parmenides51 wrote: Let $x, m, a, s$ be positive real numbers such that $x, m, a, s < 1$. Prove that $$\frac{x^4}{m^3 + a^2 + s} +\frac{m^4}{a^3 + s^2 + x} + \frac{a^4}{s^3 + x^2 + m} + \frac{s^4}{x^3 + m^2 + a} > \frac{x^3 + m^3 + a^3 + s^3}{3}$$. Because $$\sum_{cyc}\frac{x^4}{m^3 + a^2 + s}>\sum_{cyc}\frac{x^4}{m + a+ s}\geq\frac{\sum\limits_{cyc}x^3}{3}.$$

How to prove \sum_{cyc}\frac{x^4}{m + a+ s}\geq\frac{\sum\limits_{cyc}x^3}{3}.$$ ?

Want-to-study-in-NTU-MATH wrote: How to prove $$\sum_{cyc}\frac{x^4}{m + a+ s}\geq\frac{\sum\limits_{cyc}x^3}{3}.$$? By C-S and Muirhead we obtain: $$\sum_{cyc}\frac{x^4}{m+a+s}=\sum_{cyc}\frac{x^6}{x^2(m+a+s)}\geq\frac{\left(\sum\limits_{cyc}x^3\right)^2}{\sum\limits_{cyc}x^2(m+a+s)}\geq\frac{\sum\limits_{cyc}x^3}{3}.$$

Thanks you very much (what a beautiful proof) Here I provide another proof of the inequality. Lemma: for all x belongs to (0,1), we have x⁴/(1-x)≥x³/3 We can prove it by tangent line method. Now observe the inequality we desired(we want to prove that it is true for all positive numbers x,m,a,s).We can know that it is homogeneous, so we can assume that x+m+a+s=1.Then the inequality is equivalent to \sum_{cyc}\frac{x^4}{1-x}\geq\frac{\sum\limits_{cyc}x^3}{3}.$$ Now we use the lemma above, then we finish the proof. p.s. I also tried using titu's lemma, but I failed because I didn't multiply x²

Want-to-study-in-NTU-MATH wrote: p.s. I also tried using titu's lemma, but I failed because I didn't multiply x² I think, it's better to say "using Bergström''s lemma" because Harald Bergström wrote the C-S in his form before than Titu Andreescu born.