MellowMelon wrote: Let $a, b, c$ be positive reals such that $abc=1$. Show that \[\frac{1}{a^5(b+2c)^2} + \frac{1}{b^5(c+2a)^2} + \frac{1}{c^5(a+2b)^2} \ge \frac{1}{3}.\] replacing $a,b,c$ by $\frac{1}{a},\frac{1}{b},\frac{1}{c}$ yields that the inequality is equivalent to $ \sum \frac{a^5b^2c^2}{(c+2b)^2} \ge \frac{1}{3}$ or $\sum \frac{a^3}{(c+2b)^2} \ge \frac{1}{3} $ which is just Holder inequality because $\sum \frac{a^3}{(c+2b)^2}\sum (2b+c)\sum(2b+c) \ge (a+b+c)^3$ and $a+b+c \ge 3$

Let $a=x^6, b=y^6, c=z^6$ so $xyz=1$. Then it suffices to show \begin{align*} \sum\frac{1}{a^5(b+2c)^2} = \sum\frac{y^{30}z^{30}}{(y^6+2z^6)^2} &\ge \frac{(\sum{y^{15}z^{15}})^2}{\sum(y^6+2z^6)^2}\\ &= \frac{\sum{y^{30}z^{30}}+2\sum{x^{30}y^{15}z^{15}}}{5\sum{x^{12}}+4\sum{y^6z^6}}\ge\frac13,\end{align*}or, after homogenizing, \[3\sum{y^{30}z^{30}}+6\sum{x^{30}y^{15}z^{15}} \ge 5\sum{x^{28}y^{16}z^{16}}+4\sum{x^{16}y^{22}z^{22}},\]which is just Muirhead.

MellowMelon wrote: Let $a, b, c$ be positive reals such that $abc=1$. Show that \[\frac{1}{a^5(b+2c)^2} + \frac{1}{b^5(c+2a)^2} + \frac{1}{c^5(a+2b)^2} \ge \frac{1}{3}.\] Is this problem in Mathematical Reflections? Let $(x,y,z)=(a^{-1}, b{-1}, c^{-1})^$. Then the inequality to prove is $\sum \frac{x^3}{(2y+z)^2}\ge \frac{1}{3}$ Using Jensen's inequality applied to $f(t)=t^{-2}$, we have that $\sum xf(\frac{2y+z}{x})\ge (x+y+z)f(3)=\frac{x+y+z}{9}$ But the last one is greater than $\frac {1}{3}$ because $x+y+z\ge 3$

Titu Andreescu is one of the people in charge of Mathematical Reflections, isn't he? He's the one who proposed the problem, so it wouldn't surprise me if it appeared there too (although probably after the fact).

MellowMelon wrote: Titu Andreescu is one of the people in charge of Mathematical Reflections, isn't he? He's the one who proposed the problem, so it wouldn't surprised me if it appeared there too (although probably after the fact). Yes =) This problem is very similar to IMO 95/2

MellowMelon wrote: Let $a, b, c$ be positive reals such that $abc=1$. Show that \[\frac{1}{a^5(b+2c)^2} + \frac{1}{b^5(c+2a)^2} + \frac{1}{c^5(a+2b)^2} \ge \frac{1}{3}.\] Another solution: By Cauchy-Schwarz ineq, we have: $\sum\limits_{cyc} {\frac{1}{{{a^5}{{(b + 2c)}^2}}}} = \sum\limits_{cyc} {\frac{{{b^4}{c^4}}}{{a{{(b + 2c)}^2}}}} \ge \frac{{{{({a^2}{b^2} + {b^2}{c^2} + {c^2}{a^2})}^2}}}{{a{{(b + 2c)}^2} + b{{(c + 2a)}^2} + c{{(a + 2b)}^2}}}$ we need to prove that: $\frac{{{{({a^2}{b^2} + {b^2}{c^2} + {c^2}{a^2})}^2}}}{{a{{(b + 2c)}^2} + b{{(c + 2a)}^2} + c{{(a + 2b)}^2}}} \ge \frac{1}{3}$ $\Leftrightarrow 3({a^4}{b^4} + {b^4}{c^4} + {c^4}{a^4} + 2({a^2} + {b^2} + {c^2})) \ge \sum\limits_{cyc} a {b^2} + 4\sum\limits_{cyc} {{a^2}b} + 12$ By AM-GM ineq, we have: $RHS \le \frac{5}{2}({a^2}{b^2} + {b^2}{c^2} + {c^2}{a^2} + {a^2} + {b^2} + {c^2}) + 12$ And $LHS \ge 6({a^2}{b^2} + {b^2}{c^2} + {c^2}{a^2}) + 6({a^2} + {b^2} + {c^2}) - 9$ $\ge \frac{5}{2}({a^2}{b^2} + {b^2}{c^2} + {c^2}{a^2} + {a^2} + {b^2} + {c^2}) + 12$ ( because abc=1) So we have done .

