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Let $a$, $b$ and $c$ be positive numbers such that $a^3+b^3+c^3=3$. Prove that: $$\frac{a^3}{a+b}+\frac{b^3}{b+c}+\frac{c^3}{c+a}\geq\frac{3}{2}.$$

This inequality we can prove by BW with computer.

I am looking for an human proof, which we can use during competition.

For example, we need to prove that $$\sum_{cyc}\left(\frac{a^3}{a+b}-\frac{a^2}{2}\right)\geq\frac{3}{2}-\frac{a^2+b^2+c^2}{2}$$ or $$\sum_{cyc}\frac{a^2(a-b)}{a+b}\geq\sqrt[3]{3(a^3+b^3+c^3)^2}-a^2-b^2-c^2$$ or $$\sum_{cyc}\frac{a^2(a-b)}{a+b}\geq\frac{3(a^3+b^3+c^3)^2-(a^2+b^2+c^2)^3}{\sqrt[3]{9(a^3+b^3+c^3)^4}+(a^2+b^2+c^2)\sqrt[3]{3(a^3+b^3+c^3)^2}+(a^2+b^2+c^2)^2}$$ and since $\sqrt[3]{3(a^3+b^3+c^3)^2}\geq a^2+b^2+c^2$ or $3(a^3+b^3+c^3)^2\geq(a^2+b^2+c^2)^3$,

it remains to prove that $$\sum_{cyc}\frac{a^2(a-b)}{a+b}\geq\frac{3(a^3+b^3+c^3)^2-(a^2+b^2+c^2)^3}{3(a^2+b^2+c^2)^2}$$ and from here I don't see what to do.

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    What is BW ?"BW with computer" – 2017-02-14
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    @A---B See here: https://www.artofproblemsolving.com/community/c6h522084 – 2017-02-15
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    Does it help to use elementary symmetric functions? It seems a little bashy, but it might work. – 2017-03-29
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    @Isaac Browne The last inequality is cyclic and not symmetric. I think it does not help or gives something terrible. – 2017-03-29
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    It is fairly straightforward to see by standard calculus that $(1,1,1)$ is a local minimum of $\sum_{cyc}\frac{a^3}{b+c}$ with respect to $a^3+b^3+c^3=3$. It involves computing the Lagrangian of the problem and its Hessian. – 2017-03-29
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    It can be seen that generally if a^n+b^n+c^n=3 then q= a^n/(a+b) +b^n/(b+c) +c^n/(c+a) has 1/2 as its minimum value for all n=2 or n>2. So n=3 is not an exception and for any n greater than 2, q is equal or greater than 1/2. – 2017-03-29
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    I have a nice proof for $1\leq a \leq (3)^{\frac{1}{3}}$ ; $0\leq b \leq 1$ and $\frac{5}{13}\leq c \leq 1$ . Do you want to see it Michael Rozenberg ? – 2017-04-01
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    @max8128 Thank you, but it's not proof. – 2017-04-01
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    @Michael Rozenberg see [here](https://math.stackexchange.com/questions/2214343/cyclic-inequality) – 2017-04-02
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    @Michael Rozenberg why we can't use the Ev method here ? – 2018-01-07
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    I don't see how we can use this here. – 2018-01-07
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    Maybe this result can help : https://artofproblemsolving.com/community/c7h1572330_the_wonderful_result_without_convexity_ – 2018-01-09
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    Are you sure that it's the right link. – 2018-01-09
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    @MichaelRozenberg : the link to "the wonderful result ?" – 2018-01-09
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    Yes, but don't see how it helps. – 2018-01-09
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    for show in convex $C(\{x_i\}), f(a,b,c) \geq 0$ if $i=1..m, f(x_i)\geq 0$ and no $0 \in Det(df(C(\{x_i\}))$ – 2018-01-09
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    @Michael Rozenberg I have a doubt over here. $$a^3+b^3+c^3=3$$ $$\Rightarrow a^3-1+b^3-1+c^3-1=0$$ $$\Rightarrow (a-1)(a^2+a+1)+(b-1)(b^2+b+1)+ (c-1)(c^2+c+1)=0 \Rightarrow$$ The only positive values satisfying the given condition are $a=b=c=1$ Hence the the equality holds. – 2018-01-18
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    No! We can take always $a>1$, $b<1$ and for example, even $c=1$. – 2018-01-18
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    Maybe a solution here : http://www.les-mathematiques.net/phorum/read.php?4,1484510,1621128#msg-1621128 – 2018-03-03

