Great Inequality of Hong Kong!

[Only 1st part of the solution]

Assume that $a=b=c=1$ . Then it's obvious that $k \le 3$ .

Now let's prove that ${ (a+b+c) }^{ 2 }+9\ge 9\sqrt [ 3 ]{ ({ a }^{ 2 }+1)({ b }^{ 2 }+1)({ c }^{ 2 }+1) }$ for all $a,b,c$ and we are done.

We first show the following.

Claim 1 : For any positive real numbers $x, y$ with $xy \ge 1$ , we have

$\large\ \left( { x }^{ 2 } + 1 \right) \left( { y }^{ 2 } + 1 \right) \le { \left( { \left( \frac { x + y }{ 2 } \right) }^{ 2 } + 1 \right) }^{ 2 }$ . $...(2)$ .

Proof : Note that $xy \ge 1$ implies ${ \left( \frac { x + y }{ 2 } \right) }^{ 2 } - 1 \ge xy - 1 \ge 0$ . We find that

$\large\ \left( { x }^{ 2 } + 1 \right) \left( { y }^{ 2 } + 1 \right) = { \left( xy - 1 \right) }^{ 2 } + { \left( x + y \right) }^{ 2 } \le { \left( { \left( \frac { x + y }{ 2 } \right) }^{ 2 } - 1 \right) }^{ 2 } + { \left( x + y \right) }^{ 2 } = { \left( { \left( \frac { x + y }{ 2 } \right) }^{ 2 } + 1 \right) }^{ 2}$ .

WLOG assume that $a \ge b \ge c$ . This implies $a \ge 1$ . Let $d = \frac { a + b + c }{ 3 }$ , Note that

$ad = \frac { a\left( a + b + c \right) }{ 3 } \ge \frac { 1 + 1 + 1 }{ 3 } = 1$ .

Then we can apply $(2)$ to the pair $a, d$ and the pair $b, c$ . We get

$\large\ \left( { a }^{ 2 } + 1 \right) \left( d^{ 2 } + 1 \right) \left( b^{ 2 } + 1 \right) \left( c^{ 2 } + 1 \right) \le { \left( { \left( \frac { a + d }{ 2 } \right) }^{ 2 } + 1 \right) }^{ 2 }{ \left( { \left( \frac { b + c }{ 2 } \right) }^{ 2 } + 1 \right) }^{ 2 }.$ . $....(3)$

Next from

$\left( \frac { a +d }{ 2 } \right) .\left( \frac { b + c }{ 2 } \right) \ge \sqrt { ad } .\sqrt { bc } \ge 1$ ,

we can apply $(2)$ again to the pair $\left( \frac { a+d }{ 2 } ,\frac { b+c }{ 2 } \right)$ . Together with $(3)$ , we have

$\large\ \left( { a }^{ 2 }+1 \right) \left( d^{ 2 }+1 \right) \left( b^{ 2 }+1 \right) \left( c^{ 2 }+1 \right) \le { \left( { \left( \frac { a+b+c+d }{ 4 } \right) }^{ 2 } + 1 \right) }^{ 4 } = { \left( d^{ 2 } + 1 \right) }^{ 4 }.$ .

Therefore, $\left( { a }^{ 2 }+ 1 \right) \left( b^{ 2 }+1 \right) \left( c^{ 2 }+1 \right) { \le \left( d^{ 2 }+1 \right) }^{ 3 }.$ , and the below expression follows by taking the cube root of both above sides.

$\large\ \sqrt [ 3 ]{ \left( { a }^{ 2 } + 1 \right) \left( b^{ 2 } + 1 \right) \left( c^{ 2 } + 1 \right) } \le { \left( \frac { a+b+c }{ 3 } \right) }^{ 2 } + 1$ .