We will prove the following problem:
Problem. Let $a,b,c\ge 0: ab+bc+ca=3.$ Prove that$$\frac{2\sqrt{34}}{17}\le\frac{1}{\sqrt{a+bc+7}}+\frac{1}{\sqrt{b+ca+7}}+\frac{1}{\sqrt{c+ab+7}}\le 1.$$
- Proof for RHS
We prove the following lemma:
Lemma. For all $x,y,z\ge 0$ sastifying $$x^2+y^2+z^2+\frac{3}{4}(xyz)^2\le \frac{15}{4},\tag{!}$$then $0\le x+y+z\le 3.$
Equality holds at $x=y=z=1.$
Apply the lemma by $$x=\frac{3}{\sqrt{a+bc+7}};y=\frac{3}{\sqrt{b+ca+7}};z=\frac{3}{\sqrt{c+ab+7}},$$
we only need to prove $$9\sum_{cyc}\frac{1}{a+bc+7}+\frac{3}{4}.\frac{729}{(a+bc+7)(b+ca+7)(c+ab+7)}\le \frac{15}{4}.\tag{1}$$
Let $a+b+c=3u;ab+bc+ca=3v^2=3; abc=w^3.$
After some manipulations, we rewrite the $(1)$ as a $w^3$ function
$$\color{blue}{f(w^3)=5w^6+w^3\left(45u^2+39u-94\right)+438u-433\ge 0.}$$
We will prove, $$f'(w^3)=10w^3+45u^2+39u-94\ge 0.\tag{2}$$
Notice that, $f'(w^3)$ is a linear of $w^3$ which $uvw$ says that we prove the $(2)$ is true in two cases
- $w^3=0.$ Let $a=0;bc=3$ then $$g(w^3)=5(b+c)^2+13(b+c)-94>0,$$which is true by $b+c\ge 2\sqrt{bc}=2\sqrt{3}.$
- Two equal variables. Let $0<b=c=x\le \sqrt{3}; a=\dfrac{3-x^2}{2x}.$ The $(2)$ becomes $$5(3x-x^3)+5\left(2x+\frac{3-x^2}{2x}\right)^2+13\left(2x+\frac{3-x^2}{2x}\right)-94$$
$$=-5\frac{(x - 1)^2}{x^2}\left(x^3 - \frac{x^2}{4} - \frac{42x}{5} - \frac{9}{4}\right )\ge 0,\forall 0<x\le \sqrt{3}.$$
Hence, we proved the $(2).$
Now, $f(w^3)$ is increasing when $0\le w^3\le 1.$ It implies $$f(w^3)\ge f(0)=438u-433>0.$$
Hence, the $(1)$ is proven and we end proof for RHS.
- Proof for LHS
By AM-GM, it is enough to prove
$$\color{blue}{\frac{1}{2a+2bc+31}+\frac{1}{2b+2ca+31}+\frac{1}{2c+2ab+31}\ge \frac{1}{17}. }\tag{3}$$
Notice that we can rewrite the $(3)$ as $$g(w^3)=-8w^6+A(u,v^2)w^3+B(u,v^2)\ge 0.$$It is easy to see $g(w^3)$ is concave which $uvw$ says that we only need to prove the $(3)$ in two cases
- $w^3=0.$ Let $a=0;bc=3:$ the $(3)$ becomes$$\frac{1}{37}+\frac{1}{2b+31}+\frac{1}{2c+31}\ge \frac{1}{17}\iff \frac{1}{2b+31}+\frac{1}{2c+31}\ge \frac{20}{629} ,$$
$$\iff \frac{18}{629}(b+c)+\frac{19538}{629}\ge 0. $$
- Two equal variables. Let $0<b=c=x\le \sqrt{3}; a=\dfrac{3-x^2}{2x}.$ The $(3)$ is also true by computer checking.
We end proof for RHS.
In conclusion:
- Maximal value is equal to $1$ when $a=b=c=1.$
- Minimal value is equal to $\dfrac{2\sqrt{34}}{17}$ when $a=b\rightarrow 0;c\rightarrow +\infty $ and cyclic permutations.
Here is an proof for the lemma.
Assume that there is exist $(x,y,z)$ satisfying the $(!)$ and $x+y+z>3.$
We consider $(m,n,p)$ such that $0\le m=kx<x;0\le n=ky<y;0\le p=kz<z$ where $0<k=\frac{3}{x+y+z}<1.$
Hence, $m+n+p=3$ and by the $(!),$ $$m^2+n^2+p^2+\frac{3}{4}(mnp)^2\le x^2+y^2+z^2+\frac{3}{4}(xyz)^2\le \frac{15}{4}.\tag{4}$$
Now, we will prove $$m^2+n^2+p^2+\frac{3}{4}(mnp)^2\ge \frac{15}{4}.\tag{5}$$
Indeed, let $m+n+p=3=3u; mn+np+pn=3v^2; mnp=w^3.$ The $(5)$ becomes
$$f(w^3)=\frac{3}{4}w^6-6v^2+\frac{21}{4}\ge 0,$$
where $f'(w^3)=\dfrac{3}{2}w^3\ge 0.$
By $uvw$ we only need to prove the $(5)$ is true in two cases
- $w^3=0.$ Let $m=0; n+p=3$ then $$n^2+p^2\ge \frac{9}{2}>\frac{15}{4}.$$
- Two equal variables. Let $0\le t= m=n\le \dfrac{3}{2}; p=3-2t$ then $$2t^2+(3-2t)^2+\frac{3}{4}t^4(3-2t)^2\ge \frac{15}{4},$$
$$\iff (t - 1)^2 \left(t^4 - t^3 - \frac{3t^2}{4} - \frac{t}{2} + \frac{7}{4}\right)\ge 0,$$
which is true by $0\le t\le \dfrac{3}{2}.$
The $(5)$ is true and leads to contradiction again the $(4).$
Hence $0<x+y+z\le 3$ and the lemma is proven.
