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

Question

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: $a=0.9; b=0.8$

2. 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.

3. 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!

Answer 1

Proof.

Firstly, we consider the case $abc=0.$

Plugging $a=0;bc=1$ the original inequality becomes $$bc\ge \frac{b+c}{b^3+c^3} \iff b^2+c^2\ge 2.$$

From now, we use $abc>0.$

By $ab+bc+ca=1$, we'll prove $$\sqrt{a^2b^2+b^2c^2+c^2a^2}\ge \frac{ab(a+b)+bc(b+c)+ca(c+a)-3abc}{a^3+b^3+c^3+3abc}.(ab+bc+ca),$$ or $$\sqrt{a^2b^2+b^2c^2+c^2a^2}+ab+bc+ca\ge \frac{(a^2+b^2+c^2)(a+b+c)}{a^3+b^3+c^3+3abc}.(ab+bc+ca).$$

Now, we can eliminate the yield $a+b+c$ by multiplying $ab+bc+ca-\sqrt{a^2b^2+b^2c^2+c^2a^2}>0$ for both side.

Hence, it remains to prove $$2abc\left(a^3+b^3+c^3+3abc\right)\ge (a^2+b^2+c^2)(ab+bc+ca)\left(ab+bc+ca-\sqrt{a^2b^2+b^2c^2+c^2a^2}\right). \tag{1}$$

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Notice that,\begin{align*} &(ab+bc+ca)^2(a^2+b^2+c^2)-2abc\left(a^3+b^3+c^3+3abc\right)\\&=\sum_{cyc}\left(2a^2bc(b^2+c^2)+b^2c^2(b^2+c^2-a^2)\right), \end{align*} then we can rewrite $(1)$ as $$(a^2+b^2+c^2)(ab+bc+ca)\sqrt{a^2b^2+b^2c^2+c^2a^2}\ge \sum_{cyc}\left(2a^2bc(b^2+c^2)+b^2c^2(b^2+c^2-a^2)\right).$$ From now, Cauchy-Schwarz helps well.

Indeed, it's sufficient to show that$$(a^2+b^2+c^2)\sqrt{a^2b^2+b^2c^2+c^2a^2}.bc\ge 2a^2bc(b^2+c^2)+b^2c^2(b^2+c^2-a^2),$$ or $$(a^2+b^2+c^2)\sqrt{a^2b^2+b^2c^2+c^2a^2}\ge 2a^2(b^2+c^2)+bc(b^2+c^2-a^2).$$ Since $a^2+b^2+c^2=\sqrt{(b^2+c^2-a^2)^2+4a^2(b^2+c^2)},$ we apply CBS inequality as$$\sqrt{\left[(b^2+c^2-a^2)^2+4a^2(b^2+c^2)\right]\left[(bc)^2+a^2(b^2+c^2)\right]}\ge bc(b^2+c^2-a^2)+2a^2(b^2+c^2).$$

Thus, the proof is done. Equality holds at $(a,b,c)=(0,1,1).$

Answer 2

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 $$9x^5+99x^4-72x^3+36x^2-9x+1\geq99x^4-72x^3+15x^2\geq0.$$

Answer 3

Now, I found a SOS identity after homogenzing the original problem. $$(a^2b^2+b^2c^2+c^2a^2)(a^3+b^3+c^3+3abc)^2$$$$-(ab+bc+ca)^2(ab(a+b)+bc(b+c)+ca(c+a)-3abc)^2=$$ $$\left( \sum_{\mathrm{cyc}}{a} \right) \sum_{\mathrm{cyc}}{\left[ \left( a^5b^2+a^5c^2+\frac{a^3b^4}{2}+\frac{a^3c^4}{2} \right) (b+c-a)^2+(b-c)^2\left( \frac{a^4b}{2}+\frac{a^4c}{2}+\frac{1}{2}a^3bc+\frac{3}{2}ab^2c^2 \right) (b+c-a)^2 \right]}\ge 0$$

Answer 4

Remark: In your approach 2 (the pqr method), we can use the concave function rather than squaring both sides.

Proof.

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

From $q^2 \ge 3pr$, we have $r \le \frac{1}{3p}$. From degree four Schur inequality, we have $r \ge \frac{5p^2q - p^4 - 4q^2}{6p} = \frac{5p^2 - p^4 - 4}{6p}$. Thus, we have $$\max\left(0, \frac{5p^2 - p^4 - 4}{6p}\right) \le r \le \frac{1}{3p}.$$

We need to prove that $$f(r) := \sqrt{1-2pr} - \frac{p-6r}{p^3-3p+6r} \ge 0.$$ It is easy to prove that $f(r)$ is concave.

If $p \ge 2$, we have $f(0) = \frac{p^2 - 4}{p^2 - 3} \ge 0$ and $f(\frac{1}{3p}) > 0$. Thus, $f(r) \ge 0$ on $[0, \frac{1}{3p}]$.

If $\sqrt 3 \le p < 2$, we have $f(\frac{5p^2 - p^4 - 4}{6p}) \ge 0 $ and $f(\frac{1}{3p}) > 0$. Thus, $f(r) \ge 0$ on $[\frac{5p^2 - p^4 - 4}{6p}, \frac{1}{3p}]$.

We are done.