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 $$A = xy(x+y)^2 + yz(y+z)^2 + zx(z+x)^2$$
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 .
Best Answer
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.