For $x=\frac{1}{a}, y=\frac{1}{b}, z=\frac{1}{c},$ we get \[LHS=\frac{1}{a^{5}(b+2c)^{2}}+\frac{1}{b^{5}(c+2a)^{2}}+\frac{1}{c^{5}(a+2b)^{2}}=(xyz)^2\cdot[\frac{x^3}{(2y+z)^2}+\frac{y^3}{(2z+x)^2}+\frac{z^3}{(2x+y)^2}]\stackrel{(xyz=1)}{=}\] \[=\frac{x^3}{(2y+z)^2}+\frac{y^3}{(2z+x)^2}+\frac{z^3}{(2x+y)^2}\stackrel{(Holder's\ inequality)}{\geq}\frac{(x+y+z)^3}{(2y+z+2z+x+2x+y)^2}=\frac{x+y+z}{9}\stackrel{(AM-GM)}{\geq}\] \[\geq\frac{\sqrt[3]{xyz}}{3}\stackrel{(xyz=1)}{=}\frac{1}{3}=RHS\] IMPORTANT REMARK Have a look at my old problem here. You'll see that my problem is a generalization of IMO 1995 inequality and for this one (which is O161, from Mathematical Reflections 3/2010, proposed by Titu Andreescu).

By am-gm : $\frac{(bc)^3}{(ab+2ac)^2}+\frac{ab+2ac}{27}+\frac{ab+2ac}{27} \ge \frac{bc}{3}$ $\Leftrightarrow $ $LHS \ge \frac{ab+bc+ac}{9} \ge \frac{3\sqrt[3]{(abc)^2}}{9}=1/3$

MellowMelon wrote: Let $a, b, c$ be positive reals such that $abc=1$. Show that \[\frac{1}{a^5(b+2c)^2} + \frac{1}{b^5(c+2a)^2} + \frac{1}{c^5(a+2b)^2} \ge \frac{1}{3}.\] Put $a=\frac{1}{x}, b=\frac{1}{y}, c=\frac{1}{z}$ then: $x,y,z >0, xyz=1$ $\frac{1}{a^5(b+2c)^2}+\frac{1}{b^5(c+2a)^2} + \frac{1}{c^5(a+2b)^2} \ge \frac{1}{3}$ $\Leftrightarrow \sum\frac{x^3}{(2y+z)^2}\ge \frac{1}{3}$ By AG-GM: $ \sum[\frac{x^3}{(2y+z)^2}+\frac{2y+z}{27}+\frac{2y+z}{27}]\ge \sum[\frac{x}{3}]$ $\Leftrightarrow \sum\frac{x^3}{(2y+z)^2}\ge \frac{1}{9}(x+y+z)\ge \frac{1}{3}$ equati if only if $a=b=c=1$

I just recently thought about this problem, so I figured I post my solution (which doesn't involve Holder). Note that if we do the substitutions $a\to \frac 1x$ and so on, the inequality becomes \[ \sum \frac {x^3}{(2y+z)^2} \geq \frac 13 . \] A natural idea is to try Cauchy-Schwarz, so we do so: \[ \sum \frac {x^3}{(2y+z)^2} \cdot \sum x\geq \left( \sum \frac {x^2}{2y+z} \right)^2 \] However, we know that we can apply Cauchy-Schwarz for the sum in the RHS above, so \[ \sum \frac {x^2}{2y+z} \cdot \sum (2y+z) \geq \left( \sum x \right)^2 \ \Rightarrow \ \sum \frac {x^2}{2y+z} \geq \frac 13 \sum x . \] Putting all this together we get \[ \sum \frac {x^3}{(2y+z)^2} \geq \frac 19 \sum x \geq \frac 19 \cdot 3 = \frac 13 , \] via AM-GM because $xyz=1$.

