We have x³ + y³ \geq x²y + xy² for all x, y > 0 by rearrangement. So \sum 1/(a³ + b³ + abc) \leq \sum 1/(a²b + ab² + abc) = 1/abc by simple computation.

$a^3+b^3\geq ab^2+a^2b$, so we get that: $\sum \frac{abc}{a^3+b^3+abc}\leq \sum\frac{abc}{ab(a+b+c)} = \sum\frac{c}{a+b+c}=1$

Ok, thanks.

Out of curiosity, why would you even want to let $abc=1$? You can apply Jensen's in the exact same fashion without de-homogenizing.

That's a good point. I don't know why; I was just working with it and decided to set $abc=1$. Then I saw the solution with Jensen's after. So it was merely a detour in my thought process. Thanks for pointing that out. It's probably a clearer solution that way.

this has been discussed a couple times I want to say it's USAMO 97, not sure. the idea is to say a^3 + b^3 >= ab(a+b) to make each denominator smaller. [Moderator edit: For this problem, see also http://www.artofproblemsolving.com/Forum/viewtopic.php?t=58140 http://www.artofproblemsolving.com/Forum/viewtopic.php?t=19469 http://www.artofproblemsolving.com/Forum/viewtopic.php?t=69177 http://www.artofproblemsolving.com/Forum/viewtopic.php?t=55677 http://www.artofproblemsolving.com/Forum/viewtopic.php?t=49889 ]

I came across this thread looking for practice USAMO inequalities so I apologize for reviving such an old thread, but here's a slightly different solution using symmetry and bunching:

Some motivation. The summands are just so 'nice'. This makes me feel the urge to separate right hand side into $\frac{a}{abc(a+b+c)},$ etc. Solution. And miraculously it works. Of course by muirhead and $[3,0] \succ [2,1]$ one has $a^3 + b^3 \geq a^2b + ab^2.$ Then, the left hand side is at most $\sum_{cyc} \frac{a}{abc(a+b+c)} = \frac{1}{abc}.$

Does this count as a one liner ?

Well I will add another one using Cauchy- Schwarz Inequality We have using that $\dfrac{a^2}{bc}+\dfrac{b^2}{ca}\ge \dfrac{(a+b)^2}{c(a+b)}\Rightarrow \dfrac{a^2}{bc}+\dfrac{b^2}{ca}+1\ge \dfrac{a+b+c}{c}$ $\Rightarrow \dfrac{1}{\dfrac{a^2}{bc}+\dfrac{b^2}{ca}+1}\le \dfrac{c}{a+b+c}\Rightarrow \sum_{cyc}\dfrac{1}{\dfrac{a^2}{bc}+\dfrac{b^2}{ca}+1}\le 1$ Equality holds iff $a=b=c$.

Maverick wrote: Prove that, for all positive real numbers $ a$, $ b$, $ c$, the inequality \[ \frac {1}{a^3 + b^3 + abc} + \frac {1}{b^3 + c^3 + abc} + \frac {1}{c^3 + a^3 + abc} \leq \frac {1}{abc} \]holds. Prove that, for all positive real numbers $ a$, $ b$, $ c$, the inequality \[ \frac {1}{a^3 + b^3 + abc} + \frac {1}{b^3 + c^3 + abc} + \frac {1}{c^3 + a^3 + abc} \geq \frac {5}{2abc+a^3+b^3+c^3}\]holds.

