$a,b,c\ge 0: ab+bc+ca=1$. Prove that: $\sqrt{a^2b^2+b^2c^2+c^2a^2}\ge \frac{a+b+c-6abc}{a^3+b^3+c^3+3abc}$

inequalitysymmetric-polynomialsuvw

Problem. Let $a,b,c\ge 0: ab+bc+ca=1$. Prove that:
$$\sqrt{a^2b^2+b^2c^2+c^2a^2}\ge \frac{a+b+c-6abc}{a^3+b^3+c^3+3abc}$$
I posted it in AOPS.

Here is my attempt:

  1. Approach 1:

It is easy by Schur inequality: $$a^3+b^3+c^3+3abc\ge ab(a+b)+bc(c+b)+ca(c+a)=a+b+c-3abc$$
and: $a+b+c-3abc\ge a+b+c-6abc \iff abc\ge 0$. Also $a+b+c=(a+b+c)(ab+bc+ca)\ge 9abc$.

Hence, we need to prove the following less ugly inequality:$$\sqrt{a^2b^2+b^2c^2+c^2a^2}\ge \frac{a+b+c-6abc}{a+b+c-3abc}$$
After checking, the above inequality is wrong in case: $a=0.9; b=0.8$

  1. Approach 2:

We can rewrite the original as pqr form: $$\sqrt{1-2pr}\ge \frac{p-6r}{p^3-3p+6r}$$
Squaring both side, we obtain: $$(1-2pr)(p^3-3p+6r)^2\ge (p-6r)^2$$
But it is quite complicated to me since the high degree yield $p^7; p^6,…$.

I continued toward with a powerful method which I have just known about it by some solutions recently. My following approach might be not good enough.

  1. Approach 3:

I used uvw method, which set: $a+b+c=3u\ge \sqrt{3}; ab+bc+ca=3v^2=1;abc=w^3$. The original problem becomes: $$\sqrt{1-6uw^3}\ge \frac{3u-6w^3}{27u^3-9u+6w^3}$$

We can write it as a function of $w^3$: $$f(w^3)=\left(9u^3-3u+2w^3\right)\sqrt{1-6uw^3}+2w^3-u\ge 0$$
I do some calculating manipulation:

$$f^{'}(w^3)=2\sqrt{1-6uw^3}+\left(9u^3-3u+2w^3\right).\left(\sqrt{1-6uw^3}\right)^{'}+2$$
And: $\left(\sqrt{1-6uw^3}\right)^{'}=\dfrac{-3u}{\sqrt{1-6uw^3}}$
which implies:$$f^{'}(w^3)=2\sqrt{1-6uw^3}-\dfrac{3u(9u^3-3u+2w^3)}{\sqrt{1-6uw^3}}+2$$
I want to show that $f(w^3)$ is concave. Unluckily, I get some troubles.

It is:$f^{'}(w^3)$ can be positive sign in some cases. Here is an example:

$$f^{'}(w^3)=2\sqrt{1-6uw^3}-\dfrac{3u(9u^3-3u+2w^3)}{\sqrt{1-6uw^3}}+2<0$$
in case: $ a=0;b=\frac{\sqrt{238}}{7};c=\frac{\sqrt{238}}{34}$

But:
$$f^{'}(w^3)=2\sqrt{1-6uw^3}-\dfrac{3u(9u^3-3u+2w^3)}{\sqrt{1-6uw^3}}+2>0 $$
in case: $a=\frac{4}{5};b=\frac{9}{10};c=\frac{14}{85}$

It is quite interesting now. I don't give up so I came up with the idea that we can divide the starting problem into 2 cases.

Firstly, we will homogenize the derivative as: $$f^{'}(w^3)=2\sqrt{9v^4-6uw^3}-\dfrac{3u(9u^3-9uv^2+2w^3)}{\sqrt{9v^4-6uw^3}}+6v^2$$

If $f^{'}(w^3)<0$, it is equivalent to:$$9\left(3u^4-2v^4+2uw^3-3u^2v^2\right)=3u(9u^3-9uv^2+2w^3)-2\left(9v^4-6uw^3\right)>6v^2\sqrt{9v^4-6uw^3}$$
Or:$$ \sqrt{9v^4-6uw^3}< \frac{3\left(3u^4-2v^4+2uw^3-3u^2v^2\right)}{2v^2}$$
In this case, $f(w^3)$ is concave. Hence, we just need to check:

