\geq$$ $$\geq\sum_{cyc}\left(u(2a+w)+\frac{a(4ua+5v^2)}{2a+w}\right)$$ or $$6u^2+5v^2-2uw-\sum_{cyc}\frac{a(4ua+5v^2)}{2a+w}+\sum_{cyc}\left(\sqrt{u(2a+w)}-\sqrt{\frac{a(4ua+5v^2)}{2a+w}}\right)^2\geq0$$ or $$\frac{3w^2(u-w)(5v^2-2uw)}{\prod\limits_{cyc}(2a+w)}+\sum_{cyc}\left(\sqrt{u(2a+w)}-\sqrt{\frac{a(4ua+5v^2)}{2a+w}}\right)^2\geq0,$$ which is true by AM-GM for $5v^2-2uw\geq0.$
Let $5v^2-2uw\leq0$.
Thus, by AM-GM again we obtain: $$12u^2+5v^2+uw\geq12u^2+5v^2+\frac{5v^2}{2}=$$ $$=\sum_{cyc}\left(2ua+\frac{4ua+5v^2}{2}\right)\geq2\sum_{cyc}\sqrt{ua(4ua+5v^2)},$$ which ends a proof.
Explanation, why $$6u^2+5v^2-2uw-\sum_{cyc}\frac{a(4ua+5v^2)}{2a+w}=\frac{3w^2(u-w)(5v^2-2uw)}{\prod\limits_{cyc}(2a+w)}.$$ $$\prod_{cyc}(2a+w)=9w^3+\sum_{cyc}(4abw+2aw^2)=9w^3+12v^2w+6uw^2;$$ $$\sum_{cyc}a(4ua+5v^2)(2b+w)(2c+w)=\sum_{cyc}(4ua^2+5v^2a)(4bc+2bw+2cw+w^2)=$$ $$=\sum_{cyc}(16ua^2bc+8uw(a^2b+a^2c)+4uw^2a^2+20v^2abc+20v^2wab+5v^2w^2a)=$$ $$=48u^2w^3+8uw(9uv^2-3w^3)+4uw^2(9u^2-6v^2)+60v^2w^3+60v^4w+15uv^2w^2=$$ $$=3w(24u^2v^2+20v^4+12u^3w-3uv^2w+16u^2w^2+20v^2w^2-8uw^3).$$ Id est, $$6u^2+5v^2-2uw-\sum_{cyc}\frac{a(4ua+5v^2)}{2a+w}=$$ $$=6u^2+5v^2-2uw-\tfrac{24u^2v^2+20v^4+12u^3w-3uv^2w+16u^2w^2+20v^2w^2-8uw^3}{3w^2+4v^2+2uw}=$$ $$=\tfrac{5uv^2w-2u^2w^2-5v^2w^2+2uw^3}{3w^2+4v^2+2uw}=\tfrac{w(u-w)(5v^2-2uw)}{3w^2+4v^2+2uw}=\tfrac{3w^2(u-w)(5v^2-2uw)}{\prod\limits_{cyc}(2a+w)}.$$