For $a,b,c>0$ and $a^3+b^3+c^3=3$ prove that $$\frac{a}{b+2}+\frac{b}{c+2}+\frac{c}{a+2}\le 1$$
We have:$$a^3+1+1+b^3+1+1+c^3+1+1\ge 3\left(a+b+c\right)$$
$$\Rightarrow 3\left(a+b+c\right)\le a^3+b^3+c^3+6\le 9\Rightarrow a+b+c\le 3$$
Use AM-GM:
$$\frac{a}{b+2}=\frac{1}{2}\left(a-\frac{ab}{b+2}\right)\ge \frac{1}{2}\left(a-\frac{ab}{\sqrt{2b}}\right)=\frac{1}{2}\left(a-\frac{a\sqrt{b}}{\sqrt{2}}\right)$$
$$\Rightarrow LHS\le \frac{1}{2}\left(a+b+c-\frac{a\sqrt{b}+b\sqrt{c}+c\sqrt{a}}{\sqrt{2}}\right)\le 1$$
Need prove:$a\sqrt{b}+b\sqrt{c}+c\sqrt{a}\ge \sqrt{2}$
Help me prove that last inequality
We need to prove that $$8+abc\geq2(a^2+b^2+c^2)+a^2c+b^2a+c^2b.$$ By Holder $$27=(1+1+1)(a^3+b^3+c^3)^2\geq(a^2+b^2+c^2)^3,$$ which says that $a^2+b^2+c^2\leq3$. Thus, it's enough to prove that $$2+abc\geq a^2c+b^2a+c^2b$$ and since $3\geq a^2+b^2+c^2,$ it's enough to prove that $$2\left(\frac{a^2+b^2+c^2}{3}\right)^{\frac{3}{2}}+abc\geq a^2c+b^2a+c^2b.$$ But the last inequality is homogeneous, which says that we can assume $a^2+b^2+c^2=3$ and we need to prove that $2+abc\geq a^2c+b^2a+c^2b.$
Now, let $\{a,b,c\}=\{x,y,z\}$ such that $x\geq y\geq z$.
Hence, by Rearrangement and AM-GM we obtain: $$a^2c+b^2a+c^2b-abc=a\cdot{ac}+b\cdot{ba}+c\cdot{cb}-xyz\leq$$ $$\leq x\cdot{xy}+y\cdot{xz}+z\cdot{yz}-xyz=y(x^2+z^2)=y(3-y^2)\leq2$$ and we are done!