Note $\sum \frac {1}{a^5(b+2c)^2}=\sum\frac {(bc)^3}{(ab+2ac)^2}$. Now, $(\sum\frac {(bc)^3}{(ab+2ac)^2})(\sum \frac {1}{bc})\geq (\sum\frac {bc}{ab+2ac})^2$. Also, $(\sum\frac {bc}{ab+2ac})(\sum \frac {bc(2c+b)}{bc})\geq \frac{\sum(bc)^2}{3(a+b+c)}$. So now it's required to show $(\sum ab)^4\geq 9(\sum a)^2$, which is obvious since, $(\sum ab)^2\geq 3\sum a^2bc=3\sum a$ , so done.

Generalization of USA TST 2010 Generalization of USA TST 2010(2)

Hölder kills it directly. \[ { \left(\sum_{cyc} a(b+2c) \right) \left(\sum_{cyc} a(b+2c) \right) \left(\sum_{cyc} \frac{1}{a^5(b+2c)^2} \right) \geq \left(\sum_{cyc}\frac{1}{a}\right)^3 =(ab+bc+ac)^3 } \] which is equivalent to \[ \left(\sum_{cyc} \frac{1}{a^5(b+2c)^2} \right) \geq \frac{ab+bc+ca}{9} \] and by AM-GM we are done.

Partially clear denominators and simplify; we want to show that \[ Q=\sum_{cyc}\frac{b^5c^5}{(b+2c)^2}\ge \frac{1}{3}.\] But by Holder, $(Q)\left( \sum_{cyc} \frac{b+2c}{bc}\right) ^2\ge (ab+bc+ca)^3.$ Since $\frac{b+2c}{bc}=ab+2ac$, we have $Q\ge \frac{ab+bc+ca}{9}\ge \frac{\sqrt[3]{(abc)^2}}{9}=\frac{1}{3}$.

$a=\frac{vw}{u^2}$, $b=\frac{wu}{v^2}$, $c=\frac{uv}{w^2}$. Assume WLOG that $u^9+v^9+w^9=1$. Inequality to prove is \[\sum\frac{u^9}{uvw(w^3+2v^3)^2}=\sum u^9f(u^3-v^3)\ge\frac13\] where $f(x)=\frac{1}{uvw(u^3+v^3+w^3-x)^2}$ is convex and increasing over the domain $u^9+v^9+w^9=1$. Thus, \[\sum u^9f(u^3-v^3)\ge f(\underbrace{u^{12}+v^{12}+w^{12}-u^9v^3-v^9w^3-w^9u^3}_{\ge0})\ge f(0)=\frac1{uvw(u^3+v^3+w^3)^2}\ge\frac13\] by AM-GM

my solution = let $S=\frac{1}{a^5(b+2c)^2} + \frac{1}{b^5(c+2a)^2} + \frac{1}{c^5(a+2b)^2}$ than by holder's we have $(S^{1/3})(a(b+2c)+b(c+2a)+c(a+2b))^{2/3} \ge \frac{1}{a}+\frac{1}{b}+\frac{1}{c}=ab+bc+ca$ which is equivalent to $S(3(ab+bc+ca))^2\ge (ab+bc+ca)^3$ thus it remains to prove that $ab+bc+ca\ge 3$ which is trivial by AM-GM as $ab+bc+ca\ge 3(abc){2/3} = 3$ as $abc=1$ so we are done

