Putting the problem into LaTeX: Problem. Let a, b, c be three positive reals such that ab + bc + ca = 1. Prove the inequality $\sqrt[3]{\frac{1}{a}+6b}+\sqrt[3]{\frac{1}{b}+6c}+\sqrt[3]{\frac{1}{c}+6a}\leq\frac{1}{abc}$. This was the 2nd problem in the 2nd German TST 2005, so you can guess what kind of list it's from . On the exam, I messed up the problem, trying to apply all imaginable hardcore strategies and methods but failing to realize that every method one learns in a standard course (Jensen, AM-GM, power mean, ...) will lead to the solution (for instance, I just couldn't believe that one could get rid of the $\sqrt[3]{...}$ roots using power means and the inequality would remain correct - we on ML are used to much sharper inequalities...). Anyway, here is the solution I found some hours after the exam, being confronted with the fact that it was just too straightforward for me: Since ab + bc + ca = 1, we have $\frac{1}{a}+6b=\frac{ab+bc+ca}{a}+6b=\left(b+\frac{bc}{a}+c\right)+6b=7b+c+\frac{bc}{a}$, and similarly $\frac{1}{b}+6c=7c+a+\frac{ca}{b}$; $\frac{1}{c}+6a=7a+b+\frac{ab}{c}$, so that $\left(\frac{1}{a}+6b\right)+\left(\frac{1}{b}+6c\right)+\left(\frac{1}{c}+6a\right)$ $=\left(7b+c+\frac{bc}{a}\right)+\left(7c+a+\frac{ca}{b}\right)+\left(7a+b+\frac{ab}{c}\right)$ $=8\left(a+b+c\right)+\left(\frac{bc}{a}+\frac{ca}{b}+\frac{ab}{c}\right)$. But since $\left(\frac{bc}{a}+\frac{ca}{b}+\frac{ab}{c}\right)-\left(a+b+c\right)=\frac{\left(ca-ab\right)^2+\left(ab-bc\right)^2+\left(bc-ca\right)^2}{2abc}\geq 0$, we have $a+b+c\leq \frac{bc}{a}+\frac{ca}{b}+\frac{ab}{c}$, and thus $8\left(a+b+c\right)+\left(\frac{bc}{a}+\frac{ca}{b}+\frac{ab}{c}\right)=\left(6\left(a+b+c\right)+\left(\frac{bc}{a}+\frac{ca}{b}+\frac{ab}{c}\right)\right)+2\left(a+b+c\right)$ $\leq \left(6\left(a+b+c\right)+\left(\frac{bc}{a}+\frac{ca}{b}+\frac{ab}{c}\right)\right)+2\left(\frac{bc}{a}+\frac{ca}{b}+\frac{ab}{c}\right)$ $=6\left(a+b+c\right)+3\left(\frac{bc}{a}+\frac{ca}{b}+\frac{ab}{c}\right)=3\frac{\left(ab+bc+ca\right)^2}{abc}=3\frac{1}{abc}$ (since ab + bc + ca = 1). Hence, $\left(\frac{1}{a}+6b\right)+\left(\frac{1}{b}+6c\right)+\left(\frac{1}{c}+6a\right)\leq3\frac{1}{abc}$. Hence, $\frac{\left(\frac{1}{a}+6b\right)+\left(\frac{1}{b}+6c\right)+\left(\frac{1}{c}+6a\right)}{3}\leq\frac{1}{abc}$. Now, by the power-mean inequality (more precisely, by the fact that the $\frac13$-power mean is $\leq$ to the arithmetic mean), we have $\left(\frac{\sqrt[3]{\frac{1}{a}+6b}+\sqrt[3]{\frac{1}{b}+6c}+\sqrt[3]{\frac{1}{c}+6a}}{3}\right)^3\leq\frac{\left(\frac{1}{a}+6b\right)+\left(\frac{1}{b}+6c\right)+\left(\frac{1}{c}+6a\right)}{3}$, so that $\left(\frac{\sqrt[3]{\frac{1}{a}+6b}+\sqrt[3]{\frac{1}{b}+6c}+\sqrt[3]{\frac{1}{c}+6a}}{3}\right)^3\leq\frac{1}{abc}$. Taking the cubic root of this, we get $\frac{\sqrt[3]{\frac{1}{a}+6b}+\sqrt[3]{\frac{1}{b}+6c}+\sqrt[3]{\frac{1}{c}+6a}}{3}\leq\frac{1}{\sqrt[3]{abc}}$, so that $\sqrt[3]{\frac{1}{a}+6b}+\sqrt[3]{\frac{1}{b}+6c}+\sqrt[3]{\frac{1}{c}+6a}\leq\frac{3}{\sqrt[3]{abc}}$. Now, by the AM-GM inequality, applied to the numbers ab, bc and ca, we have $\sqrt[3]{ab\cdot bc\cdot ca}\leq\frac{ab+bc+ca}{3}$. Since $ab\cdot bc\cdot ca=\left(abc\right)^2$ and ab + bc + ca = 1, this becomes $\sqrt[3]{\left(abc\right)^2}\leq\frac13$, so that $3\sqrt[3]{\left(abc\right)^2}\leq 1$. Thus, $\frac{3}{\sqrt[3]{abc}}=\frac{3\sqrt[3]{\left(abc\right)^2}}{\sqrt[3]{\left(abc\right)^3}}=\frac{3\sqrt[3]{\left(abc\right)^2}}{abc}\leq\frac{1}{abc}$. Altogether, $\sqrt[3]{\frac{1}{a}+6b}+\sqrt[3]{\frac{1}{b}+6c}+\sqrt[3]{\frac{1}{c}+6a}\leq\frac{3}{\sqrt[3]{abc}}\leq\frac{1}{abc}$. And the problem is solved. Darij

