Taking $a=\frac{x}{y},\ b=\frac{z}{x},\ c=\frac{y}{z}$ we have by C-S and Vasc (in cyclic sums) $$\sum\frac{a}{1+b^3}=\sum\frac{x^4}{y\left(x^3+z^3\right)}\ge\frac{\left(\sum x^2\right)^2}{\sum x^3y+\sum z^3y}\ge\frac{3}{2}$$

AM-GM also works by writing $\frac{a}{1+b^3}+\frac{b+1}{2}+(b^2-b+1)+2a^3\geq 4a$ but cauchy is always more beautiful and more fun

arshiya381 wrote: AM-GM also works by writing $\frac{a}{1+b^3}+\frac{b+1}{2}+(b^2-b+1)+2a^3\geq 4a$ but cauchy is always more beautiful and more fun how do you finish your proof?

Pinko wrote: For all real numbers $a,b,c>0$ such that $abc=1$, prove that $$\frac{a}{1+b^3}+\frac{b}{1+c^3}+\frac{c}{1+a^3}\geq \frac{3}{2}$$ For all real numbers $a,b,c>0 $, prove that $$\frac{a}{1+a^3}+\frac{b}{1+b^3}+\frac{c}{1+c^3}\leq \frac{5}{4}+ \frac{1}{4abc}$$For all real numbers $a,b,c>0$ such that $abc=1$, prove that $$\frac{a}{2+a^4}+\frac{b}{2+b^4}+\frac{c}{2+c^4}\leq1$$$$\frac{a}{1+2b+a^3}+\frac{b}{1+2c+b^3}+\frac{c}{1+2a+c^3}\leq \frac{3}{4}$$$$\frac{a}{1+a^3}+\frac{b}{1+b^3}+\frac{c}{1+c^3}\leq \frac{3}{2}$$p/6452054383

For all real numbers $a,b,c>0$ . Prove that $$\frac{a}{1+a^3}+\frac{b}{1+b^3}+\frac{c}{1+c^3}\leq \frac{a}{a+b^2}+\frac{b}{b+c^2}+\frac{c}{c+a^2}$$$$\frac{a}{1+a^3}+\frac{b}{1+b^3}+\frac{c}{1+c^3}\leq \frac{1}{4abc}+\frac{5}{4}$$$$\frac{a}{(1+a)^3}+\frac{b}{(1+b)^3}+\frac{c}{(1+c)^3}\leq \frac{3}{8}$$(Jichen) p/6456362095

sqing wrote: For all real numbers $a,b,c>0$ such that $abc=1$, prove that $$\frac{a}{2+a^4}+\frac{b}{2+b^4}+\frac{c}{2+c^4}\leq1$$$$\frac{a}{1+a^3}+\frac{b}{1+b^3}+\frac{c}{1+c^3}\leq \frac{3}{2}$$ For all real numbers $a,b,c,d>0$ such that $abcd=1$, prove that $$\frac{a}{3+a^3}+\frac{b}{3+b^3}+\frac{c}{3+c^3}+\frac{d}{3+d^3}\leq1$$

sqing wrote: For all real numbers $a,b,c>0$ such that $abc=1$, prove that $$\frac{a}{1+a^3}+\frac{b}{1+b^3}+\frac{c}{1+c^3}\leq \frac{3}{2}$$ For all real numbers $a,b,c>0$ such that $abc=1$, prove that $$\frac{1}{a+3}+\frac{1}{b+3}+\frac{1}{c+3}\leq \frac{3}{4}$$$$\frac{1}{a+5}+\frac{1}{b+5}+\frac{1}{c+5}\leq \frac{1}{2}$$

