after checking through computer , the least value of such $M$ is $\frac{9}{16\sqrt{2}}$ with equality case $a: b: c=\frac{3\sqrt{2}+2}{2}: \frac{2-3\sqrt{2}}{2}: 1$ and its permutation . So who can prove this ?

My friend Arman's solution (Mathlinks ID : armanf) The left hand side factors like this: \[(a-b)(b-c)(c-a)(a+b+c)\] So just let $x=a-b, y=b-c, z=c-a, s=a+b+c$ then the inequality becomes \[|xyzs|\leq \frac{M}{9}(x^{2}+y^{2}+z^{2}+s^{2})^{2}\] With the property that $x+y+z=0$ Two of $x,y,z$ have the same sign. Let $x,y$ be that two. Now putting their arithmetic mean instead of them makes the inequality stricter (l.h.s. gets bigger and r.h.s. gets smaller) So assume that $x=y$. Now $z=-2x$ and the inequality becomes: \[|2x^{3}s|\leq \frac{M}{9}(s^{2}+6x^{2})^{2}\] Now using the AM-GM inequality for $s^{2},2x^{2},2x^{2},2x^{2}$ we get \[(s^{2}+6x^{2})^{2}\geq 16\sqrt{8s^{2}x^{6}}=32\sqrt 2sx^{3}\] So $M=\frac 9{16\sqrt 2}$ works.

For real numbers $p,q,$ we have $pq\leq \frac{p^{2}+q^{2}}{2}.$ By this $|ab(a^{2}-b^{2})|=|ab(a+b)(a-b)|\leq |ab|\cdot \frac{|a+b|^{2}+|a-b|^{2}}{2}$ $=|ab|(a^{2}+b^{2})\leq \frac{|a|^{2}+|b|^{2}}{2}(a^{2}+b^{2})=\frac{(a^{2}+b^{2})^{2}}{2}.$ Thus use triangle inequality, $LHS\leq \frac{1}{2}\{(a^{2}+b^{2})^{2}+(b^{2}+c^{2})^{2}+(c^{2}+a^{2})^{2}\}.$

kunny wrote: For real numbers $p,q,$ we have $pq\leq \frac{p^{2}+q^{2}}{2}.$ By this $|ab(a^{2}-b^{2})|=|ab(a+b)(a-b)|\leq |ab|\cdot \frac{|a+b|^{2}+|a-b|^{2}}{2}$ $=|ab|(a^{2}+b^{2})\leq \frac{|a|^{2}+|b|^{2}}{2}(a^{2}+b^{2})=\frac{(a^{2}+b^{2})^{2}}{2}.$ Thus $LHS\leq \frac{1}{2}\{(a^{2}+b^{2})^{2}+(b^{2}+c^{2})^{2}+(c^{2}+a^{2})^{2}\}.$ if AM-GM is applied in such way , then the equality case I mention above wont satisfy

Hi gays,I solved this problem during the competition,and my solution is similiar to Nimas' solution,and shyong the equality holds not only in that case .

Nima Ahmadi Pour wrote: My friend Arman's solution (Mathlinks ID : armanf) Two of $x,y,z$ have the same sign. Let $x,y$ be that two. Now putting their arithmetic mean instead of them makes the inequality stricter (l.h.s. gets bigger and r.h.s. gets smaller) What is it? With $x+y=const$ ?

I guess this might be a more direct way of seeing it (essentially the same argument): Define $x,y,z,s$ as before and suppose $x,y$ have the same sign. WLOG, they're both positive (otherwise we can replace $a,b,c$ with $-a,-b,-c$.) Let $m = \frac{x+y}{2}$, so $z =-2m$. Now, $xy \leq m^{2}\leq \frac{x^{2}+y^{2}}{2}$ so $|(a-b)(b-c)(c-a)(a+b+c)|$ $= |xyzs| \leq |2m^{3}s|$ $\leq \frac{1}{16\sqrt{2}}\cdot (2m^{2}+2m^{2}+2m^{2}+s^{2})^{2}$ $= \frac{1}{16\sqrt{2}}\cdot (2m^{2}+z^{2}+s^{2})^{2}$ $\leq \frac{1}{16\sqrt{2}}\cdot (x^{2}+y^{2}+z^{2}+s^{2})^{2}$ $= \frac{9}{16\sqrt{2}}\cdot (a^{2}+b^{2}+c^{2})^{2}$ (and as before, we can check the equality case)

