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

Question

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!

Answer 1

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: $$\sum_{cyc}\left(\frac{5}{3}a^4+\frac{5}{3}a^3b+\frac{5}{3}a^3c+4a^2b^2+9a^2bc\right)-6\sum_{cyc}(a^3+2a^2b+2a^2c+abc)+$$ $$+9\sum_{cyc}(2a^2+ab)-18(a+b+c)+27\geq0$$ and since $$\sum_{cyc}(a^2+2ab)-6(a+b+c)+9=(a+b+c-3)^2\geq0,$$ it's enough to prove: $$\sum_{cyc}\left(\frac{5}{3}a^4+\frac{5}{3}a^3b+\frac{5}{3}a^3c+4a^2b^2+9a^2bc\right)-6\sum_{cyc}(a^3+2a^2b+2a^2c+abc)+$$ $$+\sum_{cyc}(15a^2+3ab)\geq0,$$ for which it's enough to prove: $$9\left(\sum_{cyc}(a^3+2a^2b+2a^2c+abc)\right)^2\leq\sum_{cyc}(5a^4+5a^3b+5a^3c+12a^2b^2+27a^2bc)\sum_{cyc}(5a^2+ab),$$ which we can prove by uvw, but a full expanding is better, I think: $$\sum_{sym}(4a^6-3a^5b+9a^4b^2-7a^3b^3+6a^4bc-6a^3b^2c-3a^2b^2c^2)\geq0,$$ which is true by Schur and Muirhead: $$\sum_{sym}(4a^6-3a^5b+9a^4b^2-7a^3b^3+6a^4bc-6a^3b^2c-3a^2b^2c^2)=$$ $$=\sum_{sym}(4a^6-8a^5b+4a^4bc+5a^5b+9a^4b^2-7a^3b^3+2a^4bc-6a^3b^2c-3a^2b^2c^2)\geq0,$$ which finishes the solution.

Answer 2

** 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.

Answer 3

Some thoughts.

We use the pqr method. Let $p := a + b + c, q := ab + bc + ca, r := abc$.

After clearing the denominators, the desired inequality is written as \begin{align*} &3{p}^{4}-{p}^{3}r+{p}^{2}{q}^{2}-6{p}^{3}-9{p}^{2}q-{q}^{3}+18 {p}^{2}+6pq+9{q}^{2}-18p-27q+27 \\ \ge{}& 2\left( {a}^{2}b+{c}^{2}a+{ b}^{2}c \right) {p}^{2}. \end{align*}

We use the following upper bound for $a^2b + b^2c + c^2a$: $$a^2b + b^2c + c^2a \le \frac{8p^3q-27p^2r-15pq^2+54qr}{6p^2}.$$ (Proof. WLOG, assume that $c = \min(a, b, c)$. Letting $a = c + s, b = c + t$ for $s, t \ge 0$, all coefficients of the resulting polynomial in $c$ are non-negative.)

It suffices to prove that \begin{align*} &3{p}^{4}-{p}^{3}r+{p}^{2}{q}^{2}-6{p}^{3}-9{p}^{2}q-{q}^{3}+18 {p}^{2}+6pq+9{q}^{2}-18p-27q+27 \\ \ge{}& \frac{8p^3q-27p^2r-15pq^2+54qr}{3}, \end{align*} or \begin{align*} &3p^4-27q+p^2q^2-\frac83 p^3q-9p^2q+5pq^2+18p^2-6p^3+9q^2-q^3+6pq-18p+27\\ \ge{}& (p^3-9p^2+18q)r. \end{align*}

If $p^3-9p^2+18q > 0$, using $q^2 \ge 3pr$, it suffices to prove that \begin{align*} &3p^4-27q+p^2q^2-\frac83 p^3q-9p^2q+5pq^2+18p^2-6p^3+9q^2-q^3+6pq-18p+27\\ \ge{}& (p^3-9p^2+18q)\cdot \frac{q^2}{3p} \end{align*} which is true for all $p, q \ge 0$ with $p^2 \ge 3q$ (verified by Mathematica - a Computer Algebra System (CAS)).

If $p^3-9p^2+18q \le 0$, using degree three Schur inequality $r\ge\frac{4pq-p^3}{9}$, it suffices to prove that \begin{align*} &3p^4-27q+p^2q^2-\frac83 p^3q-9p^2q+5pq^2+18p^2-6p^3+9q^2-q^3+6pq-18p+27\\ \ge{}& (p^3-9p^2+18q)\cdot \frac{4pq - p^3}{9} \end{align*} which is true for all $p, q \ge 0$ with $p^2 \ge 3q$ (verified by Mathematica - a Computer Algebra System (CAS)).

$\phantom{2}$

Remark 1. When $a+b+c=1$, the form $c_1 q^2 + c_2 q + c_3 r + c_4 qr$ upper bound for $a^2b + b^2c + c^2a$ was found before. Here we need to find appropriate coefficients $c_1, c_2, c_3, c_4$.

