Inequalities with Strange Equality Conditions

This page serves to debunk the myth that

An expression attains its maximum or minimum when all (some) of the variables are equal.

If $a, b$ are real numbers such that $a + b = 4$ , what is the minimum value of

$\big[(a-1) ( b-1)\big]^2?$

Clearly, since the expression is a perfect square, the minimum that it can attain is 0. This can indeed be attained, say with $a = 1, b=3$ . The minimum does not occur when $a = b = \frac{4}{2}$ , even though the expression is symmetric. $_\square$

Some inequalities are less obvious and they do not have equality case when all variables are equal.

If $x,y,z\ge 0$ satisfy $x+y+z=3$ , find the maximum value of $x^2y+y^2z+z^2x$ .

Setting $x=y=z=1$ gives $x^2y+y^2z+z^2x=3$ . However, is this the maximum value?

In fact, setting $x=0, y=2, z=1$ gives $x^2y+y^2z+z^2x=4,$ which is the actual maximum value. $_\square$

Try your hand at the list of problems below. Be warned that these problems are hard to (properly) solve, and could require a lot of ingenuity.

Let $x, y, z$ be non-negative real numbers satisfying the condition $x+y+z = 1$ . The maximum possible value of

$x^3y^3 + y^3z^3 + z^3x^3$

has the form $\frac {a} {b} ,$ where $a$ and $b$ are positive, coprime integers. What is the value of $a+b$ ?

The correct answer is: 65

Without loss of generality, we may assume that $x \geq y \geq z \geq 0$ . Since $y$ and $z$ are non-negative real numbers, $x = 1 - y - z \leq 1$ . Furthermore, since squares are non-negative,

$\left(x - \frac {1}{2} \right)^2 \geq 0 \Rightarrow x^2 - x + \frac {1}{4} \geq 0 \Rightarrow \frac {1}{4} \geq x(1-x).$

We can cube both sides of the inequality since $x (1-x) \geq 0$ , and both terms are positive. This gives $\frac {1}{64} \geq x^3(1-x)^3 = x^3(y+z)^3 =x^3 y^3 + 3x^3 y^2 z + 3 x^3 y z^2 + x^3 z^3$ . Observe that the right hand side is very close to the quantity that we want to maximize. In fact, since $0 \leq y, z \leq x$ , we have $3 x^3 y z^2 \geq 3 (y^2 z) y z^2 =3y^3 z^3 \geq y^3z^3$ . Thus, $\frac {1}{64} \geq x^3 y^3 + 3x^3 y^2 z + 3 x^3 y z^2 + x^3 z^3 \geq x^3 y^3 + y^3 z^3 + z^3 x^3$ .

In order for equality to hold throughout, we need $x = \frac {1} {2}$ in the first inequality, and $z=0$ in the last inequality. We can check that the maximium $\frac {1} {64}$ is attained at $(x, y, z) = \left( \frac {1}{2}, \frac {1}{2}, 0 \right)$ . Hence, $a + b = 1 + 64 = 65$ .

Note: This inequality is tricky because the maximum is attained at $\left( \frac {1}{2}, \frac {1}{2}, 0 \right)$ and its permutations, hence standard approaches will tend to fail to work.

If a,ba, ba,b are real numbers such that a+b=4 a + b = 4 a+b=4, what is the minimum value of

If $a, b$ are real numbers such that $a + b = 4$ , what is the minimum value of