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Let $x^2+y^2+z^2\leq27$ and $P = x+y+z+xy+yz+zx$. Find the value of $x, y, z$ such that $P$ is the maximum value and minimum value.

My attempt :

$$(x-y)^2 + (y-z)^2 + (z-x)^2 \geq 0$$

$$27 \geq x^2+y^2+z^2 \geq xy+yz+zx\tag{1}$$

$$(x+y+z)^2 \leq 3(x^2+y^2+z^2) \le 3 \cdot 27$$

$$(x+y+z)^2 \leq 81$$

$$x+y+z \leq 9\tag{2}$$

From $(1), (2)$, $ x+y+z+xy+yz+zx \leq 36$, so $P_{\text{max}} = 36$ with equality hold at $x=y=z=3$.

Please suggest how to find $P_{\text{min}}$.

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  • $\begingroup$ You can label equations using \tag{label}. $\endgroup$
    – Shaun ♦
    Commented Apr 28, 2017 at 13:53
  • $\begingroup$ Would you explain what facts you're using to get the different lines in your attempt, please? It's not clear what you've done. $\endgroup$
    – Shaun ♦
    Commented Apr 28, 2017 at 14:02
  • $\begingroup$ @Shaun The OP first finds an upper bound for $P$ and shows that this bound is sharp. It looks fine to me. $\endgroup$
    – user1551
    Commented Apr 28, 2017 at 14:11
  • $\begingroup$ @user1551 Ah, I see now. Thank you for clarifying it. $\endgroup$
    – Shaun ♦
    Commented Apr 28, 2017 at 14:24
  • $\begingroup$ You can probably bruteforce the problem with Lagrange multipliers (lookup KKT theorem) $\endgroup$
    – Gabriel Romon
    Commented Apr 28, 2017 at 14:55

2 Answers 2

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You may use the same method to find the minimum. First, we obtain a lower bound for $P$: $$ \begin{align} P&=x+y+z+xy+yz+zx\\ &=\frac12 [ (x+y+z+1)^2 - (x^2+y^2+z^2) - 1 ]\\ &\ge\frac12 (0 - 27 - 1)\tag{1}\\ &= -14. \end{align} $$ Next, note that at $\left(\frac{\sqrt{53}-1}2,-\frac{\sqrt{53}+1}2,0\right)$, we have $x+y+z+1=0$ and $x^2+y^2+z^2=27$. Hence tie can occur in $(1)$ and the lower bound $-14$ is attainable.

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  • $\begingroup$ $P=x+y+z+xy+yz+zx=\frac{1}{2} [ (x+y+z+1)^2 -(x^2+y^2+z^2) - 1 ]$ How is this derived ? $\endgroup$
    – user403160
    Commented Apr 28, 2017 at 16:44
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    $\begingroup$ @carat Expand the brackets and collect terms. $\endgroup$
    – user1551
    Commented Apr 28, 2017 at 16:47
  • $\begingroup$ I got it. Thank you very much, user1551. $\endgroup$
    – user403160
    Commented Apr 28, 2017 at 16:58
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A minimum is $-14$.

Prove that $$(x+y+z)\sqrt{\frac{x^2+y^2+z^2}{27}}+xy+xz+yz+\frac{14}{27}(x^2+y^2+z^2)\geq0$$

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    $\begingroup$ Please give me hint on how to prove that. $\endgroup$
    – user403160
    Commented Apr 28, 2017 at 15:58
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    $\begingroup$ @carat Let $x^2+y^2+z^2=k(xy+xz+yz)$. Finely we need to prove that $(13k+27)^2\geq0$. $\endgroup$
    – Michael Rozenberg
    Commented Apr 29, 2017 at 6:52
  • $\begingroup$ Thank you very much, Michael Rozenberg. I'll try. $\endgroup$
    – user403160
    Commented Apr 29, 2017 at 13:31

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