Amin12 wrote: Let $a,b$ and $c$ be positive real numbers. Prove that $$\sum_{cyc} \frac {a^3b}{(3a+2b)^3} \ge \sum_{cyc} \frac {a^2bc}{(2a+2b+c)^3} $$ Prove inquality in below for positive $a , b , c .$ $${\frac{(a+b+c)^2}{c(3a+2b)^3+b(3c+2a)^3+a(3b+2c)^3}}\geq{ \frac{a}{(2a + 2b + c)^3}+\frac{b}{(2b + 2c + a)^3}+\frac{c}{(2c + 2a + b)^3}}$$

Any solution?

Any ideas?

sqing wrote: Amin12 wrote: Let $a,b$ and $c$ be positive real numbers. Prove that $$\sum_{cyc} \frac {a^3b}{(3a+2b)^3} \ge \sum_{cyc} \frac {a^2bc}{(2a+2b+c)^3} $$ Prove inquality in below for positive $a , b , c .$ $${\frac{(a+b+c)^2}{c(3a+2b)^3+b(3c+2a)^3+a(3b+2c)^3}}\geq{ \frac{a}{(2a + 2b + c)^3}+\frac{b}{(2b + 2c + a)^3}+\frac{c}{(2c + 2a + b)^3}}$$ Can you give a proof to this inequality? Thanks

bumping ...

Amin12 wrote: Let $a,b$ and $c$ be positive real numbers. Prove that $$\sum_{cyc} \frac {a^3b}{(3a+2b)^3} \ge \sum_{cyc} \frac {a^2bc}{(2a+2b+c)^3} $$ I think the following is much more better. https://artofproblemsolving.com/community/c6h1722836p11264633

sqing wrote: Amin12 wrote: Let $a,b$ and $c$ be positive real numbers. Prove that $$\sum_{cyc} \frac {a^3b}{(3a+2b)^3} \ge \sum_{cyc} \frac {a^2bc}{(2a+2b+c)^3} $$ Prove inquality in below for positive $a , b , c .$ $${\frac{(a+b+c)^2}{c(3a+2b)^3+b(3c+2a)^3+a(3b+2c)^3}}\geq{ \frac{a}{(2a + 2b + c)^3}+\frac{b}{(2b + 2c + a)^3}+\frac{c}{(2c + 2a + b)^3}}$$ Bump for this for $4$ years...