# Prove that $\sum\limits_{cyc}\frac{a}{a^2+ab+b^2+3}\leq\frac{1}{2}$

Let $a$, $b$ and $c$ be non-negative numbers. Prove that: $$\frac{a}{a^2+ab+b^2+3}+\frac{b}{b^2+bc+c^2+3}+\frac{c}{c^2+ca+a^2+3}\leq\frac{1}{2}$$

I think this inequality is very interesting because most of the contest's inequalities are homogeneous, but this inequality is non-homogeneous.

Testing for $c=0$ gives $$\frac{a}{a^2+ab+b^2+3}+\frac{b}{b^2+3}\leq0.455...<\frac{1}{2}.$$ For $b=c=0$ we obtain something obvious.

One of the standard ways to prove these inequalities is to try to make a homogenization.

By the way, trying of homogenization gives a wrong inequality: $$\sum\limits_{cyc}\frac{a}{a^2+ab+b^2+3}\leq\sum\limits_{cyc}\frac{a}{2\sqrt{3(a^2+ab+b^2}}.$$ Thus, it remains to prove that $$\sum\limits_{cyc}\frac{a}{\sqrt{3(a^2+ab+b^2)}}\leq1,$$ which is wrong for $c=0$ and $a\rightarrow+\infty$.

Also we can try the following.

We know that $\sum\limits_{cyc}\frac{x}{2x+y}\le1$ for positives $x$, $y$ and $z$.

Indeed, by C-S we obtain: $$1-\sum_{cyc}\frac{x}{2x+y}=1-\frac{3}{2}-\sum_{cyc}\left(\frac{x}{2x+y}-\frac{1}{2}\right)=\frac{1}{2}\sum_{cyc}\frac{y}{2x+y}-\frac{1}{2}=$$ $$=\frac{1}{2}\sum_{cyc}\frac{y^2}{2xy+y^2}-\frac{1}{2}\geq\frac{1}{2}\frac{(x+y+z)^2}{\sum\limits_{cyc}(y^2+2xy)}=\frac{1}{2}-\frac{1}{2}=0.$$ Thus, it's enough to prove that $$a^2+ab+b^2+3\geq2(2a+\sqrt{ab})$$ because if it's true so $$\sum_{cyc}\frac{a}{a^2+ab+b^2+3}\leq\sum_{cyc}\frac{a}{2(2a+\sqrt{ab})}=\frac{1}{2}\sum_{cyc}\frac{\sqrt{a}}{2\sqrt{a}+\sqrt{b}}\leq\frac{1}{2}.$$ But the inequality $a^2+ab+b^2+3\geq2(2a+\sqrt{ab})$ is wrong! Try $a=2$ and $b=\frac{1}{4}$

Also we can try to use a full expanding (I tried!) and to hope to use AM-GM,

but I think this way is very ugly and it's probably nothing.

Any hint would be desirable.

Thank you!

• Can someone explain me, why we see two persons, which want to close this topic? Thank you! Can this person (which want to close) explain me, why you do it? – Michael Rozenberg Mar 9 '17 at 8:55
• Hello Michael I have a very simple proof for all the cases except $a,b,c\geq 1$.My question is have you a proof in this case ? – user448747 Nov 6 '17 at 13:25
• I did not think about this because I am looking for a full solution. – Michael Rozenberg Nov 6 '17 at 13:27
• Okay no problem Can I pick up some of your ideas ? – user448747 Nov 6 '17 at 13:30
• A full expanding gives something interesting, but I still don't see how to end the proof. – Michael Rozenberg Nov 6 '17 at 13:32

** Partial result **

OK - as announced in the comments, I hope that @FatsWallers will post his proof for all cases except $a,b,c≥1$ (he said he has one). That would be great, because I have a proof only for the case $a,b,c≥1$ which I post here. It's rather straightforward.

In the original question, resubstitute $a \to 1+a$, $b \to 1+b$, $c \to 1+c$.

From the original $a,b,c≥1$ (the case I wish to prove), you have now $a,b,c\geq 0$ and you need to show

$$\frac12 - \sum_{cyclic}\frac{1+a}{a^2 + a b + 3 a + b^2 + 3 b + 6} \geq 0$$

Full expansion (with the usual help of Wolframalpha) gives

$$\frac{N}{2 (a^2 + a b + 3 a + b^2 + 3 b + 6) (a^2 + a c + 3 a + c^2 + 3 c + 6) (b^2 + b c + 3 b + c^2 + 3 c + 6)} \geq 0$$