3 Answers 3

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Assume $$ (H):\;\; a,b,c>0, \; a^3+b^3+c^3=3,$$ and prove that

$$(I_0): \dfrac{a^3}{a+b}+\dfrac{c^3}{a+c}+\dfrac{b^3}{b+c}\geq\dfrac{3}{2}$$

Proof Step 1

By Cauchy-Schwarz we can see that the following inequality $(I_1)$ $$ (I_1):\;\; s(a,b,c) =a^3 (a+b)+c^3 (a+c)+b^3 (b+c)\leq 6. $$ implies ($I_0$). Indeed we have $$9=(\sum_{cycl} a^3)^2 \leq \sum_{cycl} \dfrac{a^3}{a+b}\times \sum_{cycl} a^3(a+b).$$ Thus, I will show $(I_1)$

Step 2. Suppose $\max (a,b,c)< \sqrt{2}.$

Let $x\geq 0.$ Young ineg. allows to write:
$$ a^3 x \leq \dfrac{1}{2} a^3 +\dfrac{1}{2} a^3 x^2.$$ Applying again Young ineg. in the the previous ineq. (second term of the right hand side) of the previous ineq. it follows $$ a^3 x \leq \dfrac{1}{2} a^3 +\dfrac{1}{2}(\dfrac{1}{2} a^3+ {\dfrac{1}{2} a^3 x^4})= (\dfrac{1}{2}+ \dfrac{1}{2^2})a^3+ \dfrac{1}{2^2} a^3 x^4 ,$$

and by iteration
$$a^3 x \leq \sum_{j=1}^n \dfrac{1}{2^j} a^3+ \dfrac{1}{2^n} x^{2^n}=(1-2^{-n}) a^3 + \dfrac{1}{2^n}a^3 x^{2^n}.$$

I apply that to $s(a,b,c)$ and get: $$ s(a,b,c)\leq (1-2^{-n})\times 2(a^3+b^3+c^3) +2 \times 3 \times \dfrac{1}{2^n}(a^{2^n}+b^{2^n}+c^{2^n})\leq 6,$$ passing to the limit :

There is a mistake in my proof (step 2). There was a missprint ($2n$ instead of $2^n$) which is now corrected. Now the inequality is exact but ineffective.

Step 3. I suppsose here that $\max (a,b,c)\geq \sqrt{2}.$

without loss of generality I suppose that $a\geq \sqrt{2}.$

We also have $a\leq 3^{1/3}$ and $b,c \leq (3-\sqrt{2}^3)^{1/3}.$

First we can easily see that if $b$ or $c$ vanishes then
$$s(a,b,c)<6.$$ It is left to the reader.

So I suppose that $b>0$ and $c>0.$

To finish the proof we will see with Lagrange multipliers method that the functional $s(a,b,c)$ has no critical point for (a,bc) in the domain of this step. Indeed let $$h(a,b,c)=s(a,b,c)-\lambda (a^3+b^3+c^3-3).$$

We need to solve $$\dfrac{\partial h}{\partial a}=a^3+3 a^2(a+b)+c^3-3 a^2 \lambda =0 $$ and $$\dfrac{\partial h} {\partial b}= a^3 + b^3 + 3 b^2 (b + c) - 3 b^2 \lambda. $$