$\displaystyle\sum_{\text{cyc}} \displaystyle\frac{1}{a^5(b+2c)^2} =\displaystyle\sum_{\text{cyc}} \displaystyle\frac{(1/a)^3}{((1/c)+(2/b))^2} \ge \displaystyle\frac{\left( \displaystyle\sum_{\text{cyc}} (1/a) \right)^3}{\left( \displaystyle\sum_{\text{cyc}} (1/c)+(2/b) \right)^2} = \displaystyle\frac{T^3}{(3T)^2} = \displaystyle\frac{T}{9} \ge 1/3$ with $T=ab+bc+ca \ge 3$. By Holder in fraction form and $p=3,q=3/2$

A not-so-elegant solution: First, we try to simplify the denominators on the LHS. We have, \[\frac{1}{a^5(b+2c)^2}=\frac{1}{a^5(b^2+4bc+4c^2)}\geq\frac{1}{a^5(3b^2+6c^2)},\] so it is enough to show that \[\sum_{cyc}\frac{1}{a^5(b^2+2c^2)}\geq1.\] We know that, by $abc=1$, we have \[\sum_{cyc}\frac{1}{a^5(b^2+2c^2)}=\sum_{cyc}\frac{b^6c^6}{bc(b^2+2c^2)}=\sum_{cyc}\frac{(b^3c^3)^2}{b^3c+2bc^3},\] so we use Titu on the LHS after which it is enough to show that \[\frac{(a^3b^3+b^3c^3+c^3a^3)^2}{\sum_{cyc}(a^3b+2ab^3)}\geq1.\] By weighted AM-GM, we have \[\frac{1}{9}a^6b^6+\frac{1}{9}a^3b^6c^3+\frac{7}{9}a^6b^3c^3\geq a^{\frac{17}{3}}b^{\frac{11}{3}}c^{\frac{8}{3}}=a^3b,\] so \[\frac{1}{9}\sum_{cyc}a^6b^6+\frac{8}{9}\sum_{cyc}a^6b^3c^3\geq\sum_{cyc}a^3b.\] Similarly, \[\frac{2}{9}\sum_{cyc}a^6b^6+\frac{16}{9}\sum_{cyc}a^6b^3c^3\geq2\sum_{cyc}ab^3,\] so \[\frac{1}{3}\sum_{cyc}a^6b^6+\frac{8}{3}\sum_{cyc}a^6b^3c^3\geq\sum_{cyc}(a^3b+2ab^3).\] However, the LHS is obviously at most $\sum_{cyc}a^6b^6+2\sum_{cyc}a^6b^3c^3$, so \[\sum_{cyc}a^6b^6+2\sum_{cyc}a^6b^3c^3\geq\sum_{cyc}(a^3b+2ab^3).\] The LHS above is the square of $a^3b^3+b^3c^3+c^3a^3$ and the RHS is positive, so we have \[\frac{(a^3b^3+b^3c^3+c^3a^3)^2}{\sum_{cyc}(a^3b+2ab^3)}\geq1,\] and we are done.

My solution, which is very similar to aditya21's: By Holder's inequality, $(\sum_{cyc}a(b+2c))^{\frac{2}{3}} (\sum_{cyc}\frac{1}{a^5(b+2c)^2})^{\frac{1}{3}} \ge \sum_{cyc} \frac{1}{a} = ab+bc+ca$. Thus, $\sum_{cyc}\frac{1}{a^5(b+2c)^2} \ge \frac{ab+bc+ca}{9}$. By AM-GM, $\frac{ab+bc+ca}{9} \ge \frac{1}{3}$, as requested.

Replace $a, b, c$ with $\frac{1}{x}, \frac{1}{y}, \frac{1}{z}$, respectively; the inequality becomes $\sum_{\text{cyc}}\frac{x^3}{(z+2y)^2}\geq \frac{1}{3}.$

Note that from AM-GM we have $x+y+z\geq 3\sqrt[3]{xyz}=3$, so from our lemma we have \[\sum_{\text{cyc}}\frac{x^3}{(z+2y)^2}\geq \frac{(x+y+z)^3}{(3x+3y+3z)^2}=\frac{x+y+z}{9}\geq \frac{1}{3},\]as desired.

