parmenides51 wrote: Let $a, b, c, d$ be non-negative real numbers satisfying $a + b + c + d = 3$. Prove that $$\frac{a}{1 + 2b^3} + \frac{b}{1 + 2c^3} +\frac{c}{1 + 2d^3} +\frac{d}{1 + 2a^3} \ge \frac{a^2 + b^2 + c^2 + d^2}{3}$$When does the equality hold? Let $$a= 3x , \quad b=3y , \quad c=3z , \quad d = 3w$$ Then, we have $$x+y+z+w = 1 $$ and we need to prove that $$\frac{x}{1+54y^3}+ \frac{y}{1+54z^3}+ \frac{z}{1+54w^3}+ \frac{w}{1+54x^3} \geq x^2+y^2+z^2+w^2$$ Observe that, $$\frac{x}{1+54y^3}+ \frac{y}{1+54z^3}+ \frac{z}{1+54w^3}+ \frac{w}{1+54x^3} = \frac{x^4}{x^3+54x^3y^3}+ \frac{y^4}{1y^3+54y^3z^3}+ \frac{z^4}{z^3+54z^3w^3}+ \frac{w^4}{w^3+54w^3x^3}$$ Now, we apply Cauchy-Schwarz inequality $\frac{x^4}{x^3+54x^3y^3}+ \frac{y^4}{1y^3+54y^3z^3}+ \frac{z^4}{z^3+54z^3w^3}+ \frac{w^4}{w^3+54w^3x^3} \geq $