Thus
$$ (Eq_0): \lambda=\dfrac{4 a}{3}+b+\dfrac{c^3}{3 a^2}=\dfrac{a^3}{3 b^2}+\dfrac{4 b}{3}+c.$$

But

$$\dfrac{4 a}{3}+b+\dfrac{c^3}{3 a^2}\leq \dfrac{4 \times3^{1/3}}{3}+(3-\sqrt{2}^3)^{1/3}+\dfrac{(3-\sqrt{2}^3)}{3 \sqrt{2}^2}\approx 2.50616$$ and $$\dfrac{a^3}{3 b^2}+\dfrac{4 b}{3}+c> \dfrac{a^3}{3 b^2}\geq \dfrac{1}{(3 - 2 \sqrt{2})^{2/3}}\approx 3.23868.$$ Hence, the identity $(Eq_0)$ is not possible. c.q.f.d

New Step 3. (New version without Lagrange multipliers.)

I suppsose here that $\max (a,b,c)\geq \sqrt{2}.$ suppsose here that $\max (a,b,c)\geq \sqrt{2}.$

without restriction to the generality I suppose that $a\geq \sqrt{2}.$

We also have $a\leq 3^{1/3}$ and $b,c \leq (3-\sqrt{2}^3)^{1/3}.$

Thus $\max(a+b,a+c,a+c)\leq 3^{1/3}+(3-\sqrt{2}^3)^{1/3}<2$

Hence $$a^3(a+b)+b^3(b+c)+c^3(a+c)< 2(a^3+b^3+c^3)=6.$$ c.q.f.d

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    Indeed, the inequality $\sum\limits_{cyc}(a^4+a^3b)\leq6$ is true. Must be a nice proof for this. Thank you for your interest! – 2017-08-17
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    Il y a une erreur : http://www.les-mathematiques.net/phorum/read.php?4,1484510,1510840#msg-1510840 pourquoi ne l'as-tu pas signalé ? – 2017-08-23
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    Reponds à l'objection de Sylvie que j'ai mis en lien, alors. – 2017-08-23
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    Dattier, I do not understand your language (is it Spanish, Italian, French, ...?) but I upvoted anyway because it is unique :) – 2017-10-14
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I propose a simple alternative of the third step of my proof in order to avoïd Lagrange multipliers.

New Step 3. (version without Lagrange multipliers.)

I suppose here that $\max (a,b,c)\geq \sqrt{2}.$ without loss of generality I suppose that $a\geq \sqrt{2}.$

We also have $a\leq 3^{1/3}$ and $b,c \leq (3-\sqrt{2}^3)^{1/3}.$

Thus $\max(a+b,a+c,a+c)\leq 3^{1/3}+(3-\sqrt{2}^3)^{1/3}<2$

Hence $$a^3(a+b)+b^3(b+c)+c^3(a+c)< 2(a^3+b^3+c^3)=6.$$ c.q.f.d

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    A master put. Nice. – 2017-08-11
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    @Aviateur: This should have been an addition to your other answer (in the form of an edit), not a standalone answer. – 2017-09-28
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Bonjour Step 1 We have already proved that using Cauchy-Schawrz (or AM>HM ) it suffices to prove the following inequality. $a^3(a+b)+b^3(b+c)+c^3(c+a)\leq 6.$ I'll show this inequality. Step 2 In order to relax the constraint $a^3+b^3+c^3=3$, we remark that the inequality is: $a^3(a+b)+b^3(b+c)+c^3(c+a)\leq 6 (\dfrac{a^3+b^3+c^3}{3})^{4/3}$

Taking the third power of the right and left side we get an equivalent inequality which has the form; $$p(a,b,c)\geq 0$$ where $p$ is a homogeneous polynomial of degree 12.given by $$p(a,b,c)=8 \left(a^3+b^3+c^3\right)^4-3 \left(a^3 (a+b)+c^3 (a+c)+b^3 (b+c)\right)^3.$$

The sequel consists to show that $p(a,b,c)\geq 0, \forall a,b,c>0$ vanishing iff $a=b=c$ and this will lead to the result.

