62208 and M

I was going to solve this conventionally, but then I noticed a nice "shortcut." I'm probably going to get a lot of criticism for this method, but oh well, it works. Since $a+b+c+d=0$ , we can let $d=-a-b-c$ , so that $62208(a^7+b^7+c^7+d^7)^2=2^8\cdot 3^5\cdot \left(a^7+b^7+c^7-(a+b+c)^7\right)^2\le M(a^2+b^2+c^2+d^2)^7=M\left(a^2+b^2+c^2+(a+b+c)^2\right)^7$ $2^8\cdot 3^5\cdot\left(a^7+b^7+c^7-(a+b+c)^7\right)^2\le M\left(a^2+b^2+c^2+(a+b+c)^2\right)^7$ $f(a,b,c)=M\left(a^2+b^2+c^2+(a+b+c)^2\right)^7-2^8\cdot 3^5\cdot\left(a^7+b^7+c^7-(a+b+c)^7\right)^2\ge 0$ Notice that $f(a,b,c)$ is symmetric, meaning that $f(a,b,c)=f(b,a,c)=f(c,b,a)=\cdots$ for all permutations of $(a,b,c)$ . If we consider optimizing $f(a,b,c)$ by taking the partial derivative with respect to each of the variables, we would get a symmetric equation, which would imply that $a=b=c$ when the function is optimized. Thus, we obtain $2^8\cdot 3^5\cdot \left(3a^7-(3a)^7\right)^2=M\left(3a^2+(3a)^2\right)^7$ $2^8\cdot 3^7\cdot \left(1-3^6\right)^2\cdot a^{14}=12^7\cdot M\cdot a^{14}=2^{14}\cdot 3^7\cdot M\cdot a^{14}$ $M=\frac{2^8\cdot 3^7\cdot (3^6-1)^2}{2^{14}\cdot 3^7}=\frac{((3^3-1)(3^3+1))^2}{2^6}=13^2\cdot 7^2=\boxed{8281}$

Since the problem is homogeneous, we need to extremize $F = a^7 + b^7 + c^7 + d^7$ subject to the constraints $a+b+c+d = 0$ and $a^2 + b^2 + c^2 + d^2 = 1$ . Putting $d = -(a+b+c)$ , this is the same as extremizing $F = a^7 + b^7 + c^7 - (a+b+c)^7$ subject to the constraint $a^2 + b^2 + c^2 + (a+b+c)^2 = 1$ .

If we now put $X = b+c$ , $Y = a+c$ , $Z = a+b$ , this is the same as extremizing $\begin{aligned} F & = \; 2^{-7}\big[(Y+Z-X)^7 + (X+Z-Y)^7 + (X+Y-Z)^7 - (X+Y+Z)^7\big] \\ & = \; -\tfrac{7}{16}XYZ\big[3(X^4 + Y^4 + Z^4) + 10(X^2Y^2 + X^2Z^2 + Y^2Z^2)\big] \; = \; -\tfrac{7}{16}XYZ\big[5(X^2 + Y^2 + Z^2)^2 - 2(X^4 + Y^4 + Z^4)\big] \end{aligned}$ subject to the constraint $X^2 + Y^2 + Z^2 = 1$ . Thus we want to extremize $F \; = \; -\tfrac{7}{16}XYZ\big[5 - 2(X^4 + Y^4 + Z^4)\big]$ subject to the constraint $X^2 + Y^2 + Z^2 = 1$ .

The Cauchy-Schwarz Inequality tells us that $(X^2 + Y^2 + Z^2)^2 \le 3(X^4 + Y^4 + Z^4)$ and so $X^4 + Y^4 + Z^4 \ge \tfrac13$ provided that $X^2 + Y^2 + Z^2 = 1$ (with equality when $|X| = |Y| = |Z| = \tfrac{1}{\sqrt{3}}$ ). The AM/GM Inequality tells us that $|XYZ|^{\frac23} \le \tfrac13(X^2 + Y^2 + Z^2)$ , so that $|XYZ| \le \tfrac{1}{3\sqrt{3}}$ provided that $X^2 + Y^2 + Z^2 = 1$ (with equality when $|X| = |Y| = |Z| = \tfrac{1}{\sqrt{3}}$ ). Thus, subject to the constraint $X^2 + Y^2 + Z^2 = 1$ , we have $|F| \le \tfrac{7}{16} \times \tfrac{1}{3\sqrt{3}} \times \big[5 - \tfrac23\big] \; = \; \tfrac{91}{144\sqrt{3}}$ with equality when $|X| = |Y| = |Z| = \tfrac{1}{\sqrt{3}}$ .

