I came up with the inequality accidentally so there is no original proof so far. It would be great if you can give some useful help to prove it.
Problem. Given non-negative real numbers $a,b,c$ satisfying $a+b+c+abc=4.$ Prove that $$\sqrt{31a+b+c} +\sqrt{31b+a+c} +\sqrt{31c+b+a}\le 3\sqrt{3}\cdot\sqrt{a+b+c+8}. $$
Equality occurs when $a=b=c=1$ or $a=b=2;c=0$ and its permuations.
I've tried to use Cauchy-Schwarz inequality without success.
Indeed,$$\sum_{cyc}\sqrt{31a+b+c}=\sum_{cyc}\sqrt{\frac{31a+b+c}{b+c+xa}\cdot(b+c+xa)}\le \sqrt{\sum_{cyc}\frac{31a+b+c}{b+c+xa}\cdot (x+2)(a+b+c).}$$ We'll prove $$\sum_{cyc}\frac{31a+b+c}{b+c+xa}\le \frac{27}{x+2}\cdot\frac{a+b+c+8}{a+b+c}.$$ Choose $x$ such that equality holds at $a=b=2;c=0$ $$1+2\cdot\frac{64}{2(x+1)}=\frac{81}{x+2}\iff x=7.$$Now, we need to prove $$\color{black}{\sum_{cyc}\frac{31a+b+c}{b+c+7a}\le 3\cdot\frac{a+b+c+8}{a+b+c}}$$ which is not true for $a=b=\dfrac{1}{2}.$
Also, we can rewrite the OP as $$\color{black}{\sqrt{31a+b+c} +\sqrt{31b+a+c} +\sqrt{31c+b+a}\le 3\sqrt{3}\cdot\sqrt{3a+3b+3c+2abc}. }$$
By using Cauchy-Schwarz inequality $$\sum_{cyc}\sqrt{31a+b+c}=\sum_{cyc}\sqrt{\frac{31a+b+c}{b+c+xabc}\cdot(b+c+xabc)}\le \sqrt{\sum_{cyc}\frac{31a+b+c}{b+c+xabc}\cdot 2(a+b+c)+3xabc.}$$ We'll prove $$\sum_{cyc}\frac{31a+b+c}{b+c+xabc}\le 27\cdot\frac{a+b+c+8}{2(a+b+c)+3xabc}.$$ It is not valid for $x$ value for equality occuring $a=b=2;c=0$ since $$1+2\cdot\frac{64}{2}>27\cdot\frac{12}{8}.$$
I truly hope we can find some nice ideas for my inequality. Thank you for sharing.
This is an extremally nice problem!
By C-S $$\sum_{cyc}\sqrt{31a+b+c}\leq\sqrt{\sum_{cyc}\frac{31a+b+c}{3a+2}\sum_{cyc}(3a+2)}$$ and it's enough to prove that: $$\sum_{cyc}\frac{31a+b+c}{3a+2}\leq\frac{9(a+b+c+8)}{a+b+c+2}.$$ Now, let $a+b+c=3u$, $ab+ac+bc=3v^2$ and $abc=w^3$.
Thus, we need to prove a linear inequality of $v^2$, which by $uvw$ says that it's enough to prove the last inequality for equality case of two variables.
Let $b=a$.
Thus, the condition gives $c=\frac{4-2a}{1+a^2}$ and after these substitutions we need to prove that: $$(a-1)^2(2-a)(a^6+18a^5-19a^4-12a^3+24a^2-36a+36)\geq0,$$ which is true because by AM-GM
$$a^6+18a^5-19a^4-12a^3+24a^2-36a+36=$$ $$=a^6+9(a-2)^2+a^2(18a^3-19a^2-12a+15)\geq$$ $$\geq a^2(12(a^3-a^2-a+1)+6a^3-7a^2+3)\geq$$ $$\geq a^2(3a^3+3a^3+3-7a^2)\geq $$ $$\geq a^2\left(3\sqrt[3]{\left(3a^3\right)^2\cdot3}-7a^2\right)=2a^4\geq0$$