Let $a, b, c$ be positive reals with $abc = 1$. Prove the inequality $$\frac{a^5}{a^3 + 1}+\frac{b^5}{b^3 + 1}+\frac{c^5}{c^3 + 1} \ge \frac32$$and determine all values of a, b, c for which equality is attained
Problem
Source: Thailand Mathematical Olympiad 2015 p2
Tags: algebra, inequalities
16.08.2020 23:10
solved here and here
07.12.2021 19:42
Since $abc=1$, then $$ \frac{a^5}{a^3+1}+\frac{b^5}{b^3+1}+\frac{c^5}{c^3+1}=\frac{a^5}{\sqrt[3]{(abc)^2}\cdot (a^3+abc)}+\frac{b^5}{\sqrt[3]{(abc)^2}\cdot (b^3+abc)}+\frac{c^5}{\sqrt[3]{(abc)^2}\cdot (c^3+abc)}= $$$$ =\frac{1}{\sqrt[3]{(abc)^2}}\cdot \left(\frac{a^5}{a^3+abc}+\frac{b^5}{b^3+abc}+\frac{c^5}{c^3+abc}\right)=\frac{1}{\sqrt[3]{(abc)^2}}\cdot \left(\frac{a^4}{a^2+bc}+\frac{b^4}{b^2+ca}+\frac{c^4}{c^2+ab}\right) $$On one hand, by AM-GM one gets $$ \frac{1}{\sqrt[3]{(abc)^2}}\geq \frac{1}{\frac{a^2+b^2+c^2}{3}}=\frac{3}{a^2+b^2+c^2} $$On another hand, by Cauchy Schwartz in its Engel form, it holds $$ \frac{a^4}{a^2+bc}+\frac{b^4}{b^2+ca}+\frac{c^4}{c^2+ab}\geq \frac{(a^2+b^2+c^2)^2}{a^2+b^2+c^2+ab+bc+ca} $$Therefore, $$ \frac{1}{\sqrt[3]{(abc)^2}}\cdot \left(\frac{a^4}{a^2+bc}+\frac{b^4}{b^2+ca}+\frac{c^4}{c^2+ab}\right)\geq \frac{3\cdot (a^2+b^2+c^2)}{a^2+b^2+c^2+ab+bc+ca} $$Finally, $$ \frac{3\cdot (a^2+b^2+c^2)}{a^2+b^2+c^2+ab+bc+ca} \geq \frac{3}{2} \iff a^2+b^2+c^2\geq ab+bc+ca \, , $$which holds by rearrangement.
08.12.2021 05:10
Let $a, b, c$ be positive reals with $abc = 1$. Prove that $$\frac{a}{a^3+b^3+1}+\frac{b}{b^3+c^3+1}+\frac{c}{c^3+a^3+1} \leq 1$$ Let $a,b,c$ be positive reals such that $abc=1.$ Prove that $$\frac{a}{a^2+b^2+1}+\frac{b}{b^2+c^2+1}+\frac{c}{c^2+a^2+1} \leq 1 $$