AlgebraFC wrote: Replace $a, b, c$ with $\frac{1}{x}, \frac{1}{y}, \frac{1}{z}$, respectively; the inequality becomes $\sum_{\text{cyc}}\frac{x^3}{(z+2y)^2}\geq \frac{1}{3}.$

Note that from AM-GM we have $x+y+z\geq 3\sqrt[3]{xyz}=3$, so from our lemma we have \[\sum_{\text{cyc}}\frac{x^3}{(z+2y)^2}\geq \frac{(x+y+z)^3}{(3x+3y+3z)^2}=\frac{x+y+z}{9}\geq \frac{1}{3},\]as desired. The Lemma is a direct consequence of Holder.

posting the solution JFF \[f(x)=\frac{1}{x^2} \quad \textit{the second derivative is positive so it is convex}\]\[x=1/a,y=1/b=z=1/c\]\[\sum \frac{x^3}{(z+2y)^2}=\sum xf(\frac{z+2y}{x^2}) \ge (\sum x)f(3) \ge \frac13\]\[\iff x+y+z \ge 3\sqrt[3]{xyz} =3\]

By Holder, $$\left(\sum_{\text{cyc}}\frac{1}{a^5(b+2c)^2}\right)\left(\sum_{\text{cyc}}a(b+2c)\right)^2\ge\left(\sum_{\text{cyc}}\frac{1}{a}\right)^3,$$so $$\sum_{\text{cyc}}\frac{1}{a^5(b+2c)^2}\ge\frac{(ab+bc+ca)^3}{(3ab+3bc+3ca)^2}\ge\frac{1}{3}$$by AM-GM. $\square$

We multiply by $(abc)^5$ to get $$\sum_{cyc} \frac{b^5c^5}{(b+2c)^2}\geq \frac{1}{3}.$$We will then homogenize: $$\sum_{cyc}\frac{b^5c^5}{(abc)^\frac{8}{3}(b+2c)^2}\geq\frac{1}{3}$$Due to the condition $abc=1$, since this is homogenous, we can 6th power each variable (essentially the substitution $a\rightarrow a'^6$ etc.) so that this becomes $$\sum_{cyc}\frac{b^{30}c^{30}}{(abc)^{16}(b^6+2c^6)^2}\geq\frac{1}{3}$$We can then expand: $$\sum_{cyc}\frac{b^{30}c^{30}}{a^{28}b^{16}c^{16}+4a^{16}b^{22}c^{22}+4a^{16}b^{16}c^{28}}\geq\frac{1}{3}$$By Titu's, we have $$\sum_{cyc}\frac{b^{30}c^{30}}{a^{28}b^{16}c^{16}+4a^{16}b^{22}c^{22}+4a^{16}b^{16}c^{28}}\geq \frac{(a^{15}b^{15}+b^{15}c^{15}+c^{15}a^{15})^2}{5cyc(28,16,16)+4cyc(22,22,16)}.$$Therefore, it remains to show that $$\frac{(a^{15}b^{15}+b^{15}c^{15}+c^{15}a^{15})^2}{5cyc(28,16,16)+4cyc(22,22,16)}\geq\frac{1}{3},$$which becomes $$3cyc(30,30,0)+6cyc(30,15,15)\geq 5cyc(28,16,16)+4cyc(22,22,16),$$which follows by Muirhead's.

