The Inequality Inspired by Joel Tan 2

Algebra Level 5

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

Find the maximum possible value of

$x (x+y)(y+z)^6(x+z)^2$

The answer can be written as $\dfrac {a^x}{b^y}$ for positive integers $a, b, x, y$ , where $a, b$ are as small as possible. Find $a+b+x+y$ .

Inspiration

The answer is 34.

2 solutions

Daniel Liu
Jun 24, 2015

By AM-GM, $\dfrac{1}{5}=\dfrac{x+\dfrac{4(y+z)}{4}}{5}\ge \sqrt[5]{\dfrac{1}{4^4}x(y+z)^4} \implies x(y+z)^4\le \dfrac{4^4}{5^5}=\dfrac{2^8}{5^5}$ With equality case $x=\dfrac{y+z}{4}$

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

Multiplying the above two inequalities gives $x(x+y)(y+z)^6(x+z)^2\le \dfrac{2^{17}}{5^{10}}$ But can this be reached?

Solving for the equality case, we see that $(x,y,z)=\left(\dfrac{1}{5}, \dfrac{1}{5}, \dfrac{3}{5}\right)$ satisfies both.

This means our answer is $2+17+5+10=\boxed{34}$ and we're done.

I don't quite like splitting out the AM-GM into 2 inequalities and multiplying them, as it is hard to justify the initial steps from a problem solver POV, instead of a problem creator POV. See my note on your other solution to see how all of these "magical constants" can be determined.

Calvin Lin Staff - 5 years, 11 months ago

$Why\quad x+4\cfrac { \left( y+z \right) }{ 4 } \quad instead\quad of\quad any\quad other\quad x+a\cfrac { \left( y+z \right) }{ a } \quad ?$

Ayush Verma - 5 years, 11 months ago

So that he can attain the respective indices.

sandeep Rathod - 5 years, 11 months ago

It turns out that $a=4$ is the "magic constant" that makes both equality cases for both AM-GM's work out. See Calvin's comment on the solution here to see how to solve for these "magic constants".

Daniel Liu - 5 years, 11 months ago

Joshua Prettyman
Jul 2, 2015

Write in terms of $x$ and $y$ using $z=1-x-y$ then take the gradient of the expression in the $x,y$ plane and equate to zero. Ignore the possible solutions which give $x=0$ , $x=1$ or $y=1$ since these give a value of zero and are thus obviously not maxima. After simplifying, we get the system: $1-11x-3y+21xy+16x^2 = 0$ $1-2x = 3y$ which gives the solution $x=1/5, y=1/5$ in the region $[0,1]^2$ . These values give the expression the value $\frac{2^{17}}{5^{10}}$

Interesting solution. Can you explain in more detail what you did, since I'm not very familiar with "taking the gradient" (or calculus in general). Thanks!

Daniel Liu - 5 years, 11 months ago

To take the gradient of a function of a single variable (lets use $x^2$ you just differentiate the expression. Take my word for it that $\frac{d}{dx}x^2 = 2x$ , this reads "the derivative of $x^2$ with respect to $x$ (that's the $\frac{d}{dx}$ bit) equals $2x$ ". I don't have space here to tell you how to take derivatives, you can find this out elsewhere, but lets just say that almost every function you are familiar with (polynomials, sine, cosine, exponentials, etc.) have derivatives and the methods for finding derivative are well known in these cases. This tells you that if you draw the function $x^2$ you and you look at the gradient of the curve at, for example, the point where $x=3$ , you will find that the gradient is $2x=6$ , the 'steepness' of a line that just touches the curve at the point $x=3$ will be 6. You should also know that if the gradient is $0$ then the function is either at a maximum or a minimum. For example if the gradient of $x^2$ is zero then we have $2x=0$ which means $x=0$ , if you know what the graph of $x^2$ looks like then you will know there is a minimum at the point $x=0$ .

Now in 2D we might have a function of $x$ and $y$ . In this question the function is $x(x+y)^2(y+z)^6(z+x)^2$ (it looks like a function of three variable but we can substitute $z=1-x-y$ ). lets keep it simple and instead look at the function $x^2(y^2+1)$ . The gradient of this is $\nabla x^2(y^2+1) = \left( \begin{array}{c} \partial_x \\ \partial_y \end{array} \right) x^2(y^2+1) = \left( \begin{array}{c} \frac{d}{dx}x^2(y^2+1) \\ \frac{d}{dy} x^2(y^2+1)\end{array} \right) = \left( \begin{array}{c} 2x(y^2+1) \\ 2x^2y\end{array} \right)$ You might notice that this is a vector (the number on top tells us about the x-direction and the number on the bottom tells us about the y-direction). If the function is at a maximum then $\left( \begin{array}{c} 2x(y^2+1) \\ 2x^2y\end{array} \right) = 0$ , that is, both the top and bottom things are zero. This is a bit of a stupid example because as you can see this just gives $x=0$ and $y=$ anything, the function is the max/min value on the whole line $x=0$ , in fact the function is zero all the way along this line, like a valley. In the function in the question however we have just a single value (at x=1/5, y=1/5), not a whole line, so this point is what we're looking for.

Joshua Prettyman - 5 years, 10 months ago

The Inequality Inspired by Joel Tan 2

The answer is 34.

2 solutions

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