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

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.

• The question seems to be in spirit of AM-GM and somewhat partial fractions. Try manipulation with those. Commented Sep 23, 2023 at 15:20
• Can you show full the ans, I can't understand it! Thanks! Commented Sep 23, 2023 at 15:31
• Isn't there a misprint in the numerator of the LHS of $(2)?$ shouldn't it be $a+b$ instead of $a+1?$ Commented Sep 23, 2023 at 16:28
• Oh I fixed it!! Commented Sep 24, 2023 at 1:07
• It seems (2) is not true for $a = 1, b = 3/2, c = 1/2$? Commented Sep 24, 2023 at 9:45

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 $$\color{red}{a+b+c=3\iff a=3-b-c\iff b=3-a-c\iff c=3-b-a}$$.
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