Finding $\small{\min\limits_{a+b+c=2(ab+bc+ca)}\sum_{cyc}\sqrt{\dfrac{1}{ab}+\dfrac{1}{bc}+1}.}$

Question

If $a,b,c>0 : a+b+c=2(ab+bc+ca),$ then find the minimal value$$P=\sqrt{\dfrac{1}{ab}+\dfrac{1}{bc}+1}+\sqrt{\dfrac{1}{bc}+\dfrac{1}{ca}+1}+\sqrt{\dfrac{1}{ca}+\dfrac{1}{ab}+1}.$$ By set $a=b=c=1/2,$ I got $P\ge 9.$

I've tried to use classical inequality based on this equality case. For example, $$4.\frac{1}{4ab}+4.\frac{1}{4cb}+1\ge 9\sqrt[9]{\frac{1}{(4ab)^4.(4bc)^4}}$$ Similarly, it's enough to prove $$\sum_{cyc}\sqrt[18]{\frac{1}{(4ab)^4.(4bc)^4}}\ge 3$$ Also by AM-GM, it remains to prove $abc\le \dfrac{1}{8}.$

I can't go further because I am still stuck to show $abc\le \dfrac{1}{8}$ is true or false.

Hope you can help me continue my idea. Thank you.

Answer 1

Let $ab=z$, $ac=y$ and $bc=x$.

Thus, the condition gives $$\frac{xy+xz+yz}{\sqrt{xyz}}=2(x+y+z),$$ which gives: $$\sum_{cyc}\sqrt{\frac{1}{ab}+\frac{1}{ac}+1}=\sum_{cyc}\sqrt{\left(\frac{1}{z}+\frac{1}{y}\right)\cdot\frac{4(xyz)(x+y+z)^2}{(xy+xz+yz)^2}+1}$$ and we need to prove that: $$\sum_{cyc}\sqrt{\frac{4x(y+z)(x+y+z)^2}{(xy+xz+yz)^2}+1}\geq9,$$ which is homogeneous already, which says that we can assume $x+y+z=xy+xz+yz.$

Indeed, let $k(x+y+z)=xy+xz+yz,$ $x=kp$, $y=kq$ and $z=kr$.

Thus, $p+q+r=pq+pr+qr$ and we need to prove that: $$\sum_{cyc}\sqrt{\frac{4x(y+z)(x+y+z)^2}{(xy+xz+yz)^2}+1}\geq9$$ or $$\sum_{cyc}\sqrt{\frac{4kp(kq+kr)(kp+kq+kr)^2}{(k^2pq+k^2pr+k^2pr)^2}+1}\geq9$$ or $$\sum_{cyc}\sqrt{\frac{4p(q+r)(p+q+r)^2}{(pq+pr+qr)^2}+1}\geq9,$$ which is the same inequality.

Thus, by C-S $$\sum_{cyc}\sqrt{\frac{4x(y+z)(x+y+z)^2}{(xy+xz+yz)^2}+1}=\sum_{cyc}\sqrt{4x(y+z)+1}=$$ $$=\frac{\sum\limits_{cyc}\sqrt{(y+z-x)^2+4x(y+z))(1+4x(y+z))}}{x+y+z}\geq$$ $$\geq\frac{\sum\limits_{cyc}(y+z-x+4x(y+z))}{x+y+z}=9.$$

Answer 2

Proof.

By your idea to prove $P\ge 9,$ we can apply AM-GM inequality as $$2\sqrt{a+b+abc}\le \sqrt{\frac{ab}{c}}(c+1)+\frac{a+b+abc}{c+1}\sqrt{\frac{c}{ab}}=\frac{ab+bc+ca}{(c+1)\sqrt{abc}}+2\sqrt{abc}.$$ It implies that $$\frac{ab+bc+ca}{(c+1)\sqrt{abc}}\ge \frac{2(a+b)}{\sqrt{a+b+abc}+\sqrt{abc}},$$or $$\sqrt{\dfrac{1}{ab}+\dfrac{1}{bc}+1} \ge \dfrac{2a+2b+ca+cb-ab}{ab+bc+ca} \tag{1}.$$ Take cyclic sum on $(1),$ we get \begin{align*} &\sqrt{\dfrac{1}{ab}+\dfrac{1}{bc}+1}+\sqrt{\dfrac{1}{bc}+\dfrac{1}{ca}+1}+\sqrt{\dfrac{1}{ca}+\dfrac{1}{ab}+1}\\ &\ge \dfrac{4(a+b+c)+ab+bc+ca}{ab+bc+ca}\\ &\ge \dfrac{9(ab+bc+ca)}{ab+bc+ca}=9 \end{align*} Hence, the proof is done.

In conclusion: Minimum $P=9$ at $a=b=c=\dfrac{1}{2}.$

Answer 3

Counter example for $abc \leq \frac{1}{8}$ is simple. Observe that $\left(a,b,c\right) = \left(\frac{1}{4}, 3, \frac{7}{22}\right)$ is a solution and $abc = \frac{21}{88}$ which is greater than $\frac{1}{8}$.