kakalotta123 wrote: Nima Ahmadi Pour wrote: My friend Arman's solution (Mathlinks ID : armanf) Two of $x,y,z$ have the same sign. Let $x,y$ be that two. Now putting their arithmetic mean instead of them makes the inequality stricter (l.h.s. gets bigger and r.h.s. gets smaller) What is it? With $x+y=const$ ? I didn't get it exactly, just put their arithmetic mean instead of them, I didn't say to continously change them or something like that.

Let me help Nima Ahmadi Pour explain a little more clearly. He is replacing $x$ and $y$ with $x^{\prime}= \frac{x+y}{2}$ and $x^{\prime}= \frac{x+y}{2}$ respectively. So looking at our inequality \[\mid xyzs \mid \leq \frac{M}{9}(x^{2}+y^{2}+z^{2}+s^{2})^{2}\] after our substitution, if $x \neq y$, we get $\mid xyzs \mid < \mid x^{\prime}y^{\prime}zs \mid$ and $\frac{M}{9}(x^{2}+y^{2}+z^{2}+s^{2})^{2}> \frac{M}{9}(x^{\prime}^{2}+y^{\prime}^{2}+z^{2}+s^{2})^{2}$ by checking. (Remember that $x+y+z = 0$ and that $x,y,z$ are defined in terms of $a,b,c$.) Of course, when $x=y$, we have $x = x^{\prime}$ and $y = y^{\prime}$, and there's no change. Argh!! I don't like the keyboards at Slovenia. The labels on the keyboards are very misleading. The '&' key is actually '^', the z and y are interchanged, and many more symbols are interchanged. I have to not look at the keyboard to type this.

It seems to be pretty doable using spherical coordinates, as well.

What do you mean by spherical coordinates?

Spherical coordinates: one radius coordinate, two angle coordinates. I did this problem by first noting that the value we seek is the maximum of $ab(a^{2}-b^{2})+bc(b^{2}-c^{2})+ca(c^{2}-a^{2})$ on the unit sphere, and using Lagrange multipliers. After a great deal of algebra, I found the critical points; most led to zero values, leaving only one possibility (and permutations) for the maximum. It was really pretty routine, except for the many guessed factorizations while finding critical points and a mistake in calculation. I really should have used the coordinate system in Nima's post- I probably would have thought of them in my high school days, since I remember using them on a MOP homework problem.

According to the UK Leader, an Iranian observer found a solution to this question using Lagrange multiplier bashing.

This problem can be rephrased as follows. Let $x,y,z$ be reals such that $x^{2}+y^{2}+z^{2}=1.$ Find the minimum value $M$ such that $|xy(x^{2}-y^{2})+yz(y^{2}-z^{2})+zx(z^{2}-x^{2})|\leq M.$

Centy wrote: According to the UK Leader, an Iranian observer found a solution to this question using Lagrange multiplier bashing. I did that ... BRITISH!!! although my solution wasn't straight-bashing; there were some nice algebraic manipulations

Can somebody post a Lagrange multiplier based solution? I would be interested to see how they work in this situation (I've only used them before in very simple cases as exercises).

Here's mine (hidden for length)

is it possible to generalise for n variables? :

Thanks jmerry.