Ok ,here is my solution : Applying Bunhiacopski inequality we have : ( 1/a + 1/b + 1/c )[(6ab +1) + (6bc +1 ) +(6ca+1) ] (1+1+1) >= [ :Sigma: (1/a +6b) ^1/3 ] :^3: so 27/abc >= [:Sigma:(1/a+6b)^1/3] :^3: Now we need to prove that 27/abc <= 1/ (abc):^3: which is equavalent to (abc):^2: <= 27 Using Cauchy : 1=ab+ac+bc >= 3(abc)^2/3 we comlete our proof

namanhams wrote: Ok ,here is my solution : Applying Bunhiacopski inequality we have : ( 1/a + 1/b + 1/c )[(6ab +1) + (6bc +1 ) +(6ca+1) ] (1+1+1) >= [ :Sigma: (1/a +6b) ^1/3 ] :^3: so 27/abc >= [:Sigma:(1/a+6b)^1/3] :^3: Now we need to prove that 27/abc <= 1/ (abc):^3: which is equavalent to (abc):^2: <= 27 Using Cauchy : 1=ab+ac+bc >= 3(abc)^2/3 we comlete our proof Hmmm I didn't know Cauchy-Bunhiacopski can be used for three rows of numbers. :-?

Loser wrote: namanhams wrote: Ok ,here is my solution : Applying Bunhiacopski inequality we have : ( 1/a + 1/b + 1/c )[(6ab +1) + (6bc +1 ) +(6ca+1) ] (1+1+1) >= [ :Sigma: (1/a +6b) ^1/3 ] :^3: so 27/abc >= [:Sigma:(1/a+6b)^1/3] :^3: Now we need to prove that 27/abc <= 1/ (abc):^3: which is equavalent to (abc):^2: <= 27 Using Cauchy : 1=ab+ac+bc >= 3(abc)^2/3 we comlete our proof Hmmm I didn't know Cauchy-Bunhiacopski can be used for three rows of numbers. :-? that's holder.

