The Inequality Inspired by Joel Tan

Algebra Level 5

Let $x, y, z\geq 0$ be reals such that $x+y+z=1$ .

Find the maximum possible value of

$x (x+y)^5(y+z)^4(x+z)^4$

The answer can be written as $\dfrac {a^i}{b^jc^k}$ for positive integers $a, b, c, i, j, k$ , where $a, b, c$ are as small as possible. Find $a+b+c+i+j+k$ .

Inspiration

The answer is 30.

1 solution

Daniel Liu
Jun 24, 2015

By AM-GM, $\dfrac{1}{2}=\dfrac{x+y+z}{2}\ge \sqrt{x(y+z)} \implies x(y+z)\le \dfrac{1}{4}$ With equality case $x=y+z$

By AM-GM, $\begin{aligned} \dfrac{1}{6}&=\dfrac{\dfrac{5(x+y)}{5}+\dfrac{3(y+z)}{3}+\dfrac{4(x+z)}{4}}{12}\\ &\ge \sqrt[12]{\dfrac{1}{5^5\cdot 3^3\cdot 4^4}(x+y)^5(y+z)^3(x+z)^4}\\ &\implies (x+y)^5(y+z)^3(x+z)^4\le \dfrac{5^5\cdot 3^3\cdot 4^4}{6^{12}} =\dfrac{5^5}{3^9\cdot 2^4} \end{aligned}$ with equality case $\dfrac{x+y}{5}=\dfrac{y+z}{3}=\dfrac{x+z}{4}$

Multiplying the above two inequalities gives $x(x+y)^5(y+z)^4(x+z)^4\le \dfrac{5^5}{3^9\cdot 2^6}$ But can this be reached?

Solving for the equality case, we see that $(x,y,z)=\left(\dfrac{1}{2}, \dfrac{1}{3}, \dfrac{1}{6}\right)$ satisfies both.

This means our answer is $5+5+3+9+2+6=\boxed{30}$ and we're done.

Moderator note:

To de-mystify this solution slightly, let's see how we can "guess" at these constants. We want to apply AM-GM to

$x A^5 \left( \frac{ 1-z} { A} \right)^5 B^4 \left( \frac{ 1-x} { B} \right)^4 C^4 \left( \frac{ 1-y} { C} \right)^5 \\ \leq A^5 B^4 C^4 ( \frac{x + 5 \frac{1-z}A + 4 \frac{1-x} B + 4 \frac{ 1-y} { C }}{1+5+4+4} ) ^ {1 + 5 + 4 + 4 }$

We have several conditions:
1) original condition : $x +y + z = 1$
2) AM-GM equality condition: $x = \frac{1-z}A = \frac{1-x} B = \frac{1-y}C$
3) RHS is a constant condition: $1 - \frac{4}{B} = - \frac{4}{C} = - \frac{ 5}{A}$

From 3, we get $B = \frac{4A}{A+5}, C = \frac{4A}{5}$ .
Substitute into 2, we get $x = \frac{A+5} { 5 ( A + 1) }, y = \frac{ -4A^2 + 5A + 25 } { 25 (A+1) }, z = \frac{ 5-A^2} { 5 ( A + 1) }$ .
Substitute into 1, we get $9 A^2 + 5 A - 50 = 0$ .

Thus, we get $A = \frac{5}{3} , - \frac{10}{3}$ . Reject the negative value. This gives us $B = 1, C = \frac{4}{3}$ , $x = \frac{1}{2} , y = \frac{ 1}{3} , z = \frac{1}{6}$ . Now, substituting these constants into the inequality at the top, we get:

$x (1-z) ^ 5 ( 1-x) ^ 4 ( 1 - y) ^ 4 \leq \left( \frac{ 5}{3} \right)^5 1 ^ 4 \left( \frac{4}{3} \right) ^ 4 \left( \frac{7}{14} \right) ^ {14} = \frac{5^5} { 3^9 2^6 }$

with equality case at $( \frac{1}{2}, \frac{1}{3}, \frac{1}{6} )$ .

A great question and may I say that it's beautiful that the values of $(x,y,z)$ turn out to be $(\frac{1}{2},\frac{1}{3},\frac{1}{6})$ . Gorgeous.

Garrett Clarke - 5 years, 11 months ago

Does the equality case only works for $(\frac{1}{2}, \frac{1}{3}, \frac{1}{6}$ )? I solved for the equality case, and it turns out that the equality is attained iff $x = \frac{3y}{2} = 3z$ .

Reineir Duran - 5 years, 5 months ago

The Inequality Inspired by Joel Tan

The answer is 30.

1 solution

Moderator note:

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