Hi ; It Can Be Solved By $uvw$ Method Best Regard

Firstly we have to note the well known factorization:$|ab(a^2-b^2)+bc(b^2-c^2)+ca(c^2-a^2)|=|(a-b)(b-c)(a-c)(a+b+c)|$.On the other hand one can think of(even though its IMO problem) to write RHS in some terms of factors that are in LHS,and then use AM-GM.Indeed it wont take long to find the following equality:$$(3(a^2+b^2+c^2))^2=(2(a-b)^2+2(a-c)(b-c)+(a+b+c)^2)^2\geq (4^2\sqrt{2(a-b)(a-c)(b-c)(a+b+c)})=16\sqrt{2}|(a-b)(b-c)(a-c)(a+b+c)|$$hence we have$(a^2+b^2+c^2)^2\frac{9}{16\sqrt{2}}\geq |ab(a^2-b^2)+bc(b^2-c^2)+ca(c^2-a^2)|$ and we know this is the smallest $M$ because we can find equliaty case. Note that without checking equality case one can easily find the wrong constant,for example if we rewrite as $3(a^2+b^2+c^2)=(a+b+c)^2+(a-b)^2+(b-c)^2+(c-a)^2$ with simple AM-GM we will only get $\frac{9}{16}$ which is by far to big.

blackbird97 wrote: Firstly we have to note the well known factorization:$|ab(a^2-b^2)+bc(b^2-c^2)+ca(c^2-a^2)|=|(a-b)(b-c)(a-c)(a+b+c)|$.On the other hand one can think of(even though its IMO problem) to write RHS in some terms of factors that are in LHS,and then use AM-GM.Indeed it wont take long to find the following equality:$$(3(a^2+b^2+c^2))^2=(2(a-b)^2+2(a-b)(b-c)+(a+b+c)^2)^2\geq (4^2\sqrt{2(a-b)(a-c)(b-c)(a+b+c)})=16\sqrt{2}|(a-b)(b-c)(a-c)(a+b+c)|$$hence we have$(a^2+b^2+c^2)^2\frac{9}{16\sqrt{2}}\geq |ab(a^2-b^2)+bc(b^2-c^2)+ca(c^2-a^2)|$ and we know this is the smallest $M$ because we can find equliaty case. Note that without checking equality case one can easily find the wrong constant,for example if we rewrite as $3(a^2+b^2+c^2)=(a+b+c)^2+(a-b)^2+(b-c)^2+(c-a)^2$ with simple AM-GM we will only get $\frac{9}{16}$ which is by far to big. It should be $(3(a^2+b^2+c^2))^2=(2(a-b)^2+2(a-c)(b-c)+(a+b+c)^2)^2$ And you should suppose $min\{a,b,c\}=c$ to use that and also you forgot the square after $AM-GM$ you did applied. But this is most beautiful solution I ever seen to this problem it's just awesome congratulation.

An interesting fact is that if we include the condition $a,b,c \ge 0, M=\frac{1}{4}$ and is possible for $a : b : c = 0 : 1 : (1+\sqrt {2})$.

Very interesting

The answer is $\tfrac{9\sqrt{2}}{32}$, achieved at $(a,b,c)=(\sqrt{2}-3, \sqrt{2}, \sqrt{2}+3)$. To prove the bound, WLOG $a\geq b\geq c$ and note the following bounds. \begin{align*} (a-b)(b-c) &\leq (a-b)(b-c) + \frac 38 (a-2b+c)^2 = X\\ \frac{1}{2\sqrt{2}} (a-c)(a+b+c) &\leq \frac 18 (a-c)^2 + \frac 14 (a+b+c)^2 =Y\\ \end{align*}However, after a full expansion, $$X+Y = \frac 34(a^2+b^2+c^2).$$Hence, \begin{align*}\left\lvert\sum_{\mathrm{cyc}} ab(a^2-b^2)\right\rvert &= (a+b+c)(a-b)(a-c)(b-c) \\ &\leq 2\sqrt{2}\cdot XY \\ &\leq 2\sqrt{2}\left(\frac{X+Y}{2}\right)^2 \\ &= \frac{9\sqrt{2}}{32}(a^2+b^2+c^2)^2 \end{align*}as desired.