xzlbq wrote: Prove that, for all positive real numbers $ a$, $ b$, $ c$, the inequality \[ \frac {1}{a^3 + b^3 + abc} + \frac {1}{b^3 + c^3 + abc} + \frac {1}{c^3 + a^3 + abc} \geq \frac {5}{2abc+a^3+b^3+c^3}\]holds. <=> \[{\frac {27\,{u}^{6}-54\,{v}^{2}{u}^{4}+10\,{w}^{3}{u}^{3}+27\,{v}^{4}{ u}^{2}-11\,{v}^{2}{w}^{3}u+{v}^{6}+{w}^{6}}{27\,{w}^{3}{u}^{6}-81\,{v} ^{2}{w}^{3}{u}^{4}+27\,{v}^{6}{u}^{3}+11\,{w}^{6}{u}^{3}+54\,{v}^{4}{w }^{3}{u}^{2}-15\,{v}^{2}{w}^{6}u-27\,{v}^{8}u+{w}^{9}+4\,{v}^{6}{w}^{3 }}}\geq {\frac {5}{27\,{u}^{3}-27\,u{v}^{2}+5\,{w}^{3}}}\] <=> \[27\,{w}^{3} \left( 22\,{w}^{3}{u}^{3}-7\,{v}^{2}{w}^{3}u+270\,{u}^{6}- 432\,{v}^{2}{u}^{4}-15\,{v}^{6}+162\,{v}^{4}{u}^{2} \right) +729\,u \left( u-v \right) \left( u+v \right) \left( 3\,{u}^{2}-4\,{v}^{2} \right) \left( -{v}^{2}+3\,{u}^{2} \right) ^{2}\geq 0\] Need to prove \[22\,{w}^{3}{u}^{3}-7\,{v}^{2}{w}^{3}u+270\,{u}^{6}-432\,{v}^{2}{u}^{4} -15\,{v}^{6}+162\,{v}^{4}{u}^{2}\geq 0\]

Maverick wrote: Prove that, for all positive real numbers $ a$, $ b$, $ c$, the inequality \[ \frac {1}{a^3 + b^3 + abc} + \frac {1}{b^3 + c^3 + abc} + \frac {1}{c^3 + a^3 + abc} \leq \frac {1}{abc} \]holds. Note \[{a}^{3}+{b}^{3}+abc=ab \left( a+b+c \right) + \left( a+b \right) \left( a-b \right) ^{2}\] \[{\frac {1}{ab \left( a+b+c \right) }}+{\frac {1}{bc \left( a+b+c \right) }}+{\frac {1}{ca \left( a+b+c \right) }}={\frac {1}{abc}} \] Done.

Not sure if this is what splintercell990 meant by bunching: Clearing out the denominators and rearranging yields: $\sum_{sym} a^6b^3 \geq 2\sum_{sym} a^5b^2c^2 $ We can split this into two inequalities: $\sum_{cyc} a^6b^3 \geq \sum_{sym} a^5b^2c^2 $ and $\sum_{cyc} a^6c^3 \geq \sum_{sym} a^5b^2c^2 $ Both follow from weighted am-gm.

Maverick wrote: Prove that, for all positive real numbers $ a$, $ b$, $ c$, the inequality \[ \frac {1}{a^3 + b^3 + abc} + \frac {1}{b^3 + c^3 + abc} + \frac {1}{c^3 + a^3 + abc} \leq \frac {1}{abc} \]holds. stronger \[{\frac { \left( {x}^{2}+{y}^{2}+{z}^{2} \right) ^{2}}{ \left( x+y+z \right) \left( {x}^{3}+{y}^{3}+{z}^{3} \right) }}\geq {\frac {xyz}{{x}^{ 3}+{y}^{3}+xyz}}+{\frac {xyz}{{y}^{3}+{z}^{3}+xyz}}+{\frac {xyz}{{z}^{ 3}+{x}^{3}+xyz}}\] \[{\frac { \left( {x}^{3}+{y}^{3}+{z}^{3} \right) ^{2}}{ \left( {x}^{2}+ {y}^{2}+{z}^{2} \right) \left( {x}^{4}+{y}^{4}+{z}^{4} \right) }}\geq { \frac {xyz}{{x}^{3}+{y}^{3}+xyz}}+{\frac {xyz}{{y}^{3}+{z}^{3}+xyz}}+{ \frac {xyz}{{z}^{3}+{x}^{3}+xyz}}\] \[{\frac { \left( x+y+z \right) \left( {x}^{2}+{y}^{2}+{z}^{2} \right) xyz}{ \left( x{y}^{2}+y{z}^{2}+{x}^{2}z \right) \left( {x}^{2}y+{y}^{ 2}z+{z}^{2}x \right) }}\geq {\frac {xyz}{{x}^{3}+{y}^{3}+xyz}}+{\frac {xyz }{{y}^{3}+{z}^{3}+xyz}}+{\frac {xyz}{{z}^{3}+{x}^{3}+xyz}}\]