Pinko wrote: For all real numbers $a,b,c>0$ such that $abc=1$, prove that $$\frac{a}{1+b^3}+\frac{b}{1+c^3}+\frac{c}{1+a^3}\geq \frac{3}{2}$$ 2015 International Festival of Young Mathematicians Sozopol For all real numbers $a,b,c>0$ such that $abc=1$, prove that $$\frac{a}{1+b^3}+\frac{b}{1+c^3}+\frac{c}{1+a^3}\geq \frac{3}{2}\geq\frac{a}{1+a^3}+\frac{b}{1+b^3}+\frac{c}{1+c^3}$$$$\frac{a^2}{1+b^3}+\frac{b^2}{1+c^3}+\frac{c^2}{1+a^3}\geq \frac{3}{2}\geq \frac{a^2}{1+a^3}+\frac{b^2}{1+b^3}+\frac{c^2}{1+c^3}$$

Victoria_Discalceata1 wrote: Taking $a=\frac{x}{y},\ b=\frac{z}{x},\ c=\frac{y}{z}$ we have by C-S and Vasc (in cyclic sums) $$\sum\frac{a}{1+b^3}=\sum\frac{x^4}{y\left(x^3+z^3\right)}\ge\frac{\left(\sum x^2\right)^2}{\sum x^3y+\sum z^3y}\ge\frac{3}{2}$$ After expanding the last inequality, I got $2(x^4+y^4+z^4+2x^2y^2+2y^2z^2+2z^2x^2) \ge 3(x^3y+xy^3+y^3z+yz^3+z^3x+zx^3)$ which doesn't seem to hold by Muirhead.

Use the Vasc's inequality: $$(x^2+y^2+z^2)^2\geq3(x^3y+y^3z+z^3x).$$

sqing wrote: For all real numbers $a,b,c,d>0$ such that $abcd=1$, prove that $$\frac{a}{3+a^3}+\frac{b}{3+b^3}+\frac{c}{3+c^3}+\frac{d}{3+d^3}\leq1$$ For all real numbers $a,b,c,d>0$ such that $abcd=1$, prove that $$\frac{a}{1+a^3}+\frac{b}{1+b^3}+\frac{c}{1+c^3}+\frac{d}{1+d^3}\leq 2$$ Pinko wrote: For all real numbers $a,b,c>0$ such that $abc=1$, prove that $$\frac{a}{1+b^3}+\frac{b}{1+c^3}+\frac{c}{1+a^3}\geq \frac{3}{2}$$ For all real numbers $a,b,c>0$ such that $abc=1.$ Prove that $$\frac{a}{1+b^2}+ \frac{b}{1+c^2}+\frac{c}{1+a^2}\geq\frac{3}{2}$$here

sqing wrote: For all real numbers $a,b,c>0$ . Prove that $$\frac{a}{(1+a)^3}+\frac{b}{(1+b)^3}+\frac{c}{(1+c)^3}\leq \frac{3}{8}$$ Let $a,b,c>0 : abc=1$. Prove that: $$\dfrac{a}{(a+3)^2}+\dfrac{b}{(b+3)^2}+\dfrac{c}{(c+3)^2} \le \dfrac{3}{16}$$

Let $a,b,c>0 $ and $ abc=1$. Prove that $$\dfrac{a}{(a+2)^2}+\dfrac{b}{(b+2)^2}+\dfrac{c}{(c+2)^2} \le \dfrac{1}{3}$$$$\dfrac{a}{(a+\frac{13}{4})^2}+\dfrac{b}{(b+\frac{13}{4})^2}+\dfrac{c}{(c+\frac{13}{4})^2} \le \dfrac{48}{289}$$

sqing wrote: Let $a,b,c>0 $ and $ abc=1$. Prove that $$\dfrac{a}{(a+\frac{13}{4})^2}+\dfrac{b}{(b+\frac{13}{4})^2}+\dfrac{c}{(c+\frac{13}{4})^2} \le \dfrac{48}{289}$$ We can replace $\frac{13}{4}$ on $3.29$. The proof is the same to this proof. For $3.3$ it's wrong already.

We can replace $\frac{13}{4}$ on $3.2986$.