Prove that if $a,b,c > 0$ and $a + b + c = 1$, we have: $\frac{a^2}{a^3 + 5} + \frac{b^2}{b^3 + 5} + \frac{c^2}{c^3 + 5} \leq \frac{1}{4}$

Question

Prove that if $a,b,c > 0$ such that $a + b + c = 1$, then the following inequality holds: $$S = \frac{a^2}{a^3 + 5} + \frac{b^2}{b^3 + 5} + \frac{c^2}{c^3 + 5} \leq \frac{1}{4}$$

What I have tried so far is the following: Firstly I rewrote $S$ as: $$ S = \frac{1}{a} + \frac{1}{b} + \frac{1}{c} - 5\left[ \frac{1}{a(a^3 + 5)} + \frac{1}{b(b^3 + 5)} + \frac{1}{c(c^3 + 5)}\right]$$ Then, in order to upper bound $S$, I used the inequality: $ \frac{x^2}{u} + \frac{y^2}{v} + \frac{z^2}{w} \geq \frac{(x + y + z)^2}{u + v + w} $ for any $u,v,w > 0$. Therefore, I got: $$ S \leq \frac{1}{a} + \frac{1}{b} + \frac{1}{c} - 5 \cdot \frac{(1 + 1 + 1)^2}{a^4 + b^4 + c^4 + 5(a + b + c)} = \frac{45}{a^4 + b^4 + c^4 + 5}$$ Then, given what we want to show about $S$, this would reduce to proving that: $$ \frac{1}{a} + \frac{1}{b} + \frac{1}{c} \leq \frac{185 + a^4 + b^4 + c^4}{4(a^4 + b^4 + c^4 + 5)} $$ At which point I got stuck and I am not sure whether I started the right way.

I would be grateful for any suggestions. Many thanks!

Answer 1

Because by AM-GM we have: $$\frac{a^2}{a^3+5}\leq\frac{a}{4}$$ $$a^3+2\cdot2.5\geq3\sqrt[3]{a^3\cdot2.5^2}>4a.$$

Answer 2

Based on Michael Rozenberg's idea, we also have:

$$\frac{a^2}{a^3+5}=\frac{a}{6}-\frac{a(1-a)(5-a^2-a)}{a^3+5}\leq \frac{a}{6}$$

so that:

$$\frac{a^2}{a^3+5}+\frac{b^2}{b^3+5}+\frac{c^2}{c^3+5}\leq \frac{1}{6}(a+b+c)=\frac{1}{6}$$

Equality is attained when $(a,b,c)=(1,0,0)$ up to any cyclic permutation.

We can use the tangent line method to find the minimum:

$$ \begin{aligned} \frac{a^2}{a^3+5}&=\frac{3}{136}+\frac{2421}{18496}\left(a-\frac{1}{3}\right)+\frac{(3a-1)^2(1995-135a-269a^2)}{18496(a^3+5)}\\ &\geq \frac{3}{136}+\frac{2421}{18496}\left(a-\frac{1}{3}\right) \end{aligned} $$

so that the minimum value is $\frac{9}{136}$ when $a=b=c=\frac{1}{3}$. Therefore the entire range of the left hand side is $\left[\frac{9}{136},\frac{1}{6}\right]$.