have this: \[\frac{1}{2}\geq {\frac {{y}^{3}{z}^{3}}{{y}^{6}+{z}^{6}+4\,{x}^{2}{y}^{2}{z}^{2}}} +{\frac {{x}^{3}{z}^{3}}{{z}^{6}+{x}^{6}+4\,{x}^{2}{y}^{2}{z}^{2}}}+{ \frac {{x}^{3}{y}^{3}}{{x}^{6}+{y}^{6}+4\,{x}^{2}{y}^{2}{z}^{2}}}\]

$a,b,c$ be sides of a triangle,prove \[{\frac {{b}^{3}{c}^{3}}{{b}^{6}+{c}^{6}+29\,{a}^{2}{b}^{2}{c}^{2}}}+{ \frac {{c}^{3}{a}^{3}}{{c}^{6}+{a}^{6}+29\,{a}^{2}{b}^{2}{c}^{2}}}+{ \frac {{a}^{3}{b}^{3}}{{a}^{6}+{b}^{6}+29\,{a}^{2}{b}^{2}{c}^{2}}}\geq { \frac {3}{31}}\]

Clearing the denominator, we get $S(5,2,2) \leq S(6,3,0)$ This is just Muirhead inequality

By CS we have, \[(a^3+b^3+abc)\left(\frac1a+\frac1b+\frac{c}{ab}\right)\geq (a+b+c)^2\]\[\implies \frac{1}{a^3+b^3+abc}\leq \frac{c}{abc(a+b+c)}\]Summing the three inequalities, we get \[\sum_{cyc} \frac{1}{a^3+b^3+abc} \leq \sum_{cyc} \frac{c}{abc(a+b+c)} = \frac{1}{abc}\]

lets use the following inequality $a^3+b^3 \geq ab(a+b)$ then $\sum_{cyc} \frac{1}{a^3+b^3+abc} \leq \sum_{cyc} \frac{1}{ab(a+b)+abc} \leq \frac {1}{abc}$

Arne wrote: We have$ x^3 + y^3 \geq x^2y + xy^2$ for all $ x, y > 0$ by $ rearrangement.$ So $\sum \dfrac {1}{(a^3 + b^3 + abc)} \le \sum \dfrac {1}{(a^2b + ab^2 + abc)} = \dfrac {1}{abc} $ by simple computation. Fixed Arne's solution

$a^2–2ab+b^2\ge 0$ $\implies a^2–ab+b^2\ge ab$ $\implies (a+b)(a^2–ab+b^2)\ge (a+b)ab$ $a^3+b^3\ge ab(a+b)$ $\implies ab(a+b)+abc=ab(a+b+c)$ And thus follows I think everyone has the same conclusion.............

We know that $a, b, c > 0$. By Chebyshev's inequality we have $(\sum_{cyc}\frac{1}{a^3+b^3+abc})(2(a^3+b^3+c^3) + 3abc) \leq 9$ Notice that $(2(a^3+b^3+c^3) + 3abc) \leq 3(a^3+b^3+c^3)$ by AM GM $(\sum_{cyc}\frac{1}{a^3+b^3+abc}) \leq \frac{3}{a^3+b^3+c^3}$ By AM GM $a^3+b^3+c^3 \geq 3abc$ so it follows that $(\sum_{cyc}\frac{1}{a^3+b^3+abc}) \leq \frac{3}{a^3+b^3+c^3} \leq \frac{1}{abc}$. $\blacksquare$

Wait in the Chebyshev's Inequality part, shouldn't it be the other way around, because Cauchy Scwarz says that is greater than or equal to 9???

