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For any positive reals

$\frac{b^2+c^2}{b+c}+\frac{c^2+a^2}{a+c}+\frac{a^2+b^2}{a+b} \ge a+b+c$

After some work this equality boils down to proving $ab^3+b^3c+a^3c+bc^3+a^3b+c^3a-2abc^2-2ab^2c-2a^2bc \ge 0$ which I am struggling to prove. Any hints

Mathematical Curiosity
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    The inequality $a^3+b^3 \geq 2ab^2$ is not true. Check $(a,b)=(2,3)$. Maybe if you show the complete work, we can pinpoint exactly where you went wrong. – LHF Mar 15 '20 at 07:42
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    Although, I would first recommend to try to prove $\frac{a^2+b^2}{a+b}\geq \frac{a+b}{2}$. – LHF Mar 15 '20 at 07:44
  • @Atticus Your way is elegant but I did the long way and ended up with $ab^3+b^3c+a^3c+bc^3+a^3b+c^3a-2abc^2-2ab^2c-2a^2bc \ge 0 $. Can you show any directions from here?? – Mathematical Curiosity Mar 15 '20 at 08:31
  • Yes, I added it as an answer. – LHF Mar 15 '20 at 08:55
  • @MartinR, It's unclear to me if questions (with statements already asked before) where the asker asks some specific idea about his proof should be closed as duplicates. Say, for this question, if the OP tries a certain method but he gets stuck along the way or just wants to check something about it, should he comment in the original thread, or should he ask a separate question? Can you please clarify what's the correct course of action in such cases? Thanks. – LHF Mar 15 '20 at 09:57
  • @MartinR, Just to be clear, I don't have any issue with closing such questions as duplicates. I ask because in a similar situation I marked a question as duplicate (https://math.stackexchange.com/questions/3567557/problem-on-limit-of-a-sequence?noredirect=1&lq=1), but it wasn't closed in the end. The correct policy in such cases is not clear to me and I wasn't able to find anything on meta. – LHF Mar 15 '20 at 10:12
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    @Atticus: If a question author states “I tried this and this, but it could not finish it” and there is an identical question with good answers then I tend to close-vote as a duplicate. (Without showing that attempt the question might get closed as off-topic.) That is my personal opinion, others may disagree. (That's why it takes 5 close votes :) – Martin R Mar 15 '20 at 10:27
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    @Atticus: In that other situation you mentioned OP did a full proof, but got a nonsensical result and asked for the error in their proof. That might be slightly different, but I can see that I upvoted your comment about the possible duplicate back then. – Martin R Mar 15 '20 at 10:31
  • @MartinR, thanks. I guess in the end it's a matter of how well-written and specific the question is (which is why I asked for the complete work in the first comment) – LHF Mar 15 '20 at 10:35

4 Answers4

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Hint: Instead use Cauchy-Schwarz inequality $(a^2+b^2)(1+1)\geqslant (a+b)^2$

Macavity
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As highlighted by @Atticus in the comments we can prove that:

${x^2+y^2\over x+y}\geq {x+y\over 2}$

$2(x^2+y^2)\geq x^2+y^2+2xy$

$x^2+y^2-2xy\geq 0$

which is true since

$(x-y)^2 \geq 0$

h-squared
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As Atticus suggested, from $2(a^2+b^2)\geq a^2+b^2+2ab=(a+b)^2$

It follows that $\frac{a^2 +b^2}{a+b}\geq \frac{a+b}{2}.$ By similar arguments we can prove that $\frac{b^2 +c^2}{b+c}\geq \frac{b+c}{2}$ and $\frac{a^2 +c^2}{a+c}\geq \frac{a+c}{2}.$ Adding these three inequalities we get the desired inequality.

user159888
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To continue @Mathematical Curiosity's work, the inequality expands as:

$$ab^3+b^3c+a^3c+bc^3+a^3b+c^3a-2abc^2-2ab^2c-2a^2bc \ge 0$$

We can use AM-GM inequality:

$$ab^3+a^3b+a^3c+ac^3\geq 4\sqrt[4]{a^8b^4c^4}=4a^2bc$$

Now, sum with the equivalent inequalities for $4ab^2c$ and $4abc^2$ and divide by $2$ to arrive at the conclusion.

Note: To prove

$$ab^3+a^3b+a^3c+ac^3\geq 4a^2bc$$

we can also use AM-GM twice:

$$(ab^3+a^3b)+(a^3c+ac^3)\geq 2a^2b^2+2a^2c^2=2a^2(b^2+c^2)\geq 4a^2bc$$

From this chain we can deduce the following identity

$$ab^3+a^3b+a^3c+ac^3-4a^2bc=ab(a-b)^2+ac(a-c)^2+2a^2(b-c)^2\geq 0$$

Summing and dividing by $2$:

$$\sum_{cyc}(a^3b+ab^3)-2\sum_{cyc}a^2bc=\sum_{cyc}(ab+c^2)(a-b)^2\geq 0$$

LHF
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