Minimize $\sum_{cyc}\sqrt{\frac{a+b+1}{c+ab}}$ if $ab+bc+ca=1.$

Question

Problem. Let $a,b,c$ be non-negative real numbers satisfying $ab+bc+ca=1.$ Find the minimal value $$P= \sqrt{\frac{a+b+1}{c+ab}}+\sqrt{\frac{c+b+1}{a+cb}}+\sqrt{\frac{a+c+1}{b+ac}}.$$

I saw the problem was posted here.

By $a=b=1;c=0$ I got that $P\ge \sqrt{3}+2\sqrt{2}$ and tried to prove it is desired minimum.

My idea is using Holder inequality$$\left(\sum_{cyc}\sqrt{\frac{a+b+1}{c+ab}}\right)^2.\sum_{cyc}(c+ab)(a+b+1)^2[(\sqrt{6}-1)c+a+b]^3\ge \left[\sum_{cyc}(a+b+1)[(\sqrt{6}-1)c+a+b]\right]^3.$$ Hence, it is enough to prove \begin{align*} \left[\sum_{cyc}(a+b+1)[(\sqrt{6}-1)c+a+b]\right]^3&\ge (11+4\sqrt{6})\sum_{cyc}(c+ab)(a+b+1)^2[(\sqrt{6}-1)c+a+b]^3, \end{align*} which seems ugly but it is true by checking $b=a; c=\dfrac{1-a^2}{2a}.$

I am not sure that $uvw$ help to prove that last one.

About $uvw,$ see here.

I'd like to ask two question.

1. Is $\sqrt{3}+2\sqrt{2}$ good minimum enough ?

The reason is almost symmetrical inequalities achieve extremal value at two equal variables but this statement isn't true in some cases.

In Bob Dobbs's answer, River Li and Michael Rozenberg gave two examples which made me confusing to think of the symmetry principle application.

If you find similar problem, please free to leave it at comment part.

2. In case my answer is correct:

Is there others simpler using Holder for the OP?

Also, if you find something interesting to prove the inequality, please share it here.

All idea and comment is welcome. Thank you for interest!

Answer 1

Some thoughts.

Remarks: Dragon boy's answer using Ji Chen's lemma is nice. Holder also works.

I got the same form of Holder inequality independently. Actually, I used the same idea to obtain the form of Holder inequality here.

We use the pqr method.

Let $p = a + b + c, q = ab + bc + ca = 1, r = abc$.

The desired inequality is written as \begin{align*} &\left( 80\,p\sqrt {6}+66\,\sqrt {6}-230\,p-156 \right) {r}^{2}\\ &\qquad + \left( 2\,{p}^{4}\sqrt {6}-6\,{p}^{3}\sqrt {6}+43\,{p}^{4}+67\,{p}^{2 }\sqrt {6}+21\,{p}^{3}-64\,p\sqrt {6}\right.\\ &\qquad\qquad \left. -22\,{p}^{2}-220\,\sqrt {6}+124\, p+520 \right) r\\ &\qquad +8\,{p}^{5}\sqrt {6}+8\,{p}^{6}+{p}^{4}\sqrt {6}+{p}^{5 }-73\,{p}^{3}\sqrt {6}-46\,{p}^{4}-36\,{p}^{2}\sqrt {6}\\ &\qquad\qquad +68\,{p}^{3}+ 156\,p\sqrt {6}+156\,{p}^{2}+144\,\sqrt {6}-312\,p-352\\ &\ge 0. \end{align*} It is quadratic in $r$. So it is tractable.

Answer 2

The inequality, which you got is wrong.

For $c=0$ the condition gives $ab=1$ and we need to prove that: $$(2(\sqrt6)-1)+(\sqrt6+2)(a+b))^3\geq$$ $$\geq (11+4\sqrt6)((a+b+1)^2(a+b)^3+a(b+1)^2((\sqrt6-1)a+b)^3+b(a+1)^2((\sqrt6-1)b+a)^3),$$ which is wrong for $a\rightarrow+\infty$.

Answer 3

A sketch of proof

Apply Jichen's lemma for $$x=\dfrac{a+b+1}{c+ab};y=\dfrac{c+b+1}{a+cb};z=\dfrac{a+c+1}{b+ac};u=3; v=w=2, p=\frac{1}{2}$$ it suffices to prove

• $x+y+z\ge u+v+w=7$
• $xy+yz+zx\ge uv+vw+wu=16$
• $xyz\ge uvw=12$

We can easily prove above inequalities true by $uvw$ technique.

Notice All of them can be written as $$f(w^3)=kw^6+A(u,v^2)w^3+B(u,v^2)\ge 0,$$where $k<0.$

The rest is proving in two cases which is just calculus working.