siuhochung wrote: Loser wrote: namanhams wrote: Ok ,here is my solution : Applying Bunhiacopski inequality we have : ( 1/a + 1/b + 1/c )[(6ab +1) + (6bc +1 ) +(6ca+1) ] (1+1+1) >= [ :Sigma: (1/a +6b) ^1/3 ] :^3: so 27/abc >= [:Sigma:(1/a+6b)^1/3] :^3: Now we need to prove that 27/abc <= 1/ (abc):^3: which is equavalent to (abc):^2: <= 27 Using Cauchy : 1=ab+ac+bc >= 3(abc)^2/3 we comlete our proof Hmmm I didn't know Cauchy-Bunhiacopski can be used for three rows of numbers. :-? that's holder. I don't get it again? What do you mean by "holder"?

Holder inequality. You can find much information about it in Internet.

Ok,10x very much for helping me.

darij grinberg wrote: every method one learns in a standard course (Jensen, AM-GM, power mean, ...) will lead to the solution How to use Jensen here? Can you show?

namanhams wrote: Ok ,here is my solution : Applying Bunhiacopski inequality we have : ( 1/a + 1/b + 1/c )[(6ab +1) + (6bc +1 ) +(6ca+1) ] (1+1+1) >= [ :Sigma: (1/a +6b) ^1/3 ] :^3: so 27/abc >= [:Sigma:(1/a+6b)^1/3] :^3: Now we need to prove that 27/abc <= 1/ (abc):^3: which is equavalent to (abc):^2: <= 27 Using Cauchy : 1=ab+ac+bc >= 3(abc)^2/3 we comlete our proof should be equivalent to $(abc)^2 \leq \frac{1}{27}$ ....

Incidentally, this is also the 4th problem of the 1st TST of Taiwan in 2005.

I wrote sth really stupid in here and it looks like forum has some problems and I can't delete post..

indybar wrote: darij grinberg wrote: every method one learns in a standard course (Jensen, AM-GM, power mean, ...) will lead to the solution How to use Jensen here? Can you show? That was my solution when it appeared in a WOP exam. Apply Jensen to the concave function $f(x) = \sqrt[3]{x}$. You get a sharper inequality that can be easily Muirhead bashed after homogenizing and the substitution $x = bc, y = ca, z = ab$.

Here's my solution by Jensen Ineq on $f(x)= x^{\frac{1}{3}}$. we get, $\sum_{cy} (\frac{1}{a}+ 6b)^ {\frac{1}{3}} \leq 3 ( \frac{1}{3} ( \sum_{cy} \frac{1}{a} + 6\sum_{cy} a ) )^{\frac{1}{3}} $. So, It is enough to prove that $3 ( \frac{1}{3} ( \sum_{cy} \frac{1}{a} + 6\sum_{cy} a ) )^{\frac{1}{3}}\leq \frac{1}{abc} $. $\Leftrightarrow 9 \sum_{cy} a^3 b^3 c^2 + 54\sum_{cy} a^4 b^3 c^3 \leq 1$ Let $ab=p, bc=q, ca=r$. then $p+q+r=1$ $\Leftrightarrow 9(\sum p^2 qr)(p+q+r) + 54\sum_{cy} p^2q^2 r \leq (p+q+r)^5$ $\Leftrightarrow \sum_{cy} p^5 + 5 \sum_{sy}p^4 q+ 10\sum_{sy}p^3q^2 +11\sum_{cy}p^3qr \geq 42\sum_{cy}p^2q^2 r $. by Muirhead Ineq, This is true. $Q.E.D.$

$(\frac{1}{a} + 3b + 3b)^{1/3} \le 2b + \frac{1}{3a}$ It becomes $3\sum a \le 1/abc$ or $3\sum a \le \sum (ab+bc+ca)/a$. This becomes $\sum bc/a \ge \sum a$. By Muirhead its true.

What ineq do you receive mildorf?

we have $(x+y+z)^3 \leq 9.\sum(x^3)$ and $\sum \dfrac{xy}{z} \geq (x+y+z)$ hence : $A^3\leq 9 ( \sum \dfrac{1}{a }+6\sum a )$ ..............................$\leq\dfrac{1}{\sqrt[3]{abc}}$ -> here's my solution

$(ab)^2 + (bc)^2 + (ac)^2 \geq (ab)(ac)+(bc)(ab)+(ac)(bc)$ Add $2((ab)(ac)+(bc)(ab)+(ac)(bc))$ both sides and factor we have $1 \geq 3abc(a+b+c)$ or $\frac{1}{abc} \geq 3(a+b+c)$ Multiply by 2 on both sides of (1) and then add $\frac{1}{abc}$ to both sides yields $\frac{3}{abc} \geq 6(a+b+c) + \frac{1}{abc}$ or $\frac{1}{abc} \geq \frac{(6(a+b+c) + \frac{1}{abc})}{3}=\frac{\sum_{cyc} (\frac{1}{a}+6b)}{3}$ (1) Note that the RHS of (1) is Now by AM-GM $(\frac{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}}{3})^3 \geq \frac{1}{abc}$-----(2) Now by Power-Mean Inequality $\frac{\sum_{cyc} (\frac{1}{a}+6b)}{3} \geq \left ( \frac{\sum_{cyc} \sqrt[3]{\frac{1}{a}+6b}}{3} \right )^3$- (3) Combine (1), (2), and (3) together we have $(\frac{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}}{3})^3 \geq\left ( \frac{\sum_{cyc} \sqrt[3]{\frac{1}{a}+6b}}{3} \right )^3$ or $\sum_{cyc} \sqrt[3]{ \frac{1}{a} + 6b} \leq \frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{1}{abc}$ as desired.

We know: $ 1=ab+bc+ca\le a^2+b^2+c^2$ $ \wedge$ $ a+b+c\le \dfrac{ab}{c}+\dfrac{bc}{a}+\dfrac{ca}{b}$ Then: $ 3(a+b+c)+9\sqrt{3} \le 8(a+b+c)+4( \dfrac{ab}{c}+\dfrac{bc}{a}+\dfrac{ca}{b})$ $ 3(a+b+c)+9\sqrt{3}\le 4(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c})=\dfrac{4}{abc}$ $ \dfrac{1}{abc}+6(a+b+c)+18\sqrt{3} \le \dfrac{9}{abc}$ $ \dfrac{(\frac{1}{a}+6b)+3\sqrt{3}+3\sqrt{3}+(\frac{1}{b}+6c)+3\sqrt{3}+3\sqrt{3}+(\frac{1}{c}+6a)+3\sqrt{3}+3\sqrt{3}}{9}\le \dfrac{1}{abc}$ MA-MG: $ \dfrac{\sqrt[3]{(\dfrac{1}{a}+6b)(3\sqrt{3})^2}+\sqrt[3]{(\dfrac{1}{b}+6c)(3\sqrt{3})^2}+\sqrt[3]{(\dfrac{1}{c}+6a)(3\sqrt{3})^2}}{3} \le \dfrac{1}{abc}$ $ \boxed{\sqrt[3]{\dfrac{1}{a}+6b}+\sqrt[3]{\dfrac{1}{b}+6c}+\sqrt[3]{\dfrac{1}{c}+6a}\le \dfrac{1}{abc}}$

Another solution: Applying Jensen inequality: $ \sqrt[3]{\frac{1}{a}+6b}+\sqrt[3]{\frac{1}{b}+6c}+\sqrt[3]{\frac{1}{c}+6a}\leq\sqrt[3]{9\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)+54(a+b+c)}\leq\frac{1}{abc}$ So we need to prove that $ 9+54(a+b+c)abc\leq\frac{1}{a^2b^2c^2}$ As we know by am-gm: $ abc(a+b+c)\leq\frac{(ab+bc+ca)^2}{3}=\frac{1}{3}$,and $ 1=ab+bc+ca\geq3\sqrt[3]{a^2b^2c^2}$,hence $ \frac{1}{a^2b^2c^2}\geq 27\geq 9+54(a+b+c)abc$. $ \blacksquare$