It is equivalent to show that $$\sum_{\mathrm{cyc}} \frac{b^5c^5}{(b+2c)^2} \geq \frac{1}{3},$$or after applying Cauchy-Schwarz, $$(\sqrt{a^5b^5}+\sqrt{b^5c^5}+\sqrt{c^5a^5})^2\geq \frac{5}{3}(a^2+b^2+c^2)+\frac{4}{3}(ab+bc+ca) \iff 3[5,5,0]+6[5,\tfrac{5}{2},\tfrac{5}{2}] \geq 5[\tfrac{14}{3},\tfrac{8}{3},\tfrac{8}{3}]+4[\tfrac{11}{3},\tfrac{11}{3},\tfrac{8}{3}].$$Since $[5,\tfrac{5}{2},\tfrac{5}{2}]\geq [\tfrac{11}{3},\tfrac{11}{3},\tfrac{8}{3}]$ by Muirhead, it suffices to show that $$3[5,5,0]+2[5,\tfrac{5}{2},\tfrac{5}{2}]\geq 5[\tfrac{14}{3},\tfrac{8}{3},\tfrac{8}{3}].$$(Weighted) AM-GM on the LHS gives $3[5,5,0]+2[5,\tfrac{5}{2},\tfrac{5}{2}]\geq 5[5,4,1]$, and Muirhead finishes. $\blacksquare$

Rewrite $\frac 1{a^5(b+2c)^2} = \frac{b^3c^3}{a^2(b+2c)^2}.$ Now Holder in the form $$\left(\sum \frac{b^3c^3}{a^2(b+2c)^2}\right) \left(\sum a(b+2c)\right)^2 \geq \frac 13 (ab+bc+ca)^3$$suffices as $ab+bc+ca \geq 3$.

Substitute $a = \frac{x}{y}, b = \frac{y}{z}, c = \frac{z}{x}$. This becomes by Cauchy-Schwarz, \[ \left(\sum_{\text{cyc}} \frac{y^5z^2}{x^3(xy + 2z^2)^2}\right)\left(\sum_{\text{cyc}} \frac{x}{y}\right) \ge \left(\sum_{\text{cyc}} \frac{y^2z}{x(xy + 2z)}\right)^2 \] It remains to show that \[ \sum_{\text{cyc}} \frac{y^2z}{x(xy + 2z)} \ge 1 \] Since by C-S again it follows that \[ \left(\sum_{\text{cyc}} \frac{y^2z}{x(xy + 2z)}\right)\left(\sum_{\text{cyc}} xz(xy + 2z)\right) \ge (\sum_{\text{cyc}} yz)^2 \]this is just \[ \sum_{\text{cyc}} x^2yz + 2xz^2 \le \sum_{\text{cyc}} y^2z^2 + 2x^2yz \]which follows by muirhead's.

We substitute $(a,b,c) = \left(\frac 1x, \frac 1y, \frac 1z\right)$, where $x,y,z$ are positive reals and $xyz=1$. Then it suffices to show \[\sum_{\text{cyc}} \frac{1}{a^5(b+2c)^2} = \sum_{\text{cyc}} \frac{x^5}{\left(\frac 1y + \frac 2z\right)^2} = \sum_{\text{cyc}} \frac{x^5y^2z^2}{(2y+z)^2} = \sum_{\text{cyc}} \frac{x^3}{(2y+z)^2} \ge \frac 13.\] Note that Holder's gives us \[\sum_{\text{cyc}} \frac{x^3}{(2y+z)^2} \sum_{\text{cyc}} (2y+z) \sum_{\text{cyc}} (2y+z) \ge \left(\sum_{\text{cyc}} x\right)^3.\] Simplifying and using AM-GM gives \begin{align*} \sum_{\text{cyc}} \frac{x^3}{(2y+z)^2} &\ge \frac{(x+y+z)^3}{9(x+y+z)^2} \\ &\ge \frac 19 (x+y+z) \\ &\ge \frac 19 (3 \sqrt[3]{xyz}) \\ &\ge \frac 13.~\blacksquare \end{align*}

There was gen 1 here.

There was gen 2 here.

We first multiply the LHS$=X$ by $(abc)^3$, getting: \[X=\sum_{cyc}{\frac{b^3c^3}{a^2(b+2c)^2}}.\]Applying Holders, we get: \[X\cdot \left(\sum_{cyc}{a(b+2c)}\right)^2\geq (ab+bc+ca)^3,\]\[X\geq \frac{1}{9}(ab+bc+ca).\]It suffices to show $ab+bc+ca\geq 3$, which is true by AM-GM, as desired.

