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

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}

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

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 .

• @Rodrigo This problem is from my practice textbook . Aug 11, 2022 at 10:17
• @Sourav Yes because products of two positive number is also positive and preserve the sign of inequality Aug 11, 2022 at 10:18
• @Sourav Sorry , I will edit it. Aug 11, 2022 at 10:21
• If the question is to find the minimum value of $A$, how can the answer be an expression in $x,y,z$? I'm afraid that the intended answer doesn't make sense, the way the question is phrased. Aug 12, 2022 at 12:54
• @YiFan Yes, I agree. I treated it as different inequality
– User
Aug 12, 2022 at 13:35

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.

• I don't know much about Cauchy Schwars . Anyway thanks for a solution . But also , Please adress that why my procedure was wrong Aug 11, 2022 at 10:20
• I see you edited the question
– User
Aug 11, 2022 at 11:16
• Yes but only that x,y,z are positive numbers . Aug 11, 2022 at 11:50
• Does this mean that there is no certain minimum ? We will get different minimum values for different x,y,z? Aug 13, 2022 at 6:42
• Please strive not to post more (dupe) answers to dupes of FAQs, cf. recent site policy announcement here.Instead of answering it, it would be better to look for a good duplicate target, or help the user by posting comments.You can use this Aug 13, 2022 at 7:01

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 $$4/27$$.

• Numerical minimum value is 0 at $x,y\approx0,z\approx1$ Aug 12, 2022 at 20:55
• The minimum value is not $\dfrac{4}{7}$, and it should be $0$ as Suzu Hirose pointed out earlier. The correct question should be to find the maximum rather than the minimum. Aug 13, 2022 at 6:00
• @SuzuHirose No , it is given that x,y and z are positive . They can't be zero . Do you mean minimum value tends to zero but never equals to zero ? Aug 13, 2022 at 6:29
• @Get_Maths $\approx$ means "approximately", so $x, y=0.0000000001$, $z=1-x-y\approx1$ or something then $A\approx 0$. It never gets to exactly zero of course but it goes as close as you like, certainly smaller than $4/7$. Aug 13, 2022 at 6:33
• @SuzuHirose Does this mean that problem is wrong ? I am so confused with inequalities . I have found similar results with other problems too i.e. getting different expressions than book using correct methods. Please tell What should I do? Aug 13, 2022 at 6:35