How to prove: $ab+bc+ca+\sqrt[3]{abc}+20\ge 2\sum_{cyc}\sqrt{a(4a+ab+bc+ca)}$ if $a+b+c=5$

Question

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. Original link

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.

Answer 1

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, $$6u^2+5v^2-2uw-\sum_{cyc}\frac{a(4ua+5v^2)}{2a+w}=$$ $$=6u^2+5v^2-2uw-\tfrac{24u^2v^2+20v^4+12u^3w-3uv^2w+16u^2w^2+20v^2w^2-8uw^3}{3w^2+4v^2+2uw}=$$ $$=\tfrac{5uv^2w-2u^2w^2-5v^2w^2+2uw^3}{3w^2+4v^2+2uw}=\tfrac{w(u-w)(5v^2-2uw)}{3w^2+4v^2+2uw}=\tfrac{3w^2(u-w)(5v^2-2uw)}{\prod\limits_{cyc}(2a+w)}.$$

Answer 2

I will share my proof. Hope to see responses.

We divide into 2 cases:

$\bullet abc=0$. WLOG, assume $a=0:b+c=5$, the original inequality becomes:$$bc+20\ge 2\left(b\sqrt{(4+c)}+c\sqrt{(4+b)}\right)$$ Put $b=5-c$, we get: $$-c^2+5c+20\ge 2\left((5-c)\sqrt{(4+c)}+c\sqrt{(9-c)}\right)$$ which is true $\forall 0\le c\le 5$.

$\bullet abc>0$. Notice that: $$\sqrt{4a^2+a(ab+bc+ca)}-2a=\frac{a(ab+bc+ca)}{\sqrt{4a^2+a(ab+bc+ca)}+2a}$$ Set $x=ab+bc+ca; y=\sqrt[3]{abc}$. We will prove: $$x+y\ge 2\sum_{cyc}\frac{ax}{\sqrt{4a^2+ax}+2a}$$ By C-S: $$\left(\sqrt{4a^2+ax}+2a\right)(x+y)=2a(x+y)+\sqrt{(4a^2+ax)\left((x-y)^2+4xy\right)}$$ $$\ge 2a(x-y)+2a(x+y)+2x\sqrt{ay}=4ax+2x\sqrt{ay}$$ It implies: $$\frac{2ax}{\sqrt{4a^2+ax}+2a}\le \frac{2ax(x+y)}{4ax+2x\sqrt{ay}}=\frac{a(x+y)}{2a+\sqrt{ay}}$$ Hence, we prove: $$\frac{a}{2a+\sqrt{ay}}+\frac{b}{2b+\sqrt{by}}+\frac{c}{2c+\sqrt{cy}}\le 1$$ It is obviously true: $$\frac{\sqrt{a}}{2\sqrt{a}+\sqrt[6]{abc}}+\frac{\sqrt{b}}{2\sqrt{b}+\sqrt[6]{abc}}+\frac{\sqrt{c}}{2\sqrt{c}+\sqrt[6]{abc}}\le 1$$ Or: $$\frac{x}{2x+\sqrt[3]{xyz}}+\frac{y}{2y+\sqrt[3]{xyz}}+\frac{z}{2z+\sqrt[3]{xyz}}\le 1$$ The final inequality is well known and it ends proof. Indeed,$$\frac{\sqrt{a}}{2\sqrt{a}+\sqrt[6]{abc}}+\frac{\sqrt{b}}{2\sqrt{b}+\sqrt[6]{abc}}+\frac{\sqrt{c}}{2\sqrt{c}+\sqrt[6]{abc}}\le 1$$ we can let $abc=1$ and set $\sqrt{a}=\dfrac{m}{n};\sqrt{b}=\dfrac{n}{p};\sqrt{c}=\dfrac{p}{m}$.

The inequality becomes: $$\frac{m}{2m+n}+\frac{n}{2n+p}+\frac{p}{2p+m}\le 1$$ $$\iff \frac{n}{2m+n}+\frac{p}{2n+p}+\frac{m}{2p+m}\ge 1 $$ which is true by C-S inequality!

Answer 3

Dedicated to Jonson Thương.

We'll prove that $$\sqrt[3]{abc}+2+2(a+b+c)\ge$$ $$\geq\sqrt{a\left(4a+\sqrt[3]{abc}+4\right)}+\sqrt{b\left(4b+\sqrt[3]{abc}+4\right)}+\sqrt{c\left(4c+\sqrt[3]{abc}+4\right)}$$ for non-negatives $a$, $b$ and $c$.

Indeed,for $abc=0$ let $c=0$.