Consider applying Holders to find, \begin{align*} \left( \sum_{cyc} \frac{1}{a^5(b+2c)^2} \right) \left( \sum_{cyc} a(b+2c) \right)^2 \geq \left( \sum_{cyc} \frac{1}{a} \right)^3 = (ab+bc+ca)^3 \end{align*}Then we wish to show that, \begin{align*} \frac{(ab+bc+ca)^3}{\left(3ab+3bc+3ca \right)^2} \geq \frac{1}{3} \end{align*}Which simplifies to showing, \begin{align*} \frac{ab+bc+ca}{9} \geq \frac13 \end{align*}But AM-GM gives, \begin{align*} ab + bc + ca \geq 3 \end{align*}so we are done.

bÿ Holdër's änd äm-gm wë hävë \[\left(\frac{1}{a^5(b+2c)^2} + \frac{1}{b^5(c+2a)^2} + \frac{1}{c^5(a+2b)^2}\right)(a(b+2c)+b(c+2a)+c(a+2b))^2=\left(\frac{1}{a^5(b+2c)^2} + \frac{1}{b^5(c+2a)^2} + \frac{1}{c^5(a+2b)^2}\right)9(ab+bc+ca)^2 \ge \left(\frac 1a+\frac 1b+\frac 1c\right)^3=(ab+bc+ca)^3\ge3(ab+bc+ca)^2\]

By Holder we have: $$[\sum_{cyc}{\frac{1}{a^5(b+2c)^2}}]^{\frac{1}{3}}[\sum_{cyc}{a(b+2c)}]^{\frac{1}{3}}[\sum_{cyc}{a(b+2c)}]^{\frac{1}{3}}\geq [\sum_{cyc}{\frac{1}{a}}]$$And we can finish with AM-GM: $$\sum_{cyc}{\frac{1}{a^5(b+2c)^2}}\geq \frac{ab+bc+ca}{9}\geq \frac{1}{3}$$

Note that, by Holder's, \begin{align*} \sum_{\text{cyc}} \dfrac{1}{a^5\left(b+2c\right)^2} = \sum_{\text{cyc}} \dfrac{a^3b^3}{(ca + 2bc)^2} &\geq \dfrac{\displaystyle \sum_{\text{cyc}} \left(\dfrac{\left(a^3b^3\right)^{1/3}}{(ca+2bc)^{2/3}} \cdot \left(ca+2bc\right)^2\right)^3}{\left(\displaystyle \sum_{\text{cyc}}\left(ca+2bc\right)\right)^2}. \end{align*}This is equivalent to \begin{align*} \dfrac{\left(\displaystyle \sum_{\text{cyc}} ab\right)^3}{\left(\displaystyle \sum_{\text{cyc}}\left(ca+2bc\right)\right)^2} \geq \dfrac{(ab+ca+ca)^3}{\left(3(ab+bc+ca)\right)^2} = \dfrac{ab+bc+ca}{9} \geq \dfrac{1}{3}. \end{align*}The second inequality comes from AM-GM.

By applying Holders, we get that \[\left(\sum_{\text{cyc}} \frac{1}{a^5(b+2c)^2}\right)\left(\sum_{\text{cyc}} b+2c\right)^2 \left(\sum_{\text{cyc}} a \right) \ge \left(\sum_{\text{cyc}} a\right)^4\]\[\implies \sum_{\text{cyc}} \frac{1}{a^5(b+2c)^2} \ge \frac{a+b+c}{9} \ge \frac{1}{3}\]where the last inequality comes from AM-GM since $abc = 1$.

I have discussed this problem in my holder video on the inequalities playlist on my youtube channel "little fermat" . Here is the Link

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By Holder, we have $(\sum_{cyc} \frac{1}{a^5(b+2c)^2}) (\sum_{cyc} a(b + 2c))^2 \ge (\sum_{cyc} \frac 1a)^3$, noting that $\sum_{cyc} a(b + 2c) = 3ab + 3bc + 3ac = 3(\frac 1a + \frac 1b + \frac 1c)$, it is sufficient to prove that $\frac{1}{a} + \frac 1b + \frac 1c \ge 3$, which is obvious by AM-GM.