Daniel Liu's Hard Inequality

Hello every body.Sorry for too late.But this PROBLEM is very hard. In 1989,Jack Garfunkel recommended a inequality like this problem to Crux magazine in Canada.It's a hard problem and didn't have any solution at that time.

If we given $a,b,c$ are positive reals,the equality holds when $a=b=c$ and at that time,$k=\frac{\sqrt{3}}{\sqrt{2}}$

But $k=\frac{5}{4}$ must be better but we must let $a,b,c$ are non-negative and the equality holds when $a=b=3,c=0$ or it's permutation.

At present,there have many way to prove Jack Garfunkel's inequality. By applying full incremental variable,we have: $a+b+c+\frac{(a-b)(b-c)(a-c)(a+b+c)}{abc}\leq \frac{5\sqrt{2}}{4}\sqrt{x^2+y^2+z^2}$

Let $f(t)=\frac{(x+y+z-3t)^2}{(x-t)(y-t)(z-t)}$ is decreasing and $g(t)=\frac{5\sqrt{2}(x+y+z-3t)\sqrt{(x-t)^2+(y-t)^2+(z-t)^2}}{4}-(x+y+z-3t)^2$ is increasing in $[0;4]$. It's true because if we let $m=x-t;n=y-t,p=z-t$,$m,n,p > 0$ and $A=\sqrt{m^2+n^2+p^2};B=m+n+p$,we get: $m^2n^2p^2f'(t) \geq 0$ and $g'(t) \leq 0$. So that we get: $\frac{a}{\sqrt{a+b}}+\sqrt{b}\leq \frac{5}{4}\sqrt{a+b}$ This it's right because $RHS-LHS \geq 0$. The problem is solved. Another way is more interested:By C-S Inequality $LHS^2\leq \sum a(5a+b+9c)\sum \frac{a}{(a+b)(5a+b+9c)}$ We need to prove: $(a+b+c)\sum \frac{a}{(a+b)(5a+b+9c)}\leq \frac{5}{16}$ This inequality right because: $(RHS-LHS)(16\prod (5a+b+9c))=\sum ab(a+b)(a+9b)(a-3b)^2+243\sum a^3b^2c+835\sum a^3bc^2+232\sum a^4bc+1230a^2b^2c^2\geq 0$

The problem is solved.