Solved with nprime06 Since the inequality is homogenous, WLOG let $a+b+c = 1$ and since $a^3+b^3 \ge a^2b+b^2a$ we have \[\sum_{\text{cyc}} \frac{1}{a^3+b^3+abc} \le \sum_{\text{cyc}} \frac{1}{a^2b+b^2a+abc} = \frac{1}{a+b+c}\sum_{\text{cyc}} \frac{1}{ab} = \sum_{\text{cyc}} \frac{1}{ab} \le \frac{1}{abc} \]which is trivial by expanding. $\square$

fukano_2 wrote: Solved with nprime06 Since the inequality is homogenous, WLOG let $a+b+c = 1$ and since $a^3+b^3 \ge a^2b+b^2a$ we have \[\sum_{\text{cyc}} \frac{1}{a^3+b^3+abc} \le \sum_{\text{cyc}} \frac{1}{a^2b+b^2a+abc} = \frac{1}{a+b+c}\sum_{\text{cyc}} \frac{1}{ab} = \sum_{\text{cyc}} \frac{1}{ab} \le \frac{1}{abc} \]which is trivial by expanding. $\square$ Why do you even need to hold that ? pls correct me if I am wrong . Quote: let $a+b+c=1$ Since by Muirhead Inequality, \[\sum_{\text{cyc}} \frac{1}{a^3+b^3+abc} \le \sum_{\text{cyc}} \frac{1}{a^2b+b^2a+abc} \]$$\sum_{\text{cyc}} \frac{1}{a^2b+b^2a+abc}=\frac{1}{a^2b+b^2a+abc}+\frac{1}{b^2c+c^2b+abc}+\frac{1}{c^2a+a^2c+abc}$$$$=\frac{1}{a+b+c}\left(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}\right)$$$$\left(\frac{1}{a+b+c}\right)\left(\frac{a+b+c}{abc}\right)=\frac{1}{abc}$$So it true that, $$ \frac {1}{a^3 + b^3 + abc} + \frac {1}{b^3 + c^3 + abc} + \frac {1}{c^3 + a^3 + abc} \leq \frac {1}{abc} $$Proved $\square$

Here is mine solution just for storage and make things simple if something wrong in it please tell me. My solution: Here we can use that \[a^3+b^3\geq ab(a+b)\]this can be seen by the $(a+b)^3$ formula and proved by muirhead because we know that $(3,0)\succ(2,1)$ Now, We can see that \[\sum_{cyc}\frac{1}{a^3+b^3+abc}\leq \sum_{cyc}\frac{1}{ab(a+b)+abc}=\sum_{cyc}\frac{1}{ab(a+b+c)}\]After multiplying but $c$ in numerator and denominator we can get\[\sum_{cyc}\frac{1}{a^3+b^3+abc}\leq \sum_{cyc}\frac{c}{abc(a+b+c}=\frac{1}{abc}\sum_{cyc}\frac{c}{a+b+c}=\frac{1}{abc}\]And we are done !

By Rearrangement, $a^3+b^3\ge a^2b+ab^2$. Then we homogenize by assuming $a+b+c=1$. Finally: $$\sum_{\text{cyc}}\frac1{a^3+b^3+abc}\le\sum_{\text{cyc}}\frac1{a^2b+ab^2+abc}=\sum_{\text{cyc}}\frac1{ab}=\frac1{abc}.~\square$$

Let $ a, b, c$ be positive real numbers .Prove that $$ \frac{1}{a^3+b^3+kabc} + \frac{1}{b^3+c^3+kabc} + \frac{1}{a^3+c^3+kabc} \leq \frac{3}{(k+2)abc}$$Where $k\geq 1.$

blahblahblah wrote: $a^3+b^3\geq ab^2+a^2b$, so we get that: $\sum \frac{abc}{a^3+b^3+abc}\leq \sum\frac{abc}{ab(a+b+c)} = \sum\frac{c}{a+b+c}=1$ very effective

