Prove the following inequality $$\frac{3x^5+1}{x^4+x^3+1}+\frac{3y^5+1}{y^4+y^3+1}+\frac{3z^5+1}{z^4+z^3+1} \ge 4$$ where $x,y,z \ge 0$ and $x+y+z=3$
I don't really know how to approach such an inequality. I tried the inequality of means and Cauchy's inequality, but it doesn't seem like I'm getting close to anything. I think that maybe an inequality is being used that I haven't heard of yet.
Edit : I think I managed to prove the inequality, but I'm sure there is a more elegant proof.Here is my take : I will first prove that $\frac{3x^5+1}{x^4+x^3+1} \ge \frac{17}{9}x-\frac{5}{9} \iff \frac{9(3x^5+1)}{9(x^4+x^3+1)}-\frac{17x(x^4+x^3+1)}{9(x^4+x^3+1)} + \frac{5x(x^4+x^3+1)}{9(x^4+x^3+1)} \ge 0 \iff \frac{(x-1)^2(10x^3+8x^2+11x+14)}{9(x^4+x^3+1)} \ge 0$ which is true $\forall x \in [0,\infty)$ So we have that $\frac{3x^5+1}{x^4+x^3+1} \ge \frac{17}{9}x-\frac{5}{9}$ , $\frac{3y^5+1}{y^4+y^3+1} \ge \frac{17}{9}y-\frac{5}{9}$ and $\frac{3z^5+1}{z^4+z^3+1} \ge \frac{17}{9}z-\frac{5}{9}$ . By summing we get that $\frac{3x^5+1}{x^4+x^3+1}+\frac{3y^5+1}{y^4+y^3+1}+\frac{3z^5+1}{z^4+z^3+1} \ge \frac{17}{9}(x+y+z)-\frac{5}{3} = 4$
I'm sure there must be an easier and more elegant method. anyway, I made this solution using an online graph, so I would never have thought of that factorization using pencil and paper.I'm waiting to see what you think about my approach. If you have any idea or a solution, I'm here to listen. Thanks!
Remark: Here is an alternative proof via Chebyshev's sum inequality.
We have \begin{align*} \frac{3x^5+1}{x^4+x^3+1} - \frac43 &= \frac{9x^5 - 4x^4 - 4x^3 - 1}{3(x^4 + x^3 + 1)}\\ &= \frac{(9x^5 - 9x^4) + (4x^4 - 4x^3) + (x^4 - 1)}{3(x^4 + x^3 + 1)}\\ &= \frac{9x^4 + 4x^3 + (x+1)(x^2 + 1)}{3(x^4+x^3+1)}\cdot (x-1). \end{align*}
Since $f(x) := \frac{9x^4 + 4x^3 + (x+1)(x^2 + 1)}{3(x^4+x^3+1)}$ is strictly increasing on $[0, 3]$ (see the Remark at the end), by Chebyshev's sum inequality, we have $$\sum_{\mathrm{cyc}} \frac{3x^5+1}{x^4+x^3+1} - 4 \ge \frac13 \sum_{\mathrm{cyc}} \frac{9x^4 + 4x^3 + (x+1)(x^2 + 1)}{3(x^4+x^3+1)} \cdot (x + y + z - 3) = 0. $$
Remark: We have $$f(x) = 3 + \frac{-4x^3 + x^2 + x - 8}{3x^4 + 3x^3 + 3}.$$ We need to prove that $f'(x) > 0$. Equivalently, we need to prove that \begin{align*} &\Big(-4x^3 + x^2 + x - 8\Big)'\cdot (x^4+x^3+1)\\ >{}& \Big(-4x^3 + x^2 + x - 8\Big)\Big((x^4+x^3+1)\Big)' \end{align*} or $$(-12x^2 + 2x + 1)(x^4 + x^3 + 1) > (-4x^3 + x^2 + x - 8)(4x^3 + 3x^2)$$ or $$4x^6 - 2x^5 - 4x^4 + 30x^3 + 12x^2 + 2x + 1 > 0$$ which is true (easy by AM-GM, e.g. $3x^6 + 12x^2 \ge 12x^4$, etc).