We first note two inequalities: \[1 = ab+bc+ca \ge 3 \sqrt[3]{(abc)^2} \implies abc \leq \frac{1}{3\sqrt3},\]\[1 = (ab+bc+ca)^2 = \sum_{\text{cyc}} (ab)^2 + (ab)(bc) \implies abc(a+b+c) = \sum_{\text{cyc}} (ab)(bc) \leq \frac 13.\] We use Jensen's on concave function (over the positive reals) $f(x) = \sqrt[3]{x}$: \begin{align*} \sum_{\text{cyc}} \sqrt[3]{\frac 1a + 6b} &\leq 3 \sqrt[3]{\frac 13 \left(6(a+b+c) + \frac{1}{abc}\right)} \\ &= 3 \sqrt[3]{\frac{6abc(a+b+c)+1}{3abc}} \\ &\leq 3 \sqrt[3]{\frac{1}{abc}} \\ &\leq \frac{1}{abc}.~\blacksquare \end{align*}

By Hölder, \begin{align*} \left(\sum_{\text{cyc}} \sqrt[3]{\frac1{a}+6b}\right) &\le \left(\sum_{\text{cyc}} (1+6ab)\right)\left(\sum_{\text{cyc}} \frac1{a}\right)(1+1+1)\\ &=\frac{27}{abc}, \end{align*}so it suffices to show that \begin{align*} \left(\frac1{abc}\right)^2 &\ge 27\\ \end{align*}But by AM-GM, \begin{align*} 1=ab+bc+ca &\ge 3\sqrt[3]{(abc)^2}, \end{align*}which implies the desired result.

Generalization 1 Let $a,b,c,\theta,\lambda$ be positive reals such that $ab+bc+ca=\beta$ and $$\beta ^4\left(3\theta +\lambda \beta\right)=3^{5-n}$$equalities hold. Then prove that $$\sqrt[n]{\dfrac{\theta}{a}+\lambda b}+\sqrt[n]{\dfrac{\theta}{b}+\lambda c}+\sqrt[n]{\dfrac{\theta}{c}+\lambda a}\leq \dfrac{1}{\sqrt[3]{\left(abc\right)^n}}$$

By AM-GM $\frac 1a+3b+\left(\frac1{3b}+2a\right)\ge 3\sqrt[3]{\frac 1a+6b}$ so $\sqrt[3]{ \frac{1}{a} + 6b} + \sqrt[3]{\frac{1}{b} + 6c} + \sqrt[3]{\frac{1}{c} + 6a } \le \sum_{\text{cyc}}\left(\frac4{9a}+\frac{5a}3\right).$ Now we have $\frac 1{abc}=\frac1a+\frac1b+\frac1c$ so we want $\frac53(a+b+c)\le\frac59\left(\frac1a+\frac1b+\frac1c\right)$ which follows by Muirhead after homogenizing and simplifying.

Note that \[\sum \sqrt[3]{\frac{1}{a} + 6b} \stackrel{\text{PM}}{\leq} 3 \sqrt[3]{\frac{\sum \left(\frac{1}{a} + 6a\right)}{3}}\text{.}\]Thus it suffices to prove that \begin{align*} 3 \sqrt[3]{\frac{\sum \left(\frac{1}{a} + 6a\right)}{3}} &\leq \frac{1}{abc} \\ \sum \left(\frac{1}{a} + 6a\right) &\leq \frac{1}{9} \cdot \left(\frac{1}{abc}\right)^3 \\ \sum \left(9bc + 54a^2 bc\right) &\leq \left(\frac{1}{abc}\right)^2 \\ 9 + 54 \sum a^2 bc &\leq \frac{1}{a^2 b^2 c^2}\text{.} \end{align*}Now substitute $x = bc, y = ca, z = ab$ so we get \[9 + 54 \sum xy \leq \frac{1}{xyz}\text{.}\]Note that $x + y + z = 1$. However, \[2 \sum xy = \left(\sum x\right)^2 - \sum x^2 = 1 - \sum x^2 \stackrel{\text{PM}}{\leq} \frac{2}{3}\]so therefore \[9 + 54 \sum xy \leq 27\text{.}\]Additionally, \[27 \leq \frac{1}{xyz}\]by AM-GM so we are done.