Let $ a, b, c$ be positive real numbers .Prove that $$ \frac {a}{b + c +ka} + \frac {b}{c + a +kb}+\frac {c}{a + b +kc} \leq \frac{3}{k+2}$$Where $k\geq 1.$

The key is to notice that $$\frac{abc}{a^3+b^3+abc} \leq \frac c{a+b+c} \iff a^3+b^3 \geq a^2b + ab^2$$is true. Now just sum cyclically and divide by $abc$.

Let $a^3=m$, $b^3=n$, $c^3=p$, WLOG(homogeneous)let $mnp=1$, we need to prove that \[\frac1{m+n+1}+\frac1{n+p+1}+\frac1{p+m+1}\le1,\](expand) or $\sum(m^2n+mn^2)\ge2\sum m=2\sum m^{4/3}n^{1/3}p^{1/3}$ which is true by Murihead.

We see by Muirhead that $x^3 + y^3 \ge x^2y + xy^2$ for any positive reals $x,y$. Hence \begin{align*} \sum_{cyc} \frac{1}{a^3 + b^3 + abc} \\ \leq \sum_{cyc} \frac{1}{a^2b + ab^2 + abc} = \sum_{cyc} \frac{1}{ab(a + b + c)} \\ = \frac{1}{abc} ,\\ \end{align*}as desired.

WLOG, to homogenize, let $abc=1$. Note that since by Muirhead's, we have that $a^2b+ab^2\geq a^3+b^3$. Using this, we get that \[\sum_{\text{cyc}} \frac{1}{a^3+b^3+abc}=\sum_{\text{cyc}}\frac{abc}{a^3+b^3+abc} \leq \sum_{\text{cyc}} \frac{abc}{a^2b+ab^2+abc} = \sum_{\text{cyc}} \frac{c}{a+b+c}=1=\frac{1}{abc},\]finishing the problem.

By Murihead's inequality: \[a^3+b^3\geq a^2b+ab^2.\]Therefore, the LHS is $\leq$: \begin{align*} \sum_{\text{cyc}}\frac{1}{ab(a+b+c)}&=\frac{1}{a+b+c}\sum_{\text{cyc}}\frac{1}{ab}\\ &\leq\frac{a+b+c}{abc(a+b+c)}\\ &=\frac{1}{abc}.\\ \end{align*}

The equation is homogeneous so assume $abc=1$. Then it suffices to prove \[\sum_{\text{cyc}} \frac{1}{a^3+b^3+abc} \le 1.\] Notice that $(3,0)$ majorizes $(2,1)$ so \[x^3+y^3 \ge x^2y+xy^2.\] by Muirhead. Thus, we have \begin{align*} \sum_{\text{cyc}} \frac{1}{a^3+b^3+abc} &= \sum_{\text{cyc}} \frac{abc}{a^3+b^3+abc} \\ &\le \sum_{\text{cyc}} \frac{abc}{a^2b+ab^2+abc} \\ &= \sum_{\text{cyc}} \frac{c}{a+b+c} = 1. \ \square \end{align*}

$a^3+b^3=(a+b)(a^2-ab+b^2)\geq(a+b)(ab)$ as $a^2-2ab+b^2\geq 0$. Thus $LHS\leq \sum_{\mathrm{cyc}}\frac{1}{ab(a+b+c)}= \frac{a+b+c}{abc(a+b+c)}= RHS$

By AM-GM $a^{3}+b^{3} +abc \geq a^{2}b+ab^{2}+abc$ So $\frac{1}{ a^{2}b+ab^{2}+abc} \geq \frac{1}{a^{3}+b^{3}+abc}$ So $$\sum_{\text{cyc}} \frac{1}{a^{3}+b^{3}+abc} \leq \sum_{\text{cyc}} \frac{1}{a^{2}b+ab^{2}+abc}=\frac{1}{a+b+c}\sum_{\text{cyc}} \frac{1}{ab}= \frac{1}{abc} $$And we are done