with

$$N = a^4 b^2 + a^4 b c + a^4 b + a^4 c^2 + 3 a^4 c + 4 a^4 + a^3 b^3 + 2 a^3 b^2 c + 5 a^3 b^2 + 2 a^3 b c^2 + 8 a^3 b c + 4 a^3 b + a^3 c^3 + 7 a^3 c^2 + 16 a^3 c + 12 a^3 + a^2 b^4 + 2 a^2 b^3 c + 7 a^2 b^3 + 3 a^2 b^2 c^2 + 12 a^2 b^2 c + 21 a^2 b^2 + 2 a^2 b c^3 + 12 a^2 b c^2 + 27 a^2 b c + 15 a^2 b + a^2 c^4 + 5 a^2 c^3 + 21 a^2 c^2 + 33 a^2 c + 24 a^2 + a b^4 c + 3 a b^4 + 2 a b^3 c^2 + 8 a b^3 c + 16 a b^3 + 2 a b^2 c^3 + 12 a b^2 c^2 + 27 a b^2 c + 33 a b^2 + a b c^4 + 8 a b c^3 + 27 a b c^2 + 36 a b c + 12 a b + a c^4 + 4 a c^3 + 15 a c^2 + 12 a c + b^4 c^2 + b^4 c + 4 b^4 + b^3 c^3 + 5 b^3 c^2 + 4 b^3 c + 12 b^3 + b^2 c^4 + 7 b^2 c^3 + 21 b^2 c^2 + 15 b^2 c + 24 b^2 + 3 b c^4 + 16 b c^3 + 33 b c^2 + 12 b c + 4 c^4 + 12 c^3 + 24 c^2 \geq 0$$

to be established. Since $a,b,c\geq 0$, we are done, since all terms are nonnegative.

Remaining cases (not $a,b,c\geq 0$): waiting for @FatsWallers announced proof.

I think I get something :

Lemma

Let $f:\mathbb{R} \mapsto\mathbb{R^+}$ a real convex function for all $a,b,c,\alpha,\beta,\gamma$ positive real numbers then : $$\frac{f(a)}{f(a)^2+f(a)f(b)+f(b)^2+3}+\frac{f(b)}{f(b)^2+f(b)f(c)+f(c)^2+3}+\frac{f(c)}{f(c)^2+f(c)f(a)+f(a)^2+3}\leq\frac{f(a)+\alpha f(b)}{\alpha(f(a)^2+f(a)f(b)+f(b)^2+3)}+\frac{f(b)+\beta f(c)}{\beta(f(b)^2+f(b)f(c)+f(c)^2+3)}+\frac{f(c)+\gamma f(a)}{\gamma(f(c)^2+f(c)f(a)+f(a)^2+3)}$$

Proof :

If we compare this : $$\frac{f(a)}{f(a)^2+f(a)f(b)+f(b)^2+3}\leq \frac{f(a)+\alpha f(b)}{\alpha(f(a)^2+f(a)f(b)+f(b)^2+3)}\quad(1)$$

We get :

$$0\leq \alpha f(b)+(1-\alpha) f(a)$$

But with the definition of convexity we get :

$$0\leq f( (\alpha) (b)+(1-\alpha)(a))\leq\alpha f(b)+(1-\alpha) f(a)$$

So the inequality (1) is true , and sum to get the lemma .

Now put $\alpha=\frac{1}{2f(a)}$$\quad$$\beta=\frac{1}{2f(b)}$$\quad$$\gamma=\frac{1}{2f(c)}$

We get :

$$\frac{f(a)}{f(a)^2+f(a)f(b)+f(b)^2+3}+\frac{f(b)}{f(b)^2+f(b)f(c)+f(c)^2+3}+\frac{f(c)}{f(c)^2+f(c)f(a)+f(a)^2+3}\leq\frac{0.5+2f(a) f(b)}{(f(a)^2+f(a)f(b)+f(b)^2+3)}+\frac{0.5+f(b) f(c)}{(f(b)^2+f(b)f(c)+f(c)^2+3)}+\frac{0.5+f(c)f(a)}{(f(c)^2+f(c)f(a)+f(a)^2+3)}$$

But the minimum of : $$\frac{0.5+2f(a) f(b)}{(f(a)^2+f(a)f(b)+f(b)^2+3)}$$ is reached when $f(a)=f(b)=0$

So we get :

$$\frac{f(a)}{f(a)^2+f(a)f(b)+f(b)^2+3}+\frac{f(b)}{f(b)^2+f(b)f(c)+f(c)^2+3}+\frac{f(c)}{f(c)^2+f(c)f(a)+f(a)^2+3}\leq 0.5$$

Done !

• "the minimum of" occurs for $a=b=c=0$, but in the starting inequality the maximum occurs for $a=b=c=1$, which says that your last step is wrong. – Michael Rozenberg Oct 22 '17 at 19:21
• Now it would be correct dear Michael . – user448747 Oct 23 '17 at 10:22
• Why the maximum is $2$? We need to prove it. – Michael Rozenberg Oct 23 '17 at 11:35
• I edit what do you think about dear Michael ? – user448747 Nov 1 '17 at 16:20
• Why the first inequality is true? Your second inequality is obviously wrong. Try $b\rightarrow0^+$. – Michael Rozenberg Nov 3 '17 at 14:08