By Hölder's Inequality we have $$((\sum ab)^\frac{1}{3})((\sum \frac{1}{b})^ \frac{1}{3})((\sum 6ab+1)^\frac{1}{3}) \ge ({(\sum \frac{6ab+1}{a}) ^ \frac{1}{3}) } $$ We need to show $$\frac{1}{abc} \ge ((\sum ab)^\frac{1}{3})((\sum \frac{1}{b})^ \frac{1}{3})((\sum 6ab+1)^\frac{1}{3}) = \frac{9^ \frac{1}{3}}{(abc)^ \frac{1}{3}} \iff \frac{1}{3} \ge abc $$which is true by AM-GM on $ab,ac,bc$

By Holder, we have $$\sqrt[3]{ \frac{1}{a} + 6b} + \sqrt[3]{ \frac{1}{b} + 6c} + \sqrt[3]{ \frac{1}{c} + 6a} \le \left[ \left( \sum_{\text{cyc}} (6ab + 1) \right) \left( \sum_{\text{cyc}} \frac{1}{a} \right) \left( 1 + 1 + 1 \right) \right]^{1/3} = \frac{3}{(abc)^{1/3}}$$so it is sufficient to show $$\frac{3}{(abc)^{1/3}} \le \frac{1}{abc} \iff (abc)^2 \le \frac{1}{27} = \left( \frac{ab + bc + ca}{3} \right)^3$$which is true by AM-GM.

$(a+\frac{1}{6b})^\frac{1}{3}\leq \frac{a+b}{3ab(1-ab)}-ab+\frac{1}{3} \implies (a+\frac{1}{6b})^\frac{1}{3}\leq \frac{1}{3abc}-ab +\frac{1}{3}$ then $ \sqrt[3]{ \frac{1}{a} + 6b} + \sqrt[3]{\frac{1}{b} + 6c} + \sqrt[3]{\frac{1}{c} + 6a } \leq \frac{1}{abc}- (ab+ac+bc)+1=\frac{1}{abc}$

We use Hölder. Notice that \[\left(\sum_\text{cyc} \sqrt[3]{\frac1a + 6b}\right)^3 = \left(\sum_\text{cyc} \sqrt[3]{\frac{6ab + 1}{a}}\right)^3\]\[ \leq 3\left(\frac1a + \frac1b + \frac1c\right)\left(\sum_\text{cyc} (6ab + 1)\right) = 27\left(\frac1a + \frac1b + \frac1c\right),\]and so it suffices to show that \[27\left(\frac1a + \frac1b + \frac1c\right) \leq \frac{1}{(abc)^3}.\]But upon multiplying both sides by $abc$ and rearranging this becomes $abc \leq \frac{1}{3\sqrt3}$, which follows since \[\frac13 = \frac{ab + bc + ca}{3} \geq (abc)^\frac23\]by AM-GM. $\square$ I originally found a solution involving Jensen and a lot of annoying expansion, but I switched to this one after happily discovering that Hölder works.

I have discussed this problem as an application of Holder in my inequality YouTube tutorial on my channel "little fermat". Here is the Video

very nice solution- video!