Note that the inequality always holds for $a=b=c$, with any value of $M$. Thus, the answer to the problem is \[M = \max_{\substack{(a,b,c)\in\mathbb{R}^3\\(a,b,c)\ne(a,a,a)}}\frac{|ab(a^{2}-b^{2})+bc(b^{2}-c^{2})+ca(c^{2}-a^{2})|}{(a^2+b^2+c^2)^2}.\]Note that \[|ab(a^{2}-b^{2})+bc(b^{2}-c^{2})+ca(c^{2}-a^{2})| = |(a-b)(b-c)(c-a)(a+b+c)|.\]This motivates reparameterizing with $x=b-c$, $y=c-a$, $z=a-b$, and $t=a+b+c$, where now the only restriction is that $x+y+z=0$ and $x,y,z$ are not all equal. It is easy to check that \[(a^2+b^2+c^2)^2 = \frac{1}{81}(3t^2+\sum(x-y)^2)^2,\]so we have \[M = 81\cdot\max_{\substack{(x,y,z,t)\in\mathbb{R}^4\\x+y+z=0\\(x,y,z)\ne(x,x,x)}}\frac{|xyzt|}{[3t^2+\sum(x-y)^2]^2}.\]Fixing $(x,y,z)$, we see that it suffices to look at $t\ge 0$ since the thing we are maximizing stays the same if we negate $t$. Let $a=\sum(x-y)^2$, and let \[f(t) = \frac{t}{(3t^2+a)^2}.\]Note that $f(0)=0$ and $f(t)\to 0$ as $t\to\infty$, so the maximum of $f(t)$ is at a critical point of $f$ somewhere in $(0,\infty)$. It is easy to see that \[f'(t)=0 \iff (3t^2+a)^2 = t\cdot(2(3t^2+a))\cdot(6t)\iff 3t^2+a = 12t^2\iff t=\sqrt{a}/3.\]Thus, the maximum of $f(t)$ is \[f(\sqrt{a}/3) = \frac{\sqrt{a}/3}{16a^2/9} = \frac{3}{16}a^{-3/2},\]so the answer to the problem is \[M = \frac{243}{16}\cdot\max_{\substack{(x,y,z)\in\mathbb{R}^3\\x+y+z=0\\(x,y,z)\ne(x,x,x)}}\frac{|xyz|}{\left[\sum(x-y)^2\right]^{3/2}}.\]Note that replacinig $(x,y,z)$ to $(-x,-y,-z)$ does not change the expression within the maximum, so we can WLOG assume $x,y\ge 0$. In this case, we see that \[M = \frac{243}{16}\cdot\max_{x,y\ge 0}\frac{xy(x+y)}{[6x^2+6xy+6y^2]^{3/2}}.\]By de-homogenizing, we see that \[M = \frac{243}{16}\cdot\max_{x\ge 0}\frac{x+1/x}{[6(x^2+1/x^2-1)]^{3/2}} = \frac{243}{16}\cdot\max_{\alpha\ge 2}\frac{\alpha}{[6(\alpha^2-1)]^{3/2}},\]and it is easy to check that \[g(\alpha)=\frac{\alpha}{(\alpha^2-1)^{3/2}}\]is maximized at $\alpha=2$ if we restrict to $\alpha\in[2,\infty)$. Therefore, \[M = \frac{243}{16}\cdot\frac{2}{6^{3/2}\cdot 3^{3/2}} = \frac{243}{16}\cdot\frac{1}{27\sqrt{2}} = \boxed{\frac{9}{16\sqrt{2}}}.\]

dame dame

Here is COMPLEX VERSION of this problem. Determine the least real number $M$ such that the inequality $$|xy(x^2-y^2)+yz(y^2-z^2)+zx(z^2-x^2)|\leq M(|x|^2+|y|^2+|z|^2)$$ holds for all COMPLEX numbers $x,y,z$. (Used for CMO training?)

