Proving the inequality $\sum_{\text{cyc}} \frac{a}{a^2+b^3+c^3}\le\frac1{5abc}$

Question

Let $a,b,c> 0$ be three real numbers such that $a+b+c=1$. I want to prove that

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

My attempt: Using AM-GM on each denominator gives (here, LHS denotes the left-hand side) $$LHS\le\frac{1}{3a^{\frac12}bc}+\frac{1}{3b^{\frac12}ac}+\frac{1}{3c^{\frac12}ab}.$$

However, I think that my attempt doesn't work because the last expression can get larger than $\frac{1}{5abc}$. In fact, if we multiply with $3abc$ then the original inequality is: $$\sqrt a+\sqrt b+\sqrt c\le\frac35$$

However, as $a,b,c<1$, $$\sqrt a +\sqrt b + \sqrt c> a+b+c=1>\frac35.$$ So my upper bound is always larger than the given one.

Answer 1

Multiply both sides with $abc$ to get that your inequality is equivalent to

\begin{equation}\tag 1\label 1\sum_{\text{cyc}} \frac{a^2bc}{a^2+b^3+c^3}\le\frac15.\end{equation}

Now note that, since $a+b+c=1$, we have $$a^2+b^3+c^3=a^2\cdot(a+b+c)+b^3+c^3=a^3+b^3+c^3+a^2b+a^2c.$$

Hence, by AM-GM inequality for 5 variables, $$a^2+b^3+c^3\geq5\sqrt[5]{a^7b^4c^4}.$$

Hence, \begin{split}\sum_{\text{cyc}} \frac{a^2bc}{a^2+b^3+c^3}&\le\frac15\sum_{\text{cyc}} \frac{a^2bc}{\sqrt[5]{a^7b^4c^4}}\\&=\frac15\sum_{\text{cyc}} \sqrt[5]{a^3bc}\\&\overset{\text{AM-GM}}\le\frac1{25}\sum_{\text{cyc}}3a+b+c\\&=\frac{5(a+b+c)}{25}\\&\overset{a+b+c=1}=\frac15. \end{split}

Done!

Answer 2

Also, by AM-GM and C-S we obtain: $$\sum_{cyc}\frac{a}{a^2+b^3+c^3}=\sum_{cyc}\frac{a}{a^3+b^3+c^3+a^2b+a^2c}\leq\sum_{cyc}\frac{a}{3abc+a^2b+a^2c}=$$ $$=\sum_{cyc}\frac{1}{3bc+ab+ac}\leq\frac{1}{(3+1+1)^2}\sum_{cyc}\left(\frac{3^2}{3bc}+\frac{1^2}{ab}+\frac{1^2}{ac}\right)=\frac{1}{5}\sum_{cyc}\frac{1}{ab}=\frac{1}{5abc}.$$