By Hölder we find that\[\left(\sum_\text{cyc}\sqrt[3]{\frac1a+6b}\right)^3\left(\sum_\text{cyc}(1+6ab)^3a\right)\le\left(\sum_\text{cyc}(1+6ab)\right)^4=9^4.\]This is enough to finish upon observing that \begin{align*} \sum_\text{cyc}(1+6ab)^3a&=\sum_\text{cyc}(7a+bc+ca)^3(ab+bc+ca)a\\ &\ge\sum_\text{cyc}(9a^\frac89b^\frac89c^\frac89)^3(3a^\frac23b^\frac23c^\frac23)a=3\cdot9^3\sum_\text{cyc}a^\frac{13}3b^\frac{10}3c^\frac43\\ &\ge9^4a^3b^3c^3\end{align*}due to the usual AM-GM. $\blacksquare$

\[ \sqrt[3]{ \frac{1}{a} + 6b} + \sqrt[3]{\frac{1}{b} + 6c} + \sqrt[3]{\frac{1}{c} + 6a } \overset{?}{\leq} \frac{1}{abc}. \]Let $bc=\frac{x^2}{3},ac=\frac{y^2}{3},ab=\frac{z^2}{3}\iff a=\frac{yz}{x\sqrt3},b=\frac{xz}{y\sqrt3},c=\frac{xy}{z\sqrt3}$. We have the condition $x^2+y^2+z^2=3$. \[\sum{\sqrt[3]{\frac{x\sqrt{3}}{yz}+\frac{2\sqrt{3}xz}{y}}}\overset{?}{\leq} \frac{3\sqrt3}{xyz}\]\[\sqrt[3]{\frac{x\sqrt3}{yz}+\frac{2\sqrt3xz}{y}}=\sqrt[3]{\frac{x\sqrt3}{y}(\frac{1}{z}+2z)}\leq \frac{\frac{x\sqrt3}{y}+(\frac{1}{\sqrt3z}+\frac{2z}{\sqrt3})+\sqrt3}{3}\]Thus, we conclude that \[\sum{\sqrt[3]{\frac{x\sqrt3}{yz}+\frac{2\sqrt3xz}{y}}}\leq \sqrt{3}\sum{(\frac{z}{3y}+\frac{1}{3}+\frac{1}{9x}+\frac{2x}{9})}\overset{?}{\leq} \frac{3\sqrt3}{xyz}\]\[1+\frac{1}{3}\sum{\frac{z}{y}}+\frac{1}{9}\sum{\frac{1}{x}}+\frac{2}{9}\sum{x}\overset{?}{\leq }\frac{3}{xyz}\]\[27\overset{?}{\geq} 2xyz\sum{x}+\sum{xy}+3\sum{x^2y}+9xyz\]We have $\sum{x}\leq \sqrt{3\sum{x^2}}=3$ and $xyz\leq 1$ thus, $2xyz\sum{x}\leq 6$ and $\sum{xy}\leq \sum{x^2}=3$. \[6\overset{?}{\geq} \sum{x^2y}+3xyz\]By Schur with degree $4$, we have \[9-2\sum{x^2y^2}+\sum{x^2yz}=\sum{x^4}+\sum{x^2yz}\geq \sum_{sym}{x^3y}=\sum{xy(x^2+y^2)}\geq 2\sum{x^2y^2}\]\[9+\sum{x^2yz}\geq 4\sum{x^2y^2}\geq 8\sum{x^2y}-4\sum{x^2}=8\sum{x^2y}-12\implies \frac{21+\sum{x^2yz}}{8}\geq \sum{x^2y}\]Hence \[\sum{x^2y}+3xyz\leq \frac{21+\sum{x^2yz}}{8}+3xyz\leq 3+3=6\]As desired.$\blacksquare$

Note that by Holder's Inequality, $$\text{LHS} \leq (1+1+1)^{1/3} \left(\sum_{cyc} (6ab+1) \right)^{1/3}\left(\sum_{cyc}\frac{1}{a} \right)^{1/3} = \frac{3}{(abc)^{1/3}}.$$Thus it suffices to show that $$3(abc)^{2/3} \leq 1 = ab+bc+ca$$but this holds by AM-GM. QED