Remark 2. In @Michael Rozenberg's nice answer, the upper bound $a^2b + b^2c + c^2a \le \frac{4p^4-3p^3r+3p^2q^2-12p^2q-3q^3-18pr+15q^2}{6p^2} $ was used, then it suffices to prove that \begin{align*} &3{p}^{4}-{p}^{3}r+{p}^{2}{q}^{2}-6{p}^{3}-9{p}^{2}q-{q}^{3}+18 {p}^{2}+6pq+9{q}^{2}-18p-27q+27\\ \ge{}& \frac{4p^4-3p^3r+3p^2q^2-12p^2q-3q^3-18pr+15q^2}{3} \end{align*} which is true.

Answer 4

Some thoughts.

We need to prove that $$\frac{6a}{a^2+ab+b^2+3}+\frac{6b}{b^2+bc+c^2+3}+\frac{6c}{c^2+ca+a^2+3}\leq 3.$$

We use the so-called isolated fudging.

It suffices to prove that, for all $a, b, c \ge 0$, $$\frac{6a}{a^2+ab+b^2+3} \le \frac{5b^2 + 7c^2 - 4bc + 4ca + 26a - 20b - 6c + 32}{4a^2+4b^2+4c^2+32}. \tag{1}$$ (Note: Taking a cyclic sum on (1), we get the desired inequality using $$\sum_{\mathrm{cyc}} \frac{5b^2 + 7c^2 - 4bc + 4ca + 26a - 20b - 6c + 32}{4a^2+4b^2+4c^2+32} = 3.$$ (1) is true which is verfied by Mathematica - a Computer Algebra System (CAS).)

After clearing the denominators, it suffices to prove that $$m_2 c^2 + m_1 c + m_0 \ge 0 \tag{2}$$ where \begin{align*} m_2 &= 7a^2+7ab+7b^2-24a+21,\\ m_1 &= 2(2a - 2b - 3)(a^2+ab+b^2+3),\\ m_0 &= 5a^2b^2+5ab^3+5b^4+2a^3+6a^2b-18ab^2-20b^3+32a^2\\ &\qquad+32ab+47b^2-114a-60b+96. \end{align*}

We can prove that $m_2, m_0 \ge 0$.

If $2a - 2b - 3 \ge 0$, then $m_1 \ge 0$ and clearly (2) is true.

If $2a - 2b - 3 < 0$, it suffices to prove that $4m_2m_0 - m_1^2 \ge 0$. I don't have a nice proof.

Answer 5

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 !

Answer 6

$\color{brown}{\textbf{Substitutions.}}$

We should prove the inequality $$\dfrac a{a^2+ab+b^2+3}+\dfrac b{b^2+bc+c^2+3}+\dfrac c{c^2+ca+a^2+3}\le \dfrac12\tag1$$ for non-negative $a,b,c.$

Let us express the given inequality via non-negative variables, based on the symmetric polynomials, in the form of $$\begin{align} &s=a+b+c,\quad r=ab+bc+ca,\quad p=abc,\\[4pt] &\alpha=a^2b+b^2c+c^2a,\quad \beta=ab^2+bc^2+ca^2,\tag2 \end{align}$$ and define additional non-negative variables $$\begin{cases} q=a^2+ab+b^2+bc+c^2+ca+3=s^2-r+3,\\[4pt] u=a^2+ab+b^2+3=q-cs,\\[4pt] v=b^2+bc+c^2+3=q-as,\\[4pt] w=c^2+ca+a^2+3=q-bs,\\[4pt] x=u+v+w=3q-s^2,\\[4pt] y=uv+vw+wu=(q-cs)(q-as)+(q-as)(q-bs)+(q-bs)(q-cs),\\[4pt] z=uvw=(q-cs)(q-as)(q-bs)=q^3-s^2q^2+s^2rq-s^3p,\\[4pt] \zeta=u^2v+v^2w+w^2u=(q-cs)^2(q-as)+(q-as)^2(q-bs)+(q-bs)^2(q-cs), \end{cases}\tag3$$

wherein $$\begin{align} &y=3q^2-(c+a+a+b+b+a)\,sq^2+(ca+ab+bc)\,s^2=3q^2-2s^2q+s^2r\\[4pt] &=3q^2-3s^2q+s^2(q+r)=3q^2-3s^2q+s^2(s^2+3);\\[4pt] &\zeta=3q^3-(2c+a+2a+b+2b+c)\,sq^2+(c^2+2ac+a^2+2ab+b^2+2bc)\,s^2q\\[4pt] &-(c^2a+a^2b+b^2c)\,s^3=3q^3-3s^2q^2+s^4q-s^3\alpha,\\[4pt] \end{align}$$ $$\zeta=qy-3s^2q-s^3\alpha.\tag4$$

Before the final proving, we should get some auxiliary inequalities.

