# Maximize Σ(x^2+y)(y^2+x))/((x+y-1)^2)

• MHB
• lfdahl
In summary, the greatest real number $T$ that satisfies the given inequality is $-\frac34$. This can be proven by considering the cyclically symmetric functions $g(x,y)$, $g(y,z)$, and $g(z,x)$ and showing that their minimum value is $-\frac14$ for $x=y=z=\frac12(3\pm\sqrt7)$. Thus, the minimum value of $f(x,y,z)$ is $-\frac34$, making it the greatest real number that satisfies the given inequality.
lfdahl
Gold Member
MHB
Find the greatest real number, $T$, that satisfies the inequality:

$\frac{(x^2+y)(y^2+x)}{(x+y-1)^2}+\frac{(y^2+z)(z^2+y)}{(y+z-1)^2}+\frac{(z^2+x)(x^2+z)}{(x+z-1)^2}-2(x+y+z)\geq T$

for all real numbers $x$, $y$ and $z$, such that $x+y \ne 1$, $y+z \ne 1$ and $x+z \ne 1$.

lfdahl said:
Find the greatest real number, $T$, that satisfies the inequality:

$\frac{(x^2+y)(y^2+x)}{(x+y-1)^2}+\frac{(y^2+z)(z^2+y)}{(y+z-1)^2}+\frac{(z^2+x)(x^2+z)}{(x+z-1)^2}-2(x+y+z)\geq T$

for all real numbers $x$, $y$ and $z$, such that $x+y \ne 1$, $y+z \ne 1$ and $x+z \ne 1$.
[sp]
Let $f(x,y,z) = \frac{(x^2+y)(y^2+x)}{(x+y-1)^2}+\frac{(y^2+z)(z^2+y)}{(y+z-1)^2}+\frac{(z^2+x)(x^2+z)}{(x+z-1)^2}-2(x+y+z)$. Then $f(x,y,z)$ is the cyclically symmetric sum of three functions of two variables, one of which is $g(x,y) = \dfrac{(x^2+y)(y^2+x)}{(x+y-1)^2} - (x+y)$.

It seems like a good guess that $g(x,y)$ will be minimised when $y=x$ (and no doubt MarkFL would produce a "cyclic symmetry" argument to justify that). But $$g(x,x) = \frac{(x^2+x)^2 - 2x(2x-1)^2}{(2x-1)^2} = \frac{x^4 - 6x^3 + 9x^2 -2x}{(2x-1)^2} = \frac{\bigl(x^2 - 3x + \frac12\bigr)^2 - \frac14(2x-1)^2}{(2x-1)^2} = \biggl(\frac{x^2 - 3x + \frac12}{2x-1}\biggr)^2 - \frac14.$$ That has a lower bound of $-\frac14$, attained when $x$ is a root of $x^2 - 3x + \frac12$, namely $x = \frac12(3\pm\sqrt7).$

So the next step is to see whether $-\frac14$ is a lower bound for $g(x,y)$. In fact, $$g(x,y) + \frac14 = \frac{(x^2+y)(y^2+x) - \bigl(x+y-\frac14 \bigr)(x+y-1)^2}{(x+y-1)^2} = \frac{x^2y^2 - 3xy(x+y) + \frac94(x^2+y^2) + \frac{11}2xy - \frac32(x+y) + \frac14}{(x+y-1)^2}.$$ But polarising the expression in the above calculation for $g(x,x)$ gives exactly the numerator of that fraction, namely $$\bigl(xy - \tfrac32(x+y) + \tfrac12\bigr)^2 = x^2y^2 - 3xy(x+y) + \tfrac94(x^2+y^2) + \tfrac{11}2xy - \tfrac32(x+y) + \tfrac14.$$ Therefore $$g(x,y) = \biggl(\frac{xy - \tfrac32(x+y) + \tfrac12}{x+y-1}\biggr)^2 - \frac14,$$ so that $g(x,y)$ has lower bound $-\frac14$, attained when $x = y = \frac12(3\pm\sqrt7).$

The same argument applies to the other two cyclically symmetric functions $g(y,z)$ and $g(z,x)$. So the minimum value of $f(x,y,z) = g(x,y) + g(y,z) + g(z,x)$ is $-\frac34$, attained when $x = y = z = \frac12(3\pm\sqrt7).$

[/sp]

Last edited:
Opalg said:
[sp]
Let $f(x,y,z) = \frac{(x^2+y)(y^2+x)}{(x+y-1)^2}+\frac{(y^2+z)(z^2+y)}{(y+z-1)^2}+\frac{(z^2+x)(x^2+z)}{(x+z-1)^2}-2(x+y+z)$. Then $f(x,y,z)$ is the cyclically symmetric sum of three functions of two variables, one of which is $g(x,y) = \dfrac{(x^2+y)(y^2+x)}{(x+y-1)^2} - (x+y)$.

It seems like a good guess that $g(x,y)$ will be minimised when $y=x$ (and no doubt MarkFL would produce a "cyclic symmetry" argument to justify that). But $$g(x,x) = \frac{(x^2+x)^2 - 2x(2x-1)^2}{(2x-1)^2} = \frac{x^4 - 6x^3 + 9x^2 -2x}{(2x-1)^2} = \frac{\bigl(x^2 - 3x + \frac12\bigr)^2 - \frac14(2x-1)^2}{(2x-1)^2} = \biggl(\frac{x^2 - 3x + \frac12}{2x-1}\biggr)^2 - \frac14.$$ That has a lower bound of $-\frac14$, attained when $x$ is a root of $x^2 - 3x + \frac12$, namely $x = \frac12(3\pm\sqrt7).$

So the next step is to see whether $-\frac14$ is a lower bound for $g(x,y)$. In fact, $$g(x,y) + \frac14 = \frac{(x^2+y)(y^2+x) - \bigl(x+y-\frac14 \bigr)(x+y-1)^2}{(x+y-1)^2} = \frac{x^2y^2 - 3xy(x+y) + \frac94(x^2+y^2) + \frac{11}2xy - \frac32(x+y) + \frac14}{(x+y-1)^2}.$$ But polarising the expression in the above calculation for $g(x,x)$ gives exactly the numerator of that fraction, namely $$\bigl(xy - \tfrac32(x+y) + \tfrac12\bigr)^2 = x^2y^2 - 3xy(x+y) + \tfrac94(x^2+y^2) + \tfrac{11}2xy - \tfrac32(x+y) + \tfrac14.$$ Therefore $$g(x,y) = \biggl(\frac{xy - \tfrac32(x+y) + \tfrac12}{x+y-1}\biggr)^2 - \frac14,$$ so that $g(x,y)$ has lower bound $-\frac14$, attained when $x = y = \frac12(3\pm\sqrt7).$

The same argument applies to the other two cyclically symmetric functions $g(y,z)$ and $g(z,x)$. So the minimum value of $f(x,y,z) = g(x,y) + g(y,z) + g(z,x)$ is $-\frac34$, attained when $x = y = z = \frac12(3\pm\sqrt7).$

[/sp]

Thankyou, Opalg, for your exemplary analytical solution. You even solved the problem without using the "cyclic symmetry" argument! ;) Great!

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