Minimum three variable expression $\frac{(a^2+b^2+c^2)^3}{(a+b+c)^3|(a-b)(b-c)(c-a)|}$

Question

If $a,b,c \ge 0$ are distinct real numbers, find the minimum value of

$$\frac{(a^2+b^2+c^2)^3}{(a+b+c)^3|(a-b)(b-c)(c-a)|}$$

What I did: I used $a^2+b^2+c^2\geq \frac{1}{3}(a+b+c)^2$, but no success. A friend of mine said the minimum happens when one of the variables is $0$, but I don't really understand why or how to prove this.

Answer 1

This is not an easy question. It can be rephrased as:

Find the best constant $k$ such that the inequality

$$(a^2+b^2+c^2)^3\geq k(a+b+c)^3|(a-b)(b-c)(c-a)|$$

holds for any non-negative real numbers $a,b$ and $c$.

The product $(a-b)(b-c)(c-a)$ is cyclic, but in the module the inequality becomes symmetric, so we can assume that $a\leq b \leq c$. In this case, we want to find the best $k$ such that:

$$(a^2+b^2+c^2)^3\geq k(a+b+c)^3(c-a)(c-b)(b-a)$$

If we fix $b$ and $c$, the product $(c-a)(c-b)(b-a)$ is maximized when $a=0$. So let's set $a=0$ and $c=bt$ with $t> 1$ to see what happens. Then:

$$(b^2+b^2t^2)^3\geq k(b+bt)^3bt(bt-b)b$$

or

$$k\leq \frac{(t^2+1)^3}{t(t-1)(t+1)^3}$$

Now, define $f:(1,\infty) \to \mathbb{R}$

$$f(t)=\frac{(t^2+1)^3}{t(t-1)(t+1)^3}$$

The first derivative is

$$f'(t)=\frac{(t^2+1)^3(t^4+2t^3-10t^2+2t+1)}{t^2(t-1)^2(t+1)^4}$$

The fourth degree polynomial has four real roots:

$$t_{1,2}= \frac{1}{2}(-1+\sqrt{13}\pm \sqrt{10-2\sqrt{13}}),\ t_{3,4}=\frac{1}{2}(-1-\sqrt{13}\pm \sqrt{10+2\sqrt{13}})$$

Out of these, only $t_1=\frac{1}{2}(-1+\sqrt{13}+\sqrt{10-2\sqrt{13}}) > 1$ (it lies in $f$'s domain) and we can check that this is a local minima. Therefore:

$$k=\min_{t > 1} f(t)=f(t_1)=\frac{1}{9}\sqrt{3522-858\sqrt{13}}\approx 2.29985776707797...$$

And indeed, we can check in the initial inequality that if we set $(a,b,c)=(0,1,t_1)$, we get that the minimum value is $f(t_1)$.

Now we can start the actual proof of the inequality. We have to show that:

$$(a^2+b^2+c^2)^3 \geq \frac{(t_1^2+1)^3}{t_1(t_1-1)(t_1+1)^3}(a+b+c)^3(c-b)(c-a)(b-a)$$

Since $a^2 \geq 0$, it is enough to show that:

$$(b^2+c^2)^3 \geq \frac{(t_1^2+1)^3}{t_1(t_1-1)(t_1+1)^3}(a+b+c)^3(c-b)(c-a)(b-a)$$

Using AM-GM, we get:

$$\left(\frac{a+b+c}{t_1+1}\right)^3\cdot \frac{c-b}{t_1-1}\cdot \frac{c-a}{t_1}\cdot (b-a) \leq \frac{1}{6^6} \left(\frac{3(a+b+c)}{t_1+1}+\frac{c-b}{t_1-1}+\frac{c-a}{t_1}+b-a\right)^6$$

Thus, it's enough to show that:

$$b^2+c^2\geq \frac{t_1^2+1}{36}\left(\frac{3(a+b+c)}{t_1+1}+\frac{c-b}{t_1-1}+\frac{c-a}{t_1}+b-a\right)^2$$

Inside the parantheses, the coefficient of $a$ is

$$\frac{t_1-1-t_1^2}{t_1(t_1+1)} < 0$$

so we can ignore it altogether. Therefore, it remains to prove:

$$b^2+c^2\geq \frac{t_1^2+1}{36}\left(\frac{3(b+c)}{t_1+1}+\frac{c-b}{t_1-1}+\frac{c}{t_1}+b\right)^2$$

or

$$36(b^2+c^2)\geq \frac{t_1^2+1}{[t_1(t_1-1)(t_1+1)]^2}\left[(t_1^3+2t_1^2-5t_1)b+(5t_1^2-2t_1-1)c\right]^2$$

This follows from Cauchy-Schwarz:

$$\left[(t_1^3+2t_1^2-5t)b+(5t_1^2-2t_1-1)c\right]^2\leq \left[(t_1^3+2t_1^2-5t_1)^2+(5t_1^2-2t_1-1)^2\right](b^2+c^2)$$

and we can check that:

$$\frac{(t_1^2+1)\left[(t_1^3+2t_1^2-5t_1)^2+(5t_1^2-2t_1-1)^2\right]}{[t_1(t_1-1)(t_1+1)]^2}=36$$

completing the proof. Equality occurs when $(a:b:c) = (0:1:t_1)$ up to any cyclic permutation.

Note: I chose to write $t_1$ instead of the actual value in the entire proof, because the numbers are ugly up until the last equality.

Answer 2

Your friend was right!

Indeed, we can assume that $a<b<c$.

Thus, it's enough to prove that $$(a+b+c)^3(c-a)(c-b)(b-a)\leq bc(c-b)(b+c)^3$$ or $$a((b+c)(b^3+c^3)+(b+c)(2b^2+bc+2c^2)a-bca^2-2(b+c)a^3-a^4)\geq0,$$ which is obvious.

Also, $$(a^2+b^2+c^2)^3\geq(b^2+c^2)^3$$ and after assuming $c=xb$ it's enough to find $$\min_{x>1}\frac{(x^2+1)^3}{(x+1)^3(x-1)x},$$ The rest is smooth.