$\color{brown}{\textbf{Simplifications.}}$

Firstly, we can prove $$\begin{align} &r^2-3ps=(ab+bc+ca)^2-3abc(a+b+c)=(ab)^2+(bc)^2+(ca)^2\\[4pt] &-(ab)(bc)-(bc)(ca)-(ca)(ab) =\tfrac12\left((ab-bc)^2+(bc-ca)^2+(ca-ab)^2\right),\\[4pt] &3sp\le r^2. \end{align}$$

From $(3)$ should $\;q-s^2=3-r,\;$ and then $$\begin{align} &3z=3(q-s^2)q^2-3s^2q^2+3s^2rq-3s^3p\ge 3(3-r)q^2+3s^2rq-s^2r^2\\[4pt] &=9q^2-3(q-s^2)rq-s^2r^2=9q^2-3(3-r)rq+s^2rq-s^2r^2\\[4pt] &=9q^2-9rq+2r^2q+r^2(q-s^2)=9q^2-9rq+2r^2q+3r^2(3+r)\ge 0, \end{align}$$ $$3z\ge 9q^2-9rq+2r^2q+3r^2(3+r).\tag5$$

Also, using Cauchy-Schwarz inequality in the vector form, easily to get $$\begin{align} &s\alpha=(a+b+c)(a^2b+b^2c+c^2a) =\bigg|\big\{\sqrt b\,,\sqrt c\,,\sqrt a\,\big\}\bigg|^2 \cdot \bigg|\big\{a\sqrt b\,,b\sqrt c\,,c\sqrt a\,\big\}\bigg|^2\\[4pt] &\ge \bigg|\big\{\sqrt b\,,\sqrt c\,,\sqrt a\,\big\} \cdot \big\{a\sqrt b\,,b\sqrt c\,,c\sqrt a\,\big\}\bigg|^2 =(ab+bc+ca)^2=r^2;\\[4pt] &s\beta=(a+b+c)(ab^2+bc^2+ca^2) =\bigg|\big\{\sqrt a\,,\sqrt b\,,\sqrt c\,\big\}\bigg|^2 \cdot \bigg|\big\{b\sqrt a\,,c\sqrt b\,,a\sqrt c\,\big\}\bigg|^2\\[4pt] &\ge \bigg|\big\{\sqrt a\,,\sqrt b\,,\sqrt c\,\big\} \cdot \big\{b\sqrt a\,,c\sqrt b\,,a\sqrt c\,\big\}\bigg|^2 =(ab+bc+ca)^2=r^2;\\[4pt] &sr=(a+b+c)(ab+bc+ca)=a^2b+b^2c+c^2a+3abc+ab^2+bc^2+ca^2=\alpha+\beta+3p. \end{align}$$ Therefore, $$\alpha+\beta+3p=sr,\quad 3sp\in[0,r^2],\quad s\alpha\ge r^2,\quad s\beta\ge r^2 \quad\Rightarrow\quad s\alpha\in[r^2, sr-r^2].\tag6$$

Taking in account $(4),$ easily to get $$\zeta-qy=-3s^2q-s^3\alpha\ge -3s^2q-s^2(sr-r^2)=-s^3r+s^2(r^2-3q).\tag7$$

Besides, we can prove the known inequality $\;s^2\ge 3r:\;$ $$s^2-3r=a^2+b^2+c^2-ab-bc-ca=\tfrac12\left((a-b)^2+(b-c)^2+(c-a)^2\right)\ge 0.\tag8$$

Easily to see that all auxillary inequalities became exact equalities when $\;a=b=c,\;$ and then from the goal inequality $a=1.$

$\color{brown}{\textbf{Proof.}}$

New variables allow to present the given inequality in the forms of $$\dfrac{q-v}{u}+\dfrac{q-w}{v}+\dfrac{q-u}{w}\le \dfrac s2,\tag9$$ or $$6\big((q-v)vw+(q-w)wu+(q-u)uv\big)\le 3uvws,$$ $$3sz+6(\zeta-qy)\ge 0.$$ Taking in account the inequalities $(5),(7),(8)$ it suffices to prove the stronger inequality, instead of the original, in the form of $\;sf(q,r,s)\ge 0,\;$ where $$\begin{align} &f(q,r,s)=9q^2-9rq+2r^2q+3r^2(3+r)+6s(r^2-sr-3q);\\[4pt] &F(r,s)=f(s^2-r+3,r,s)=81-81r+33r^2+r^3\\[4pt] &-54s+18rs+6r^2s+54s^2-21rs^2+2r^2s^2-18s^3+9s^4\\[4pt] &=81-81r+33r^2+r^3-6s(9+3s^2-3r)+6r^2s+54s^2-21rs^2+2r^2s^2+9s^4\\[4pt] &\ge 81-81r+33r^2+r^3-(s^2+9)(9+3s^2-3r)+6r^2s+54s^2-21rs^2+2r^2s^2+9s^4\\[4pt] &=r^3+2r^2s^2+6r^2s+33r^2-18rs^2-54r+6s^4+18s^2\\[4pt] &=r^3+2r^2s^2+6r^2s+33r^2+6s^2(s^2-3r)+18(s^2-3r)\ge 0.\\[4pt] \end{align}$$

Thus, $$\color{green}{\mathbf{\dfrac a{a^2+ab+b^2+3}+\dfrac b{b^2+bc+c^2+3}+\dfrac c{c^2+ca+a^2+3}\le \dfrac12}}.$$