Thus we have shown that $\big(a^7 + b^7 + c^7 + d^7\big)^2 \; \le \; \left(\tfrac{91}{144\sqrt{3}}\right)^2(a^2 + b^2 + c^2 + d^2)^7 \; = \; \tfrac{8281}{62208}(a^2 + b^2 + c^2 + d^2)^7$ provided that $a + b + c + d = 0$ , with equality when (for example) $a = b = c = \tfrac{1}{2\sqrt{3}}$ and $d = -\tfrac12\sqrt{3}$ , and so we deduce that $M = \boxed{8281}$ .

$\dfrac{M}{62'208}$ is a maximum of $f(a,b,c,d)=\dfrac{(a^7+b^7+c^7+d^7)^2}{(a^2+b^2+c^2+d^2)^7}$ , when $g(a,b,c,d)=a+b+c+d=0$ .

Extrema occur when $\nabla f$ and $\nabla g$ are collinear (where $\nabla f$ is the vector of the partial derivatives of $f$ ).

With $S=(a^7+b^7+c^7+d^7)$ and $T=(a^2+b^2+c^2+d^2)$ , we have $\nabla g=(1; 1; 1; 1)$ and

$\nabla f = 14·S·T^6\big(T·a^6 - S·a\ ; \ \ T·b^6 - S·b\ ; \ \ T·c^6 - S·c\ ; \ \ T·d^6 - S·d \big)$ .

We have collinearity:

• when $S=0$ (which gives $M=0$ for equality, that is clearly a local minimum and not a maximum);

• when $T=0$ (which gives $a=b=c=d=0$ and then $M$ can be anything for equality);

• when $T·x^6-S·x\ =\ T·y^6-S·y$ , for $x$ and $y$ any of $a$ , $b$ , $c$ or $d$ .

$T·x^6-S·x\ =\ T·y^6-S·y\ \Leftrightarrow\ x=y$ or $\dfrac{x^6-y^6}{x-y}=\dfrac{S}{T}$ .

If $\dfrac{x^6-y^6}{x-y}=\dfrac ST$ and $\dfrac{x^6-z^6}{x-z}=\dfrac ST$ , then $z=y$ (because $h(y)=\dfrac{x^6-y^6}{x-y}$ is always increasing, therefore bijective after checking what occurs at $y=x$ ).

Therefore, $a$ , $b$ , $c$ and $d$ must be the same value $x$ , or be of two distinct values $x$ and $y$ .

If all are the same values, then the sum being 0 implies that all are 0 and that $T$ =0 (case already seen).

If values are ${x,x,y,y}$ , then the sum being 0 implies that $x=-y$ and therefore $S=0$ (case already seen).

If values are ${x,x,x,y}$ , then the sum being 0 implies that $y=-3x$ .

In this extreme case, we have: $\dfrac{M}{62'208}=f(x,x,x,-3x)=\dfrac{(3-3^7)^2·x^{14}}{(3+(-3)^2)^7·x^{14}}=\dfrac{2'184^2}{12^7}$

That is, $M=62'208·\dfrac{2'184^2}{12^7}=2^8·3^5·\dfrac{2^6·3^2·7^2·13^2}{2^{14}·3^7}=7^2·13^2=\boxed{8'281}$

Not a proof, but a fruitful thought. We want to maximize the fraction $f = \frac{(a^7+..)^2}{(a^2+..)^7}$ and we observe that

• with only 2 terms a and b, they would have to satisfy b=-a and the fraction would be 0
• when adding more terms, the fraction is the highest when one of them (say a) is different while the others are equal
• multiplying a, b, c and d by the same factor will not change f, so we will set a=3 and b=c=d=-1.

In our case we get $M=62208 f_{max} = 62208×\frac{(3^7-3)^2}{(3^2+3)^7}=8281$

Assume a=b=c with a+b+c+d=0 and obtain the following desired result:

62208×(1^7 + 1^7 + 1^7 + (-3)^7)^2/(1^2 + 1^2 + 1^2 + (-3)^2)^7=8281

Answer=8281