  • One variable is equal to $0$, assume: $a=0; bc=1$. We will show that:$b^3+c^3\ge b+c \iff (b+c)(b-c)^2\ge 0$

  • Two of varibles are equal, assume: $0\le b=c=x\le 1;a=\dfrac{1-x^2}{2x}$, which means we need to prove: $$\sqrt{x^4+2x^2\left(\dfrac{1-x^2}{2x}\right)^2}\ge \frac{2x+\dfrac{1-x^2}{2x}-6x^2.\dfrac{1-x^2}{2x}}{2x^3+\left(\dfrac{1-x^2}{2x}\right)^3+3x^2.\dfrac{1-x^2}{2x}}$$
    I still stuck in proving by hand. I used Checking software.

Now, it is enough to prove the OP true in case:$$ \sqrt{9v^4-6uw^3}\ge \frac{3\left(3u^4-2v^4+2uw^3-3u^2v^2\right)}{2v^2}$$
It means we will prove the following inequality to end proof:
$$\frac{3\left(3u^4-2v^4+2uw^3-3u^2v^2\right)}{2v^2}\ge \frac{(3uv^2-2w^3)v^2}{9u^3-9uv^2+2w^3}$$
I guess today isn't my lucky day because the last inequality is wrong in case: $ a=\frac{4}{5};b=\frac{9}{10};c=\frac{14}{85}$.

Some thoughts: I think the OP is not very hard but I still stuck to find a completed proof. For instance, BW and SOS work with deg 10 but I haven't tried.In any case, I will try my best to solve it. If anyone can continue with my idea (pqr approach) or find more possible ideas, please share it. Thank you!

Best Answer

In the standard $uvw$'s notation we need to prove that: $$\sqrt{9v^4-6uw^3}\geq\frac{(9uv^2-6w^3)3v^2}{27u^3-27uv^2+6w^3}$$ or $f(u)\geq0,$ where $$f(u)=(9u^3-9uv^2+2w^3)\sqrt{9v^4-6uw^3}-3(3uv^2-2w^3)v^2.$$ Now, $$f'(u)=(27u^2-9v^2)\sqrt{9v^4-6uw^3}-\frac{3w^3(9u^3-9uv^2+2w^3)}{\sqrt{9v^4-6uw^3}}-9v^4=$$ $$=3\left(\frac{81u^2v^4-27v^6-63u^3w^3+27uv^2w^3-2w^6}{\sqrt{9v^4-6uw^3}}-3v^4\right)\geq$$ $$\geq\frac{3(81u^2v^4-27v^6-63u^3w^3+27uv^2w^3-2w^6-3v^2\cdot3v^4)}{\sqrt{9v^4-6uw^3}}\geq$$ $$\geq\frac{3(81u^2v^4-36v^6-63u^3w^3+18uv^2w^3)}{\sqrt{9v^4-6uw^3}}=\frac{27(9u^2v^4-4v^6-7u^3w^3+2uv^2w^3)}{\sqrt{9v^4-6uw^3}}\geq0$$ because we'll prove now that $$9u^2v^4-4v^6-7u^3w^3+2uv^2w^3\geq0.$$ Indeed, this inequality is a linear inequality of $w^3$, which by $uvw$ says that it's enough to prove it in two following cases.

  1. $w^3=0.$ In this case it's obvious.

  2. Two variables are equal.

Since this inequality is homogeneous and for $b=c=0$ it's obviously true, we can assume $b=c=1$, which gives $$(a-1)^2(5a^2+8a+8)\geq0.$$ Id est, $f$ increases and by $uvw$ again it's enough to prove the starting inequality for equality case of two variables.

Let $b=a$. Thus, $c=\frac{1-a^2}{2a},$ where $0<a\leq1$ and we need to prove that: $$\sqrt{a^4+2a^2\left(\frac{1-a^2}{2a}\right)^2}\geq\frac{2a+\frac{1-a^2}{2a}-6a^2\cdot\frac{1-a^2}{2a}}{2a^3+\left(\frac{1-a^2}{2a}\right)^3+3a^2\cdot\frac{1-a^2}{2a}}$$ or $$(x-1)^2(3x+1)(9x^5+99x^4-72x^3+36x^2-9x+1)\geq0,$$ where $x=a^2$, which is true because $$9x^5+99x^4-72x^3+36x^2-9x+1\geq99x^4-72x^3+15x^2\geq0.$$

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