Hi, Assume $x=a/b$ $y=b/c$ $z=c/a$ the inequality becomes $\sum \frac{a^{11} b c^{20}}{a^{12} c^{20} +a^{11}b^{21} + c^{31} b} \le 1$

Could anyone please compute what i wrote and paste it below?

Here is a solution Applying Cauchy ${\frac{1}{x+y^{20}+z^{11}}}$ $\leq$ $ \frac{(x^\frac{37}{3}+\frac{1}{y^\frac{20}{3}}+z^\frac{7}{3})}{(x^\frac{20}{3}+ y^\frac{20}{3}+ z^\frac{20}{3})^2}$ After summing similar 3 inequality and clean equal sum , we need to prove $ \sum_{sym}x^\frac{37}{3}+\sum_{sym}x^\frac{7}{3} \leq \sum_{sym}x^\frac{40}{3}+ \sum_{sym}(xy)^\frac{20}{3}$ Because of $xyz=1$ , $ \sum_{sym}x^\frac{37}{3}= \sum_{sym}x^\frac{38}{3}y^\frac{1}{3}z^\frac{1}{3}$ and $ \sum_{sym}x^\frac{7}{3}=\sum_{sym}x^\frac{18}{3}y^\frac{11}{3}z^\frac{11}{3}$ By Muirhead Inequality $ \sum_{sym}x^\frac{40}{3} \geq \sum_{sym}x^\frac{38}{3}y^\frac{1}{3}z^\frac{1}{3}$ and $\sum_{sym}(xy)^\frac{20}{3}\geq\sum_{sym}x^\frac{18}{3}y^\frac{11}{3}z^\frac{11}{3}$ So inequality is true

From Cauchy ; $(x+y^{20}+z^{11})(x^{13}+(xz)^6+z^3)\geq(x^7+y^7+z^7)^2$ and it means ; $\frac {1}{x+y^{20}+z^{11}}\leq \frac {x^{13}+(xz)^6+z^3}{(x^7+y^7+z^7)^2}$ From $xyz=1$ we can write $x^{13}=x^{\frac{40}{3}}y^{\frac {1}{3}}z^{\frac{1}{3}}$ Change $(xz)^6$ and $z^3$ similarly. Inequality becomes $\frac {1}{x+y^{20}+z^{11}}\leq \frac {x^{\frac{40}{3}}y^{\frac {1}{3}}z^{\frac{1}{3}}+x^{\frac {20}{3}}y^{\frac {2}{3}}z^{\frac {20}{3}}+x^{\frac {11}{3}}y^{\frac {11}{3}}z^{\frac {20}{3}}}{(x^7+y^7+z^7)^2}$ Let's write it for $y$ and $z$. Now sum them all we get $\sum \frac {1}{x+y^{20}+z^{11}}\leq \frac{\sum x^{\frac{40}{3}}y^{\frac {1}{3}}z^{\frac{1}{3}}+x^{\frac {20}{3}}y^{\frac {2}{3}}z^{\frac {20}{3}}+x^{\frac {11}{3}}y^{\frac {11}{3}}z^{\frac {20}{3}}}{(x^7+y^7+z^7)^2}$ if we prove $\frac{\sum x^{\frac{40}{3}}y^{\frac {1}{3}}z^{\frac{1}{3}}+x^{\frac {20}{3}}y^{\frac {2}{3}}z^{\frac {20}{3}}+x^{\frac {11}{3}}y^{\frac {11}{3}}z^{\frac {20}{3}}}{(x^7+y^7+z^7)^2} \leq 1 $ The whole inequality will be proved. Just use Muirhead for prove it.

emregirgin35 wrote: From Cauchy ; $(x+y^{20}+z^{11})(x^{13}+(xz)^6+z^3)\geq(x^7+y^7+z^7)^2$ and it means ; $\frac {1}{x+y^{20}+z^{11}}\leq \frac {x^{13}+(xz)^6+z^3}{(x^7+y^7+z^7)^2}$ From $xyz=1$ we can write $x^{13}=x^{\frac{40}{3}}y^{\frac {1}{3}}z^{\frac{1}{3}}$ Change $(xz)^6$ and $z^3$ similarly. Inequality becomes $\frac {1}{x+y^{20}+z^{11}}\leq \frac {x^{\frac{40}{3}}y^{\frac {1}{3}}z^{\frac{1}{3}}+x^{\frac {20}{3}}y^{\frac {2}{3}}z^{\frac {20}{3}}+x^{\frac {11}{3}}y^{\frac {11}{3}}z^{\frac {20}{3}}}{(x^7+y^7+z^7)^2}$ Let's write it for $y$ and $z$. Now sum them all we get $\sum \frac {1}{x+y^{20}+z^{11}}\leq \frac{\sum x^{\frac{40}{3}}y^{\frac {1}{3}}z^{\frac{1}{3}}+x^{\frac {20}{3}}y^{\frac {2}{3}}z^{\frac {20}{3}}+x^{\frac {11}{3}}y^{\frac {11}{3}}z^{\frac {20}{3}}}{(x^7+y^7+z^7)^2}$ if we prove $\frac{\sum x^{\frac{40}{3}}y^{\frac {1}{3}}z^{\frac{1}{3}}+x^{\frac {20}{3}}y^{\frac {2}{3}}z^{\frac {20}{3}}+x^{\frac {11}{3}}y^{\frac {11}{3}}z^{\frac {20}{3}}}{(x^7+y^7+z^7)^2} \leq 1 $ The whole inequality will be proved. Just use Muirhead for prove it. I can rewrite this solution as follows: Using Cauchy-Schwarz: $(x+y^{20}+z^{11})(x^{13}+y^{-6}+z^3)\ge (x^7+y^7+z^7)^2$ Thus $\sum \frac{1}{x+y^{20}+z^{11}}\le \sum \frac{\sum x^{13}+\sum x^{-6}+\sum x^3}{(x^7+y^7+z^7)^2}$ And now, note that: 1, $x^{14}+y^{14}+z^{14}\ge x^{13}+y^{13}+z^{13}$ 2, $\sum_{sym}x^7y^7\ge \sum_{sym}x^6y^6\ge \sum_{sym}x^6y^3z^3=\sum_{sym}x^3$ From all of above inequalitys we have the our desire!

a cauchy and several simple muirhead and the proof is ready