Nothing new but I'll post for storage. Solved with andyxpandy99. Notice that this the desired inequality is equivalent to proving $$|ab(a^2-b^2)+bc(b^2-c^2)+ca(c^2-a^2)| = |(a-b)(b-c)(c-a)(a+b+c)| \le M(a^2+b^2+c^2)^2.$$WLOG let $a-b, b-c \ge 0$. Now, substituting $p = a-b, q = b-c, -2r = c-a, s = a+b+c$, we get $$|pq(-2r)s| \le \frac{M}{9}(p^2+q^2+(-2r)^2+s^2)^2.$$Notice that $p+q-2r = 0 \implies p+q = 2r$, so fixing the value of $c-a$, we see that we can increase the LHS and decrease the RHS by making $p = q = r$. We then have $$2r^3s \le \frac{M}{9}(6r^2+s^2)^2 = \frac{M}{9}(2r^2+2r^2+2r^2+s^2)^2$$which by AM-GM on the RHS gives $$\frac{M}{9}(2r^2+2r^2+2r^2+s^2)^2 \ge \frac{4M}{9}\left(\sqrt[4]{8r^6s^2}\right)^2 = \frac{8M\sqrt{2}}{9}r^3s \ge 2r^3s$$$\implies M \ge \frac{9\sqrt2}{32}$ with equality when $(a, b, c) = (\sqrt{2}-3, \sqrt{2}, \sqrt{2}+3)$. $\blacksquare$

The answer is $\boxed{\frac{9\sqrt{2}}{32}}$. This is equivalent to $|(a-b)(b-c)(c-a)(a+b+c)|\le M(a^2+b^2+c^2)^2$. In fact, setting $(a,b,c)=(\sqrt{2}+3,\sqrt{2},\sqrt{2}-3)$ gives $162\sqrt{2}\le M\cdot 576\implies \frac{9\sqrt{2}}{32}\le M$. Also, equality holds if $M=\frac{9\sqrt{2}}{32}$. All we have to do now is show that the inequality is true. Let $w=a-b, x=b-c, -2y=c-a, z=a+b+c$. The inequality becomes \[|wxyz|\le \frac{M}{9}(w^2+x^2+4y^2+z^2)^2.\] If one of $w,x,y,z$ are $0$, then the inequality obviously holds true. So assume they are all nonzero. WLOG $w,x,y,z>0$ (if this is not true, then changing $x$ to $-x$ keeps the inequality exactly the same). Claim: If $w\ne x$, changing $w$ and $x$ both to their arithmetic mean makes the inequality stricter (as in, the LHS increases but the RHS decreases). Proof: Note that $w+x=2y$. By AM-GM, we have \[w(2y-w)\le y^2\]so the LHS increases. To show that the RHS decreases, we have \begin{align*} w^2+(2y-w)^2 \\ =2w^2-4wy+4y^2 \\ =2(w^2-2wy+2y^2) \\ =2((w-y)^2+y^2) \\ \ge 2y^2 \\ \end{align*} Equality holds iff $w=x=y$, so the RHS decreases also. $\blacksquare$ Now we can set $w=x=y$. We get \[2y^3z\le \frac{M}{9}(2y^2+2y^2+2y^2+z^2)^2\] We have \begin{align*} 2y^2+2y^2+2y^2+z^2 \\ \ge 4\sqrt[4]{8y^6z^2} \\ \implies (y^2+y^2+4y^2+z^2)\ge 16\sqrt{8y^6z^2} \\ =32\sqrt{2}y^3z \end{align*} Multiplying by $\frac{M}{9}=\frac{\sqrt{2}}{32}$ gives $2y^3z$, as desired.

