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 .

  • $\begingroup$ @Rodrigo This problem is from my practice textbook . $\endgroup$
    – Get_ Maths
    Aug 11, 2022 at 10:17
  • $\begingroup$ @Sourav Yes because products of two positive number is also positive and preserve the sign of inequality $\endgroup$
    – Get_ Maths
    Aug 11, 2022 at 10:18
  • $\begingroup$ @Sourav Sorry , I will edit it. $\endgroup$
    – Get_ Maths
    Aug 11, 2022 at 10:21
  • $\begingroup$ 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. $\endgroup$
    – YiFan Tey
    Aug 12, 2022 at 12:54
  • $\begingroup$ @YiFan Yes, I agree. I treated it as different inequality $\endgroup$
    – User
    Aug 12, 2022 at 13:35

2 Answers 2


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.

  • $\begingroup$ I don't know much about Cauchy Schwars . Anyway thanks for a solution . But also , Please adress that why my procedure was wrong $\endgroup$
    – Get_ Maths
    Aug 11, 2022 at 10:20
  • $\begingroup$ I see you edited the question $\endgroup$
    – User
    Aug 11, 2022 at 11:16
  • $\begingroup$ Yes but only that x,y,z are positive numbers . $\endgroup$
    – Get_ Maths
    Aug 11, 2022 at 11:50
  • $\begingroup$ Does this mean that there is no certain minimum ? We will get different minimum values for different x,y,z? $\endgroup$
    – Get_ Maths
    Aug 13, 2022 at 6:42
  • $\begingroup$ 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 $\endgroup$ 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 $4/27$.

  • 1
    $\begingroup$ Numerical minimum value is 0 at $x,y\approx0,z\approx1$ $\endgroup$ Aug 12, 2022 at 20:55
  • $\begingroup$ 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. $\endgroup$
    – Wang YeFei
    Aug 13, 2022 at 6:00
  • $\begingroup$ @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 ? $\endgroup$
    – Get_ Maths
    Aug 13, 2022 at 6:29
  • $\begingroup$ @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 $4/7$. $\endgroup$ Aug 13, 2022 at 6:33
  • $\begingroup$ @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? $\endgroup$
    – Get_ Maths
    Aug 13, 2022 at 6:35

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