Thus, we need to prove that: $$2+2(a+b)\geq\sqrt{a(4a+4)}+\sqrt{b(4b+4)},$$ which is true by AM-GM: $$2+2(a+b)=a+a+1+b+b+1\geq$$ $$\geq2\sqrt{a(a+1)}+2\sqrt{b(b+1)}=\sqrt{a(4a+4)}+\sqrt{b(4b+4)}.$$ Let $abc\neq0$.

Thus, we need to prove that $$2\left(\sqrt[3]{abc}+2+2(a+b+c)\right)+\sum_{cyc}\left(\sqrt{2a+\sqrt[3]{abc}}-\sqrt{\frac{a(4a+\sqrt[3]{abc}+4)}{2a+\sqrt[3]{abc}}}\right)^2\geq$$ $$\geq\sum_{cyc}\left(2a+\sqrt[3]{abc}+\frac{a(4a+\sqrt[3]{abc}+4)}{2a+\sqrt[3]{abc}}\right)$$ or $$2\left(\sqrt[3]{abc}+2+2(a+b+c)\right)+\sum_{cyc}\left(\sqrt{2a+\sqrt[3]{abc}}-\sqrt{\frac{a(4a+\sqrt[3]{abc}+4)}{2a+\sqrt[3]{abc}}}\right)^2\geq$$ $$\geq\sum_{cyc}\left(2a+\sqrt[3]{abc}+\frac{a(4a+2\sqrt[3]{abc}-\sqrt[3]{abc}+4)}{2a+\sqrt[3]{abc}}\right)$$ or $$\left(4-\sqrt[3]{abc}\right)\left(1-\sum_{cyc}\frac{a}{2a+\sqrt[3]{abc}}\right)+\sum_{cyc}\left(\sqrt{2a+\sqrt[3]{abc}}-\sqrt{\frac{a(4a+\sqrt[3]{abc}+4)}{2a+\sqrt[3]{abc}}}\right)^2\geq0$$ and since by C-S $$1-\sum_{cyc}\frac{a}{2a+\sqrt[3]{abc}}=1-\sum_{cyc}\left(\frac{a}{2a+\sqrt[3]{abc}}-\frac{1}{2}\right)-\frac{3}{2}=$$ $$=\frac{1}{2}\left(\sum_{cyc}\frac{\sqrt[3]{abc}}{2a+\sqrt{abc}}-1\right)=\frac{1}{2}\left(\sum_{cyc}\frac{\sqrt[3]{bc}}{2\sqrt[3]{a^2}+\sqrt{bc}}-1\right)\geq$$ $$\geq\frac{1}{2}\left(\frac{\left(\sum\limits_{cyc}\sqrt[3]{bc}\right)^2}{\sum\limits_{cyc}\left(2\sqrt[3]{a^2bc}+\sqrt[3]{b^2c^2}\right)}-1\right)=0,$$ our inequality is proven for $4-\sqrt[3]{abc}\geq0.$

Let $4-\sqrt[3]{abc}\leq0.$

Thus, by AM-GM again we obtain: $$\sqrt[3]{abc}+2+2(a+b+c)=\frac{1}{4}\left(4\sqrt[3]{abc}+8+8(a+b+c)\right)\geq$$ $$\geq\frac{1}{4}\left(3\sqrt[3]{abc}+12+8(a+b+c)\right)=$$ $$=\frac{1}{4}\sum_{cyc}\left(4a+\left(4a+\sqrt[3]{abc}+4\right)\right)\geq\sum_{cyc}\sqrt{a\left(4a+\sqrt[3]{abc}+4\right)}$$

Answer 4

Alternative proof.

See my answer: Generalization.

"For $a,b,c>0,$ the constant $0\le k\le 5$ and an arbitrary negative expression $T$ represent to $a,b,c,$ then

$$\color{black}{\frac{\sqrt[3]{ab}+\sqrt[3]{ac}}{2\left(\sqrt[3]{ab}+\sqrt[3]{bc}+\sqrt[3]{ac}\right)}\cdot\left((k+1)\sqrt[3]{abc}+(5-k)T\right) +4a\ge 2\sqrt{a\left(4a+k\sqrt[3]{abc}+(5-k)T\right) }.} \tag{*}$$

" Apply this result when $k=0; T=\dfrac{ab+bc+ca}{5}$ we obtain $$\color{black}{\frac{\sqrt[3]{ab}+\sqrt[3]{ac}}{2\left(\sqrt[3]{ab}+\sqrt[3]{bc}+\sqrt[3]{ac}\right)}\cdot\left(\sqrt[3]{abc}+ab+bc+ca\right) +4a\ge 2\sqrt{a\left(4a+ab+bc+ca\right) }.} \tag{1}$$ Take cyclic sum on $(1),$ we get the desired inequality.