Alternative proofs of $a^2+b^2+c^2 \geqslant \frac{9abc}{a+b+c}+2(1+\sqrt 2)(a-b)(b-c)$

Question

Let $a,b$ and $c$ be positive real numbers. Prove that $$a^2+b^2+c^2 \geqslant \frac{9abc}{a+b+c}+2(1+\sqrt 2)(a-b)(b-c).$$

I want to see nice proofs.

My proof below is not nice.

Indeed, we need to prove the inequality where $(a−b)(b−c)⩾0.$

If $a⩾b⩾c$ then $f(x+y+c,y+c,c)⩾0.$ And if $a⩽b⩽c$ then $f(a,y+a,x+y+a)⩾0$ where $f(a,b,c)=LHS-RHS.$

Answer 1

Let $a\leq b\leq c$. Thus, $$(a+b+c)(a-b)(b-c)\leq(b+c)b(c-b)$$ it's $$a(a+c)\geq0.$$ Also, $$(a^2+b^2+c^2)(a+b+c)-9abc=\sum_{cyc}(a^3+a^2b+a^2c-3abc)=$$ $$=\sum_{cyc}(a^3-abc)+\sum_{cyc}(a^2b+a^2c-2abc)=$$ $$=\sum_{cyc}(a-b)^2\left(\frac{1}{2}(a+b+c)+c\right)=\frac{1}{2}\sum_{cyc}(a-b)^2(a+b+3c).$$ Now, if we'll replace $(a,b,c)$ on $(a-t,b-t,c-t)$,

where $0\leq t\leq a$ so the expression $\frac{1}{2}\sum\limits_{cyc}(a-b)^2(a+b+3c)$ decreases (because $(a-b)^2$ is not changes and $a+b+3c$ decreases), which says that it's enough to prove our inequality for $a=0$, which gives $$((1+\sqrt2)b-c)^2\geq0.$$ The case $a\geq b\geq c$ is the same.

Answer 2

Thank @Michael Rozenberg. Another way:

Let $f(a,b,c) := (a+b+c)(\mathrm{LHS} - \mathrm{RHS})$. Then $$f(a,b,c)-f(0,b-a,c-a) = a \cdot F(c),$$ with $$ F(c) = \frac{\left[710c+71(1+6 \sqrt 2)a-71(6 \sqrt 2+11)b\right]^2}{100820}+\frac{3(9 - 4 \sqrt 2)(a-b)^2}{20}\geqslant 0.$$