Prove the given inequality if $a+b+c=1$

Question

Let $a,b,c$ be positive real numbers such that $a+b+c=1$, then prove that

$\frac{a}{a^2 +b^3 +c^3}+\frac{b}{a^3 +b^2 +c^3}+\frac{c}{a^3 +b^3 +c^2} \leq \frac{1}{5abc}$

Please provide some hint to proceed. I have used Arithmetic Mean-Geometric Mean inequality to proceed in such questions, but I am not getting how to use it if it is to be used. Kindly provide some directon.

Usually the questions I have dealt so far were like if $a+b+c=1$, find maximum value of $a^3b^4c^2$ which I find using weighted A.M-G.M, but here I am not able to gather the thoughts to proceed

Answer 1

We need to prove that: $$\sum_{cyc}\frac{a}{a^3+b^3+c^3+a^2b+a^2c}\leq\frac{a+b+c}{5abc}$$ or $$\sum_{cyc}\left(\frac{a}{5abc}-\frac{a}{a^3+b^3+c^3+a^2b+a^2c}\right)\geq0$$ or $$\sum_{cyc}\frac{a(a^3+b^3+c^3+a^2b+a^2c-5abc)}{a^3+b^3+c^3+a^2b+a^2c}\geq0$$ or $$\sum_{cyc}\frac{a((a-b)(a^2+5ac-2ab-2b^2)-(c-a)(a^2+5ab-2ac-2c^2))}{a^3+b^3+c^3+a^2b+a^2c}\geq0$$ or $$\sum_{cyc}(a-b)\left(\frac{a^3+5a^2c-2a^2b-2b^2a}{a^3+b^3+c^3+a^2b+a^2c}-\frac{ b^3+5b^2c-2b^2a-2a^2b}{a^3+b^3+c^3+b^2c+b^2a}\right)\geq0$$ or $$\sum_{cyc}\tfrac{(a-b)^2(a^5+a^4b+5a^4b^2+4a^2b^3+ab^4+b^5+(5a^4+7a^3b+10a^2b^2+7ab^3+5b^4)c+(a^2+ab+b^2)c^3+5(a+b)c^4)}{(a^3+b^3+c^3+a^2b+a^2c)(a^3+b^3+c^3+b^2c+b^2a)}\geq0.$$ Done!

Answer 2

Also, by AM-GM $$a^3+b^3+c^3+a^2b+a^2c\geq a^3+4\sqrt[4]{b^3\cdot c^3\cdot a^2b\cdot a^2c}=a^3+4abc.$$ Thus, it's enough to prove that: $$\sum_{cyc}\frac{1}{a^2+4bc}\leq\frac{a+b+c}{5abc}$$ or $$\sum_{cyc}\left(\frac{1}{a^2+4bc}-\frac{1}{4bc}\right)\leq\frac{a+b+c}{5abc}-\frac{a+b+c}{4abc}$$ or $$\sum_{cyc}\frac{a^2}{bc(a^2+4bc)}\geq\frac{a+b+c}{5abc}$$ or $$\sum_{cyc}\frac{a^3}{a^2+4bc}\geq\frac{a+b+c}{5},$$ which is true by Holder: $$\sum_{cyc}\frac{a^3}{a^2+4bc}\geq\frac{(a+b+c)^3}{3\sum\limits_{cyc}(a^2+4ab)}\geq\frac{(a+b+c)^3}{3\sum\limits_{cyc}\frac {5}{3}(a^2+2ab)}=\frac{a+b+c}{5}.$$