Motivating a solution of $\frac{a+b}{\sqrt{2(a^3+bc)}}+\frac{b+c}{\sqrt{2(b^3+ca)}}+\frac{c+a}{\sqrt{2(c^3+ab)}}≤\frac{a^2+b^2+c^2+21}{8abc}$

Question

The following is the problem I have been working on:

Let $a,b,c>0$ and $a+b+c=3$. Prove that: $$\frac{a+b}{\sqrt{2(a^3+bc)}}+\frac{b+c}{\sqrt{2(b^3+ca)}}+\frac{c+a}{\sqrt{2(c^3+ab)}} \le\frac{a^{2}+b^{2}+c^{2}+21}{8abc}\tag{1}$$

A solution I read just uses one inequality that isn't very intuitive and doesn't come naturally to me

Solution:

Using this inequality: $$\frac{a+b}{\sqrt{2(a^3+bc)}} \leq\frac{a^2+7}{8abc}+\frac{2a-b-c}{24abc}+\frac{2bc-ca-ab}{8abc}\tag{2}$$ We prove $(1)$

I don't understand how inequality $(2)$ is derived to prove the inequality $(1)$. Any help with the derivation of $(2)$ would be appreciated.

Answer 1

Unfortunately, this is not a full answer, but it was too long in the comments. I tried proving the inequality, but I quickly burnt out. Maybe someone could try this with an online calculator like WolframAlpha.

For this problem, it is not AM-GM or partial fractions as it seems initially, but rather, substitution is $\color{red}{key}$.

The simplicity of the problem can be easily exploited when taking advantage of $\color{red}{a+b+c=3\iff a=3-b-c\iff b=3-a-c\iff c=3-b-a}$.

You could trial and error for a decent amount of time with substitutions, but here I will just show how your inequalities can be much reduced.
Substituting into $(2)$, \begin{align}\frac{a+1}{\sqrt{2(a^3+bc)}} &\leq \frac{1}{8}(\frac{a^2+7}{abc})+\frac{2a-b-c}{24abc}+\frac{2bc-ca-ab}{8abc}\\&\leq \frac{a^2+7}{8abc} + \frac{\color{blue}{2a^2-12a-6(c+b)+2ab+2ac+18+b^2+c^2}+4bc-6(b+c)+8-2a}{8abc}\\&\leq \frac{a^2+\color{blue}{b^2+c^2}+4bc-6(b+c)+13-2a}{8abc}\\&\leq \frac{a^2+\color{green}{b^2+c^2+2bc-6(b+c)+9}+4-2a+2bc}{8abc}\\&\leq \frac{a^2+\color{green}{a^2}+4-2a+2bc}{8abc}\\&\leq \frac{a^2-a+bc+2}{4abc}\end{align} With given constraints, it can proven with more substitutions that $(2)$ is true