We multiply both sides by $(a^3+b^3+abc)(a^3+c^3+abc)(b^3+c^3+abc)abc$. We cancel terms and get \[\sum_{\text{sym}} a^6b^3 \geq \sum_{\text{sym}} a^5b^2c^2\]This is obvious from Muirhead. $\blacksquare$

WLOG set $abc = 1$, so the inequality becomes \[\frac{1}{a^3 + b^3 + 1} + \frac{1}{b^3+c^3 + 1} + \frac{1}{c^3 + a^3 + 1} \leq 1.\]Expanding both sides and manipulating, the inequality simplifies (reversibly) to \begin{align*} 2(a^3 + b^3 +c^3) & \leq \sum_{\text{sym}} a^6b^3 \\ 2 \sum_{\text{cyc}}a^5 b^2 c^2 & \leq \sum_{\text{sym}} a^6b^3, \end{align*}which is true by Muirhead's theorem.

Without loss of generality, assume $abc = 1$, then we wish to show $$(a^3 + b^3 + 1)(b^3 + c^3 + 1) + (b^3 + c^3 + 1)(c^3 + a^3 + 1) + (c^3 + a^3 + 1)(a^3 + b^3 + 1) = (a^3 + b^3 + 1)(b^3 + c^3 + 1)(c^3 + a^3 + 1).$$$$\iff \sum a^6 + 3 \sum a^3b^3 + 4 \sum a^3 + 3 \le 2a^3b^3c^3 + \sum a^6b^3 + \sum a^6 + 3 \sum a^3b^3 + 2 \sum a^3 + 1$$$$\iff 2 \sum a^3 \le \sum a^6b^3$$$$\iff 2 \sum a^5 b^2 c^2 \le \sum a^6 b^3$$which follows by summing the Weighted AM-GM $$\frac{2}{3} a^6b^3 + \frac{1}{3} c^6a^3 \ge \sqrt[3]{a^{15}b^6c^6} = a^5b^2c^2$$symmetrically.

By AM-GM, $a ^ 3 + b ^ 3 + abc \geq 3a ^ {4 / 3}b ^ {4/3}c ^ {1/3}$. Thus, $\text{LHS} \leq \sum_{\text{cyc}} \frac{1}{3} a ^ {-4/3}b ^ {-4/3}c ^ {-1/3}.$ It now suffices to prove that \begin{align*} \frac{1}{3}\sum_{\text{cyc}}a ^ {-4/3}b ^ {-4/3}c ^ {-1/3} &\leq \frac{1}{abc} \\ \iff \sum_{\text{cyc}}a ^ {-4/3}b ^ {-4/3}c ^ {-1/3} &\leq \frac{3}{abc} \\ \iff \sum_{\text{sym}}a ^ {-4/3}b ^ {-4/3}c ^ {-1/3} &\leq \sum_{\text{sym}}a ^ {-1}b ^ {-1}c ^ {-1}, \end{align*}which follows from $(-1, -1, -1) \succ (-4/3, -4/3, -1/3)$ and Muirhead's Inequality.

The main idea is that $a^3+b^3\ge a^2b+b^2a.$ Note that $$\frac{1}{a^3+b^3+abc} + \frac{1}{b^3+c^3+abc} + \frac{1}{c^3+a^3+abc}$$$$\le \frac{1}{a^2b+b^2a+abc}+\frac{1}{b^2c+c^2b+abc}+\frac{1}{c^2a+a^2c+abc}$$$$ = (\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca})\cdot \frac{1}{a+b+c}$$$$ = \frac{a+b+c}{abc}\cdot \frac{1}{a+b+c} = \frac{1}{abc}.$$