After full expanding we need to prove that: $$\prod_{cyc}(a^2+ab+b^2)-2\sum_{cyc}(a^4b+2a^3b^2+a^3c^2+2a^3bc+3a^2b^2c)+$$ $$+3\sum_{cyc}(a^4+a^3b+a^3c+3a^2b^2+3a^2bc)-6\sum_{cyc}(a^3+2a^2b+2a^2c+abc)+$$ $$+9\sum_{cyc}(2a^2+ab)-18(a+b+c)+27\geq0.$$ We'll prove that: $$\prod_{cyc}(a^2+ab+b^2)+\frac{4}{3}(a+b+c)(a^3+b^3+c^3)+5\sum_{cyc}a^2b^2\geq$$ $$\geq2\sum_{cyc}(a^4b+2a^3b^2+a^3c^2+2a^3bc+3a^2b^2c).$$ Indeed, by AM-GM $$\prod_{cyc}(a^2+ab+b^2)+\frac{4}{3}(a+b+c)(a^3+b^3+c^3)+5\sum_{cyc}a^2b^2\geq$$ $$\geq2\sqrt{\prod_{cyc}(a^2+ab+b^2)\left(\frac{4}{3}(a+b+c)(a^3+b^3+c^3)+5\sum_{cyc}a^2b^2\right)}.$$ Thus, it's enough to prove here that $$\prod_{cyc}(a^2+ab+b^2)\left(\frac{4}{3}(a+b+c)(a^3+b^3+c^3)+5\sum_{cyc}a^2b^2\right)\geq$$ $$\geq\left(\sum_{cyc}(a^4b+2a^3b^2+a^3c^2+2a^3bc+3a^2b^2c)\right)^2.$$ Let $$a=\min\{a,b,c\}$$, $$b=a+u$$ and $$c=a+v$$.

Thus, the last inequality it's $$P\geq0$$, where $$P$$ a polynomial degree $$8$$ of $$a$$ with non-negative coefficients.

Id est, it's enough to prove that: $$\sum_{cyc}\left(\frac{5}{3}a^4+\frac{5}{3}a^3b+\frac{5}{3}a^3c+4a^2b^2+9a^2bc\right)-6\sum_{cyc}(a^3+2a^2b+2a^2c+abc)+$$ $$+9\sum_{cyc}(2a^2+ab)-18(a+b+c)+27\geq0.$$ Now, we'll prove that: $$\sum_{cyc}\left(\frac{5}{3}a^4+\frac{5}{3}a^3b+\frac{5}{3}a^3c+4a^2b^2+9a^2bc\right)-6\sum_{cyc}(a^3+2a^2b+2a^2c+abc)+$$ $$+6\sum_{cyc}(2a^2+ab)\geq0.$$ Indeed, by AM-GM again we obtain: $$\sum_{cyc}\left(\frac{5}{3}a^4+\frac{5}{3}a^3b+\frac{5}{3}a^3c+4a^2b^2+9a^2bc\right)+6\sum_{cyc}(2a^2+ab)\geq$$ $$\geq2\sqrt{2\sum_{cyc}\left(5a^4+5a^3b+5a^3c+12a^2b^2+27a^2bc\right)\sum_{cyc}(2a^2+ab)}$$ and it's enough to prove here that $$2\sum_{cyc}\left(5a^4+5a^3b+5a^3c+12a^2b^2+27a^2bc\right)\sum_{cyc}(2a^2+ab)\geq$$ $$\geq9\left(\sum_{cyc}(a^3+2a^2b+2a^2c+abc)\right)^2.$$ Now, let $$a+b+c=3u$$, $$ab+ac+bc=3v^2$$ and $$abc=w^3$$.

Thus, we need to prove here that $$2(5(27u^3-27uv^2+3w^3)\cdot3u+12(9v^4-6uw^3)+81uw^3)(2(9u^2-6v^2)+3v^2)\geq$$ $$\geq9\left(27u^3-27uv^2+2(9uv^2-3w^3)+3w^3\right)^2$$ or $$2(2u^2-v^2)(15u^4-15u^2v^2+4v^4+2uw^3)\geq3(3u^3-uv^2)^2,$$ which is a linear inequality of $$w^3$$, which says that it's enough to prove the last inequality for an extreme value of $$w^3$$, which happens in the following cases.

1. $$w^3=0$$.

Since the last inequality is homogeneous and symmetric, we can assume $$c=0$$ and $$b=1$$, which gives $$2(5a^4+5+5a^3+5a+12a^2)(2a^2+2+a)\geq9(a^3+1+2a^2+2a)^2.$$ Now, let $$a^2+1=2ta$$.

Thus, by AM-GM $$t\geq1$$ and we need to prove that $$2(5(4t^2-2)+5\cdot2t+12)(4t+1)\geq9(a+1)^2(a^2+a+1)^2$$ or $$2(4t+1)(10t^2+5t+1)\geq9(t+1)(2t+1)^2,$$ which is wrong :(

• @River Li I think we can fix a last inequality such that it would be true (for example we can take a coefficient is less than $\frac{4}{3}$). Also, maybe the inequality after words "Id est, it's enough to prove" is true already, but I can not check it now. – Michael Rozenberg Sep 6 at 17:14
• So, you reduced the inequality to the first inequality after "Id est"? It is true for all real numbers. – River Li Sep 7 at 3:02