1
$\begingroup$

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.

  1. 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!

$\endgroup$
12
  • $\begingroup$ @Michael Rozenberg Is there something wrong ? I got your answer but it is no longer viewable $\endgroup$
    – TATA box
    Aug 8, 2023 at 5:47
  • $\begingroup$ Sometimes it happens. The inequality is true for equality case of two variables, but is wrong in the general. $\endgroup$ Aug 8, 2023 at 5:50
  • $\begingroup$ There is a very nice proof which uvw is dominated $\endgroup$
    – Dragon boy
    Aug 8, 2023 at 5:52
  • 1
    $\begingroup$ @Dragon boy Show it. But read please before my post with counterexample. $\endgroup$ Aug 8, 2023 at 5:58
  • 2
    $\begingroup$ @MichaelRozenberg Of course. I think the problem is easy by uvw but I will show simple and better proof for later. Keep waiting! $\endgroup$
    – Dragon boy
    Aug 8, 2023 at 6:03

3 Answers 3

1
$\begingroup$

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.

$\endgroup$
0
$\begingroup$

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 $a\rightarrow+\infty$.

$\endgroup$
4
  • 1
    $\begingroup$ Ok, now the answer is viewable. Thank you! I'll check it carefully $\endgroup$
    – TATA box
    Aug 8, 2023 at 5:50
  • 2
    $\begingroup$ Yes, you are right @Michael Rozenberg. My Holder using is not appropriate. $\endgroup$
    – TATA box
    Aug 8, 2023 at 5:55
  • 1
    $\begingroup$ @TATAbox you made a serious mistake when check the estimate $\endgroup$
    – Dragon boy
    Aug 8, 2023 at 6:02
  • $\begingroup$ @Dragonboy Yes, a valuable lesson. $\endgroup$
    – TATA box
    Aug 8, 2023 at 6:08
0
$\begingroup$

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.

$\endgroup$
2
  • $\begingroup$ I checked your statements. It is true and you should complete it instead of giving just key $\endgroup$
    – TATA box
    Aug 10, 2023 at 5:31
  • $\begingroup$ @TATAbox I wrote it compactly enough. You should full it by $uvw$ $\endgroup$
    – Dragon boy
    Aug 10, 2023 at 5:37

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .