Let a,b,c >0 satisfy :a+b+c=3. Find Max P, P= $\frac{2}{3+ab+bc+ca}+\sqrt[3]{\frac{abc}{(1+a)(1+b)(1+c)}}$
I supposed that a=b=c. So that: $\sqrt[3]{\frac{abc}{(1+a)(1+b)(1+c)}}\leq \frac{1}{3}(\frac{a}{1+a}+\frac{b}{1+b}+\frac{c}{1+c})$
Then, I proved that : $\frac{a}{1+a} \leq \frac{1}{4}(a-1)+\frac{1}{2}$ so $P\leq \frac{2}{3+ab+bc+ca}+\frac{1}{2}$. After that, I used $t=ab+bc+ca \in (0;3]$. I had $f(t)=\frac{2}{3+t}+\frac{1}{2}, t\in(0;3]$. Its derivative is $\frac{-2}{({3+t})^2}<0$. From that step i can't find the maximum value of P. Can you help me to find the problem in my solution? If you have another solutions of this exercise, please show the idea to me! Thank you very much.
For $a=b=c=1$ we get $P=\frac{5}{6}.$
We'll prove that it's a maximal value.
Indeed, let $a+b+c=3u$, $ab+ac+bc=3v^2$ and $abc=w^3$.
Thus, we need to prove that $$\frac{2u^2}{3u^2+3v^2}+\sqrt[3]{\frac{w^3}{w^3+3uv^2+4u^3}}\leq\frac{5}{6}.$$ But the expression $$\frac{2u^2}{3u^2+3v^2}+\sqrt[3]{\frac{w^3}{w^3+3uv^2+4u^3}}=\frac{2u^2}{3u^2+3v^2}+\sqrt[3]{1-\frac{3uv^2+4u^3}{w^3+3uv^2+4u^3}}$$ increases as a function of $w^3$, which says that it's enough to prove our inequality for a maximal value of $w^3$, which happens for equality case of two variables. Let $b=a$ and $c=3-2a$, where $0<a<1.5$.
Thus, we need to prove that: $$\frac{2}{3+a^2+2a(3-2a)}+\sqrt[3]{\frac{a^2(3-2a)}{(1+a)^2(4-2a)}}\leq\frac{5}{6}$$ or
$$(a-1)^2(2+67a+468a^2+1395a^3-78a^4-1371a^5+688a^6-91a^7)\geq0,$$ which is obviously true for any $a\in\left(0,\frac{3}{2}\right).$
Also, we can prove the last inequality by the following way.
We need to prove that $$\sqrt[3]{\frac{a^2(3-2a)}{(1+a)^2(4-2a)}}\leq\frac{5}{6}-\frac{2}{3+a^2+2a(3-2a)}$$ or $$\sqrt[3]{\frac{a^2(3-2a)}{(1+a)^2(4-2a)}}\leq\frac{1+10a-5a^2}{6(1+2a-a^2)}$$ or $f(a)\geq0,$ where $$f(a)=\ln(1+10a-5a^2)+\frac{2}{3}\ln(1+a)+\frac{1}{3}\ln(2-a)+\frac{1}{3}\ln2-$$ $$-\ln6-\ln(1+2a-a^2)-\frac{2}{3}\ln{a}-\frac{1}{3}\ln(3-2a).$$ Now, $$f'(a)=\frac{10-10a}{1+10a-5a^2}+\frac{2}{3(1+a)}-\frac{1}{3(2-a)}-\frac{2-2a}{1+2a-a^2}-\frac{2}{3a}+\frac{2}{3(3-2a)}=$$ $$=\frac{10(1-a)}{1+10a-5a^2}+\frac{1-a}{(1+a)(2-a)}-\frac{2(1-a)}{1+2a-a^2}+\frac{2(a-1)}{a(3-2a)}=$$ $$=(a-1)\left(\frac{2}{a(3-2a)}-\frac{1}{(1+a)(2-a)}-\frac{8}{(1+10a-5a^2)(1+2a-a^2)}\right).$$ Now, since $$\frac{2}{a(3-2a)}-\frac{1}{(1+a)(2-a)}-\frac{8}{(1+10a-5a^2)(1+2a-a^2)}>0$$ for any $a\in\left(0,\frac{3}{2}\right),$ we see that $a_{\min}=1$ and we are done!
By the way, after your first step and $b=c\rightarrow0^+$ we get that $P$ is closed to a value, which is greater than $\frac{5}{6}.$
Id est, your first step is not so good.