Let $x=a-b, y=b-c, z=c-a,$ and $u=a+b+c.$ So the inequality is: $$9|xyz| \leq M(x^2+y^2+z^2+u^2)^2.$$Assume $x,y \geq 0.$ By AM-GM: $$|uxyz|=|uxy(x+y)| \leq |u| \frac{(x+y)^3}{4},~~~~~(\alpha)$$equality when $x=y.$ Let $v=x+y,$ by AM-GM: \begin{align*} 2u^2v^6=2u^2\cdot v^2\cdot v^2\cdot v^2 \leq \frac{(2u^2+3v^2)^4}{4^4} \\ \implies 4\sqrt{2}|u|v^3\leq (u^2+\frac{3}{2}v^2)^2 \leq (u^2+x^2+y^2+z^2)^2.~~~~~(\beta)\end{align*}Combination of $(\alpha)$ and $(\beta)$ imply, $$|uxyz| \leq \frac{1}{16\sqrt{2}} (u^2+x^2+y^2+z^2)^2 \implies M \geq \frac{9\sqrt{2}}{32}.$$$\boxed{M=\frac{9\sqrt{2}}{32}},$ i.e. the least value, when $(a,b,c)=(1-\frac{3}{\sqrt{2}}, 1, 1+\frac{3}{\sqrt{2}}).$ The opposite case, that is $x,y \leq 0,$ can be shown similar. $\square$

Davron wrote: Determine the least real number $M$ such that the inequality \[|ab(a^{2}-b^{2})+bc(b^{2}-c^{2})+ca(c^{2}-a^{2})| \leq M(a^{2}+b^{2}+c^{2})^{2}\]holds for all real numbers $a$, $b$ and $c$. anhduy98 wrote: $$P=|ab(a^2-b^2 )+bc(b^2-c^2 )+ca(c^2-a^2 )|$$$$P^2=(a-b)^2(b-c)^2(c-a)^2(a+b+c)^2$$$$a \ge b \ge c$$$$P^2=4(a-b)^2 \frac{(b-c)(a-c)}{2}\frac{(b-c)(a-c)}{2}(a+b+c)^2$$$$P^2 \leq 4(\frac{(a-b)^2+(b-c)(a-c)}{3})^3(a+b+c)^2$$$$P^2 \leq \frac{1}{2}(\frac{2x-2y}{3})^3(x+2y)$$$$(x=a^2+b^2+c^2,y=ab+bc+ca)$$$$P^2 \leq \frac{1}{2}(\frac{3x}{4})^4=\frac{81}{512}(a^2+b^2+c^2)^4$$$$P=|ab(a^2-b^2 )+bc(b^2-c^2 )+ca(c^2-a^2 )|\leq \frac{9}{16 \sqrt{2}}(a^2+b^2+c^2)^2$$ https://artofproblemsolving.com/community/c6h1642070p10350274 Let $a, b, c \ge 0$. Prove that $$ (a-b)(b-c)(c-a)(a+b+c) \le \frac{(a^2+b^2+c^2)^2}{4} $$

First observe that the LHS factors as $(a-b)(b-c)(a-c)(a+b+c)$. WLOG $a>b>c$. Let $f(a,b,c)$ = $M(a^2+b^2+c^2)^2-(a-b)(b-c)(a-c)(a+b+c)$. I claim the answer is $M=\frac{9\sqrt2}{32}$, which is achievable with $(a,b,c)=(\sqrt2-3,\sqrt2,\sqrt2+3)$. Note that if we increase all of $(a,b,c)$ at the same rate, the only terms that change are the $(a+b+c)$ term and the $(a^2+b^2+c^2)^2$ term. Now take the derivative with respect to $a+b+c$ (more precisely, take the limit as $h$ approaches $0$ of $\frac{f(a+h,b+h,c+h)-f(a,b,c)}{h}$. This is equal to $$A=4M(a+b+c)(a^2+b^2+c^2)-3(a-b)(b-c)(a-c)$$. In the equality case, this should be equal to zero (We know there has to be an equality case because the inequality is homogenous and the equality will not be reached by a limit, because we can simply rescale downwards. Sorry if this is a little handwavy). Thus, we substitute A=0 into our inequality and eliminate $M$, to get that, in the equality case, $4(a+b+c)^2=3(a^2+b^2+c^2)$. Thus it is sufficient to maximize $$\frac{(a-b)(b-c)(a-c)(a+b+c)}{(a^2+b^2+c^2)^2}$$subject to the constraint $4(a+b+c)^2-3(a^2+b^2+c^2)=0$. This is equivalent to optimizing $\frac{(a-b)(a-c)(b-c)}{(a+b+c)^3}$. I claim that this expression is maximized at the equality case I described above, which is equivalent to showing that (after plugging it in and verifying that it satisfies the constraint), showing that $\frac{\sqrt2}2(a+b+c)^3-(a-b)(b-c)(a-c)\ge 0$ subject to the constraint. We now apply Lagrange multipliers on $(a,b,c)$ to get $$\frac{3\sqrt{2}}{2}(a+b+c)^2-(b-c)(2a-b-c)=\lambda(a+4b+4c)$$. Cyclically summing this and rearranging gives that $\lambda=\frac{\sqrt{2}}{2}(a+b+c)$.. Plugging this value of $\lambda$ back in and subtracting it from both sides, we get that $$\frac{\sqrt{2}}{2}(a+b+c)(3a+3b+3c-a-4b-4c)-(b-c)(2a-b-c)=0$$, which rearranges to $$(\frac{\sqrt{2}}{2}(a+b+c)(2a-b-c))=(b-c)(2a-b-c)$$. If it is not the case that $(a,b,c)$ are an arithmetic progression in some order, we can divide by $(2a-b-c)$ and cyclically sum to get a contradiction. If they are in arithmetic progression, WLOG $b$ is the middle term and some simple algebra gets that this is optimized at the equality case I gave above.

