Extreme minimization

We are asked to minimize $\left(\frac{x}{y}\right)^4 + \left(\frac{y}{x+y}\right)^4 + \left(\frac{x+y}{x}\right)^4$ over all $x,y > 0$ . This expression is homogeneous in $x$ and $y$ , so this is equvalent to asking to minimize (putting $x=1$ ): $A(y) \; = \; y^{-4} + \left(\frac{y}{1+y}\right)^4 + (1+y)^4$ Now $A'(y) \; =\; \frac{4(1+y+y^2)^2B(y)}{y^5(1+y)^5}$ where $B(y) \; = \; y^9 + 6y^8 +13y^7 + 10y^6 - 2y^5 - y^4 + 3y^3 - y^2 - 3y - 1$ Since $B(y) \; = \; y(y^8 + 6y^7 + 13y^6 + 10y^5 - 2y^4 - y^3 + 2y^2 + y - 3) - 1 \; = \; (y +1)(y^8 + 5y^7 + 8y^6 + 2y^5 - 4y^4 + 3y^3 - y - 2) - 1$ we deduce that $A(y) \; = \; (y^8 + 6y^7 + 13y^6 + 10y^5 - 2y^4 - y^3 + 2y^2 + y - 3)^4 + y^4(y^8 + 5y^7 + 8y^6 + 2y^5 - 4y^4 + 3y^3 - y - 2)^4 + (1+y)^4$ for any $y$ such that $A'(y) = 0$ , namely such that $B(y) = 0$ . Finding the polynomial remainder when dividing this last polynomial in $y$ by $B(y)$ , we deduce that $A(y) \; = \; -28y^8 - 155y^7 - 291y^6 - 140y^5 + 130y^4 - 24y^3 - 65y^2 + 71y + 55$ for any $y$ such that $B(y) = 0$ . If we define $m = -28y^8 - 155y^7 - 291y^6 - 140y^5 + 130y^4 - 24y^3 - 65y^2 + 71y + 55$ , and calculate the remainder when the expression $m^3 - 7m^2 + 3m$ is divided by $B(y)$ , we obtain the constant $898$ .

Thus any extremal value of the original problem is a solution of the cubic $m^3 - 7m^2 + 3m - 898 = 0$ , and so certainly any minimum value of the problem is a solution of that cubic. Hence we see that $p = \boxed{-898}$ .