Finding $\small{\max\limits_{ab+bc+ca=abc+2}\frac{ab(2-c)}{a^2+abc+b^2}+\frac{bc(2-a)}{b^2+abc+c^2}+\frac{ca(2-b)}{c^2+abc+a^2}.}$

Question

If $a,b,c\ge 0: ab+bc+ca=abc+2.$ Find the maximum $$P=\frac{ab(2-c)}{a^2+abc+b^2}+\frac{bc(2-a)}{b^2+abc+c^2}+\frac{ca(2-b)}{c^2+abc+a^2}$$ By set $a=b=c=1,$ we can see that $P= 1$. The remain is proving $$\frac{ab(2-c)}{a^2+abc+b^2}+\frac{bc(2-a)}{b^2+abc+c^2}+\frac{ca(2-b)}{c^2+abc+a^2}\le 1\tag{*}$$ I tried to use $pqr$ notation and by the condition $q=r+2.$ I split the problem into 2 case:

• $abc=0$ which is easy since $a=0; bc=2$ and $$\frac{2bc}{b^2+c^2}\le 1\iff (b-c)^2\ge 0.$$
• $abc>0$ which $(*)$ is written as $$\frac{2-c}{\dfrac{a}{b}+\dfrac{b}{a}+1}+\frac{2-a}{\dfrac{b}{c}+\dfrac{c}{b}+1}+\frac{2-b}{\dfrac{a}{c}+\dfrac{c}{a}+1}\le 1$$ But I don't know how to continue it.

Hope to see some helps. Thank you very much.

Update. We have a problem USAMO 2001 which is relevant.

Answer 1

By C-S $$\sum_{cyc}\frac{ab(2-c)}{a^2+abc+b^2}=-3+\sum_{cyc}\frac{(a+b)^2}{a^2+abc+b^2}=$$ $$=-3+2\sum_{cyc}\frac{(a+b)^2}{a(2a+bc)+b(2b+ac)}\leq$$ $$\leq-3+2\sum_{cyc}\left(\frac{a^2}{a(2a+bc)}+\frac{b^2}{b(2b+ac)}\right)=-3+4\sum_{cyc}\frac{a}{2a+bc}.$$ Now, easy to see that for positives $x$, $y$ and $z$ $$\sum_{cyc}\frac{1}{x+2}\leq1$$ is equivalent to $$xy+xz+yz+xyz\geq4,$$ which says that for $abc\neq0$(the case $abc=0$ easy to check) $$\sum_{cyc}\frac{a}{2a+bc}\leq1$$ it's $$\sum_{cyc}\frac{1}{2+\frac{bc}{a}}\leq1$$ or $$\sum_{cyc}\left(\frac{ab}{c}\cdot\frac{ac}{b}\right)+abc\geq4$$ or $$a^2+b^2+c^2+abc\geq4$$ and our inequality would be proven if we'll prove that $a^2+b^2+c^2+abc\geq4.$

Now, let $a+b+c=3u$, $ab+ac+bc=3v^2$ and $abc=w^3$.

Thus, the condition ($ab+ac+bc=2+abc$) does not depend on $u$, which since $$a^2+b^2+c^2+abc=9u^2-6v^2+w^3,$$ gives that it's enough to prove the last inequality for the minimal value of $u$, which by $uvw$ happens for equality case of two variables.

Let $b=a$ and $c=\frac{a^2-2}{a(a-2)},$ where $a>2$ or $0<a\leq\sqrt2$ and we obtain: $$(a-1)^2(a^2-2)(3a^2-4a-2)\geq0,$$ which is true for $0<a\leq\sqrt2$.

But for $a>2$ we obtain: $$a^2+b^2+c^2+abc>4\geq4.$$

Answer 2

Proof.

Remark. Maximal value is equal to $1$ which attained at $a=b=c=1$ or $a=b=\sqrt{2};c=0$ and cyclic permutations.

The OP can be written as$$\frac{(a+b)^2}{a^2+abc+b^2}+\frac{(c+b)^2}{c^2+abc+b^2}+\frac{(a+c)^2}{a^2+abc+c^2}\le 4.$$Now, by using Cauchy-Schwarz \begin{align*} \sum_{cyc}\frac{(a+b)^2}{a^2+abc+b^2}&=2.\sum_{cyc}\frac{(a+b)^2}{2a^2+abc+2b^2+abc}\\&\le2.\sum_{cyc}\left(\frac{a^2}{2a^2+abc}+\frac{b^2}{2b^2+abc}\right)\\&=4\sum_{cyc}\frac{a}{2a+bc}. \end{align*} Thus, it is enough to prove that$$\frac{a}{2a+bc}+\frac{b}{2b+ac}+\frac{c}{2c+ab}\le 1.$$ or $$\frac{bc}{2a+bc}+\frac{ac}{2b+ac}+\frac{ab}{2c+ab}\ge 1 \tag{*}.$$ From now, we split $(*)$ into two cases

• $a+b+c\ge 3.$ The $(*)$ is true by Cauchy-Schwarz. We need to prove $$(ab+bc+ca)^2\ge \sum_{cyc}bc(2a+bc),$$or $$2abc(a+b+c-3)\ge 0.$$
• $a+b+c\le 3.$ WLOG, assume that $c=\min\{a,b,c\}$ and the condition gives $0\le c\le 1.$ Rewrite $(*)$ as $$\frac{bc}{2a+bc}+\frac{ca}{2b+ac}\ge \frac{2c}{2c+ab} \iff \frac{b}{2a+bc}+\frac{a}{2b+ac}\ge \frac{2}{2c+ab}. $$ By using Cauchy-Schwarz $$\frac{b}{2a+bc}+\frac{a}{2b+ac}\ge \frac{(a+b)^2}{4ab+c(a^2+b^2)},$$ which is equivalent to $$(a+b)^2(2c+ab)\ge 8ab+2c(a^2+b^2) $$ $$\iff (a-b)^2(2-c)+c(a+b)^2(a-1)(b-1)\ge 0. \tag{**}$$ We will consider two subcases

1. $(a-1)(b-1)\ge 0:$ the $(**)$ is obviously true.
2. $(a-1)(b-1)\ge 0:$ WLOG, assume that $a\ge 1\ge b\ge c\ge 0.$ Easily, we obtain $$\frac{(a-b)^2}{4}\ge (a-1)(1-b); 0<a+b\le 3-c.$$ Thus, it is enough to prove that $$4(2-c)\ge c(3-c)^2 \iff (1-c)(c^2-5c+8)\ge 0.$$The last inequality is true on $c\in[0;1]$ and implies $(**)$ is proven.

Hence, we proved $(*)$ is true which ends proof here.