Without restriction of the generality we assume that $a=\max(a,b,c)$

Step 3 We suppose that $b\geq c.$
Then set $c=z$, $b=y+z$,; and $a=x+y+z.$ Hence $z>0$ et $x,y\geq 0.$

After a hard computation (use a formal software) we get $p(a,b,c)=p(x+y+z,y+z,z)=q(x,y,z)$ where qis a homogeneous polynomial where all the coeeficients are >0 (For clarity I'll give the polynomial q below.) Moreover we see that $q(x,y,z)=0$ ssi $x=y=0$, i.e $a=b=c.$

Sete 4 In the case $c\geq b$ Analogous to the step 3 setting $b=z$, $c=y+z$, et $a=x+y+z.$ The proof is complete.

the polynomial q:

$$\begin{array}{l} p(a,b,c) = \\ =5 x^{12}+51 x^{11} y+222 x^{10} y^2+544 x^9 y^3+837 x^8 y^4+864 x^7 y^5+669 x^6 y^6+540 x^5 y^7+585 x^4 y^8\\ +604 x^3 y^9+444 x^2 y^{10}+201 x y^{11}+47 y^{12}+51 x^{11} z+444 x^{10} y z+1632 x^9 y^2 z\\+3339 x^8 y^3 z+4230 x^7 y^4 z+3627 x^6 y^5 z+2844 x^5 y^6 z\\ +3267 x^4 y^7 z+4032 x^3 y^8 z+3513 x^2 y^9 z+1833 x y^{10} z+483 y^{11} z+222 x^{10} z^2\\+1632 x^9 y z^2+4995 x^8 y^2 z^2+8280 x^7 y^3 z^2+8100 x^6 y^4 z^2\\ +5724 x^5 y^5 z^2+6480 x^4 y^6 z^2+10467 x^3 y^7 z^2+11556 x^2 y^8 z^2\\+7212 x y^9 z^2+2184 y^{10} z^2\\+567 x^9 z^3+3510 x^8 y z^3+8775 x^7 y^2 z^3+10971 x^6 y^3 z^3+7119 x^5 y^4 z^3+5688 x^4 y^5 z^3+13482 x^3 y^6 z^3\\ +20835 x^2 y^7 z^3+16281 x y^8 z^3+5787 y^9 z^3+1008 x^8 z^4+5220 x^7 y z^4 \\+10161 x^6 y^2 z^4+8262 x^5 y^3 z^4+3114 x^4 y^4 z^4+9117 x^3 y^5 z^4+22914 x^2 y^6 z^4+23688 x y^7 z^4+10098 y^8 z^4 \\+1359 x^7 z^5 \\+5688 x^6 y z^5 \\+7992 x^5 y^2 z^5+3573 x^4 y^3 z^5+3474 x^3 y^4 z^5+16173 x^2 y^5 z^5+23553 x y^6 z^5+12303 y^7 z^5 \\+1395 x^6 z^6+4563 x^5 y z^6+4482 x^4 y^2 z^6+1863 x^3 y^3 z^6\\+7632 x^2 y^4 z^6+16407 x y^5 z^6 \\+10719 y^6 z^6+1116 x^5 z^7+2844 x^4 y z^7+2088 x^3 y^2 z^7+2880 x^2 y^3 z^7+8064 x y^4 z^7+6660 y^5 z^7+702 x^4 z^8 \\+1296 x^3 y z^8+1242 x^2 y^2 z^8+2808 x y^3 z^8+2862 y^4 z^8+300 x^3 z^9\\+504 x^2 y z^9+684 x y^2 z^9+780 y^3 z^9+108 x^2 z^{10}+108 x y z^{10}+108 y^2 z^{10} \end{array} $$

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    We can make it with the last inequality in my post. – 2018-03-03