Finding the minimum value of the expression $xy(x+y)^2 + yz(y+z)^2 + zx(z+x)^2$

Question

From R. D. Sharma's Objective Mathematics,

Given that $x + y + z = 1$, ($x,y,z$ are positive real numbers) find the minimum value of $$A = xy(x+y)^2 + yz(y+z)^2 + zx(z+x)^2$$

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My attempt: By A.M.-G.M. inequality,

$$\begin{align} (x+y)^2 &\geq 4(xy) \\ (y+z)^2 &\geq 4(yz) \\ (z+x)^2 &\geq 4(zx) \end{align}$$

By multiplying by $xy$, $yz$, $zx$ in these equations respectively, gives

$$\begin{align} xy(x+y)^2 &\geq 4(xy)^2 \\ yz(y+z)^2 &\geq 4(yz)^2 \\ zx(z+x)^2 &\geq 4(zx)^2 \\ \end{align}$$

Adding these equations, we get

$$ A \geq 4[(xy)^2 + (yz)^2 + (zx)^2] $$

Using AM GM inequality once again on RHS, we get

$$ A \geq 12(xyz)^{\tfrac43}$$

However, provided answer key says that minimum value is $4xyz$. I do not want a solution but only wants to know that is something wrong in my procedure? Why?

EDIT : This question has been identified as a possible duplicate of another question . However, readers will believe that my problem is different if they re-read the paragraph just above .

Answer 1

Note that,

If $x,y\to 0$ with $z\to 1$, we have

$$\inf \left\{xy(x+y)^2 + yz(y+z)^2 + zx(z+x)^2\mid x+y+z=1 \wedge x>0\wedge y>0\wedge z>0\right\}=0$$

As for obtaining the inequality $A≥4xyz$, you can easily obtain this result using the Cauchy-Schwarz inequality:

$$(xy(x+y)^2 + yz(y+z)^2 + zx(z+x)^2)\left(\frac 1{xy}+\frac 1{yz}+\frac 1{xz}\right)≥4(x+y+z)^2=4$$

$$\frac A{xyz}≥4\implies A≥4xyz.$$

But, note that $xyz$ is not a constant. Therefore, $4xyz$ is not a "minimum". We only proved that,

$$xy(x+y)^2 + yz(y+z)^2 + zx(z+x)^2≥4xyz$$

The inequality you want to prove.

Answer 2

You have not used the condition of $x+y+z=1$. Since when $x=y=z=1/3$ the equality holds, your answer is the same with the $4xyz$ answer. So the textbook answer is flawed. you can get a numerical minimum value of $4/27$.