Let $a,b,c \ge 0: ab+bc+ca=3.$ Find maximal value $$P=\frac{ab}{2a+b}+\frac{bc}{2b+c}+\frac{ca}{2c+a}.$$ My tryings lead to wrong inequality.
I tried to use C-S $$\sum_{cyc}\frac{ab}{2a+b}\le \sum_{cyc}\frac{ab}{9}\left(\frac{2}{a}+\frac{1}{b}\right)=\frac{1}{3}\frac{ab+bc+ca}{abc}=\frac{1}{abc}$$But $abc\le 1$ so we got reverse inequality.
Which a good approach is good here ? I am very appreciate for your helps.
On
It is enough to consider $a,b,c>0.$ Let $ab=u$, $bc=v$, $ca=w$ where $u+v+w=3$ and $u,v,w>0$, then $$\sum_{cyc}\frac{ab}{2a+b}=\sum_{cyc}\frac{\sqrt{uvw}}{u+2v}\\\stackrel{Hölder}\leq\sqrt{3\sum_{cyc}\frac{uv}{(u+2v)^2}}\leq 1.$$ The last inequality follows from AM-GM. You may see this question and their answers.
On
The maximum is $1$ when $a = b = c = 1$.
Proof.
We use isolated fudging.
It suffices to prove that $$\frac{ab}{2a + b} \le \frac{2ab + bc}{9}. \tag{1}$$ (Note: Taking cyclic sum on (1), we get $\frac{ab}{2a+b}+\frac{bc}{2b+c}+\frac{ca}{2c+a} \le \frac{3(ab + bc + ca)}{9} = 1$.)
It suffices to prove that \begin{align*} (2a + b)(2a + c) - 9a \ge 0 \end{align*} or $$4a^2 + 2ab + bc + 2ca - 9a \ge 0$$ or (using $ab + bc + ca = 3$) $$4a^2 - 9a + 6 - bc \ge 0. \tag{2}$$
If $a \ge 1/4$, using $abc \le 1$ (by AM-GM), we have \begin{align*} a(4a^2 - 9a + 6 - bc) &= 4a^3 - 9a^2 + 6a - abc\\ &\ge 4a^3 - 9a^2 + 6a - 1\\ &= (4a - 1)(a - 1)^2\\ &\ge 0. \end{align*}
If $a < 1/4$, using $bc \le 3$, we have $$4a^2 - 9a + 6 - bc \ge -9\cdot \frac14 + 6 - 3 = \frac34 > 0. $$
We are done.
On
Proof.
We will prove $$\frac{ab}{2a+b}+\frac{bc}{2b+c}+\frac{ca}{2c+a}\le 3.$$ This inequality can be rewritten as $$\frac{2(a^2b^2+b^2c^2+c^2a^2)+7abc(a+b+c)}{(2a+b)(2b+c)(2c+a)}\le 3.$$Since $a^2b^2+b^2c^2+c^2a^2\ge abc(a+b+c),$ it suffices to prove $$\frac{3(ab+bc+ca)^2}{(2a+b)(2b+c)(2c+a)}\le 3,$$or $$(2a+b)(2b+c)(2c+a)\ge 27,$$ $$\iff 4\sum_{cyc}a^2b+2\sum_{cyc}ab^2+9abc\ge 27.\tag{*}$$ It is easily to show $$a+b+c\ge 3,$$$$\sum_{cyc}a^2b+\sum_{cyc}ab^2=3(a+b+c)-3abc,$$and $$(a-b)(b-c)(c-a)=\sum_{cyc}ab^2-\sum_{cyc}a^2b.$$ Hence, we will prove stronger inequality according $(*)$ $$6(a+b+c)\ge 18+(a-b)(b-c)(c-a).\tag{**}$$ WLOG, assume that $a=\text{max}\{a,b,c.\}$
We can split $(**)$ into two cases
$a\ge b\ge c.$ It means $(a-b)(b-c)(c-a)\le 0$ and $a+b+c\ge 3,$ so $(**)$ is true.
$a\ge c\ge b.$ By the condition, $3=ab+bc+ca\ge 3b^2 \implies b\in [0;1].$
Now, rewrite $(**)$ as a function according $b$ $$\color{blue}{f(b)=(a-c)b^2+b(c^2-a^2-6)+ca^2-c^2a-6(a+c)+18\le 0.}$$Do a little with manipulation on function, the result follows.
Can you end it now ?
Indeed, we easily to see that$$f'(b)=2b(a-c)+c^2-a^2-6=(a-c)(2b-a-c)-6< 0.$$ It implies $f(b)$ is decreasing on $[0;1].$
Let $b=0; ca=3$, it is sufficient to prove $$f(0)\le 0 \iff ca(a-c)-6(a+c)+18\le 0.$$ $$\iff 3(a-c)-6(a+c)+18\le 0\iff a+3c\ge 6,$$which is true by AM-GM $a+3c\ge 2\sqrt{3ac}=6.$
It means $(**)$ is proven and we end proof here.
For $a=b=c=1$ we obtain a value $1$.
We'll prove that it's a maximal value.
Indeed, let $a+b+c=3u$, $ab+ac+bc=3v^2$, where $v>0$, $abc=w^3$ and $u=tv$.
Thus, $v=1$, $t\geq1$ and we need to prove $$\sum_{cyc}ab(2b+c)(2c+a)\leq\prod_{cyc}(2a+b)$$ or $$\sum_{cyc}(2a^2b^2+7a^2bc)\leq v\sum_{cyc}(3abc+4a^2b+2a^2c)$$ or $$\sum_{cyc}(2a^2b^2+7a^2bc)+v\sum_{cyc}(a^2c-a^2b)\leq v\sum_{cyc}(3abc+3a^2b+3a^2c)$$ or $$18v^4-12uw^3+21uw^3+v(a-b)(b-c)(c-a)\leq27uv^3$$ or $$9(3uv^3-2v^4-uw^3)\geq v(a-b)(b-c)(c-a).$$ But by Maclaurin($u\geq v\geq w$) we see that $$3uv^3-2v^4-uw^3\geq0,$$ which says that it's enough to prove that: $$81(3uv^3-2v^4-uw^3)^2\geq v^2\prod_{cyc}(a-b)^2$$ or $$3(3uv^3-2v^4-uw^3)^2\geq v^2(3u^2v^4-4v^6-4u^3w^3+6uv^2w^3-w^6)$$ or $$(3u^2+v^2)w^6+2uv^2(2u^2-9uv+3v^2)+4v^6(6u^2-9uv+4v^2)\geq0,$$ which is obvious for $2u^2-9uv+3v^2\geq0.$
But for $2u^2-9uv+3v^2\leq0$ or $2t^2-9t+3\leq0$ it's enough to prove that: $$t^2(2t^2-9t+3)^2-4(3t^2+1)(6t^2-9t+4)\leq0$$ or $$(t-1)^2(16-4t+39t^2+28t^3-4t^4)\geq0,$$ which is true because $$16-4t+39t^2+28t^3-4t^4\geq16-4t+45t^2+10t^3+2t^2(-2t^2+9t-3)\geq0.$$