The left hand side factors to $|(a-b)(b-c)(c-a)(a+b+c)|$. Let $p$, $q$, $r$, $s$ be $a-b$, $b-c$, $c-a$, and $a+b+c$, respectively. Without loss of generality, let $s\ge 0$. Then, since \[p^2+q^2+r^2+s^2=3(a^2+b^2+c^2)\]we have the right-hand side as $\tfrac{M}{9} (p^2+q^2+r^2+s^2)^2$. We claim that the minimum value of $M$ is $\tfrac{9\sqrt2}{32}$. Suppose that the following is a local minimum with respect to $s$: \[f(s)=\frac{(p^2+q^2+r^2+s^2)^2}{9s|pqr|}=\frac{(x+s^2)^2}{ys}\]Then, we have \[f(s)=\frac{s^4+2s^2y+x^2}{ys}=\frac1y(s^3+2sx+x^2s^-1)\]\[f'(s)=\frac1y(3s^2+2x-x^2s^-2)=\frac1{s^2y}(3s^2-x)(s^2+x)\]Clearly, $s^2+x>0$ we when $f$ is at a local minimum, $3s^2=x=p^2+q^2+r^2$. $~$ We claim that given $p+q+r=0$ and $p^2+q^2+r^2=3s^2$, we have $|pqr|\le \frac{s^3}{\sqrt2}$. Let $P(x)$ be a monic cubic with roots $p$, $q$, $r$. Evidently, $P(x)$ has three roots and $P(x)=x^3-\tfrac32 s^2x-pqr$, so if $(x_1,y_1)$ and $(x_2,y_2)$ are $P$'s two local extrema, then $y_1y_2<0$. Since $P'(x)=3x^2-\tfrac32 s^2$, we have $x_{1,2}=\pm \tfrac{s}{\sqrt2}$ so \[y_{1,2}=\pm \frac{s^3}{\sqrt2}-pqr\]Therefore, $(pqr)^2-(\tfrac{s^3}{\sqrt2})^2\le 0$, which implies $|pqr|\le \tfrac{s^3}{\sqrt2}$ as desired. Thus, $f(s)\ge \tfrac{16s^4}{9s^4/\sqrt2}=\tfrac{32}{9\sqrt2}$. Thus, $M\le \tfrac{9\sqrt{2}}{32}$. Our construction: \[(a,b,c)=(\tfrac{\sqrt{2}}{3}+1,\frac{\sqrt{2}}{3}),\frac{\sqrt{2}}{3})+1\]Yields $(p,q,r,s)=(1,1,-2,\sqrt2)$ which yields the desired equality.