# Show $2(x+y+z)-xyz\leq 10$ if $x^2+y^2+z^2=9$

If $x,y,z$ are real and $x^2+y^2+z^2=9$, how can we prove that $2(x+y+z)-xyz\leq 10$?

Please provide a solution without the use of calculus. I know the solution in that way.

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Does the solution you know uses Lagrange multipliers ? – Thibaut Dumont Dec 6 '12 at 9:43
@ThibautDumont: I think that's what he means by "without the use of calculus". – Mhenni Benghorbal Dec 6 '12 at 9:48
I might investigate expansions of $(x-1)(y-1)(z-1)$ and $(x+y+z)(x+y+z)$. All the terms that you want will show up, and you might be able to put bounds on the extraneous terms. – orlandpm Dec 7 '12 at 2:00
@orlandpm I already tried it that way.Doesn't seem to produce anything fruitful. – Sai Krishna Deep Dec 7 '12 at 6:57

Case 1: At least one of $x,y,z < 0$. Let's say $z < 0$. If we let $f(x,y,z) = 2(x+y+z) - xyz$, it's then clear that $f(x,y,z) \leq f(t,t,z)$, where $t = \sqrt{\frac{x^2+y^2}{2}}$. It then suffices to check $f(t,t,z) \leq 10$. Let $z = -u$, so that $u$ is positive. Then we need to show $$f(t,t,-u) = 2(2t - u) + t^2 u \leq 10$$ on the condition $2t^2 + u^2 = 9$. We can try to express everything in terms of $u$: $$2(2t - u) + t^2 u \leq 10 \\ \Leftrightarrow 4(2t - u) + (9 - u^2)u \leq 20 \\ \Leftrightarrow u^3 - 5u + 20 \ge 8t = 8 \sqrt{\frac{9 - u^2}{2}} \\ \Leftrightarrow (u^3 - 5u + 20)^2 - 32 (9-u^2) \ge 0 \\ \Leftrightarrow (u-1)^2(u^4 + 2u^3 - 7u^2 + 24u + 112) \ge 0$$

The last inequality is clearly true, since $u^4 + 49 \ge 14u^2 > 7u^2$. So this case holds, with equality $z = -u = -1$, $x = y = t = 2$.

Case 2: $x,y,z \ge 0$. WLOG assume $x \ge y \ge z$.

Case 2a: $x^2 \leq \frac{9 + \sqrt{65}}{2}$. ($x$ is "small")

We check that $f(x,y,z) \leq f(x, \sqrt{y^2+z^2}, 0)$. $$2(x+y+z) - xyz \leq 2(x+\sqrt{y^2+z^2}) \\ \Leftrightarrow 2(y+z - \sqrt{y^2+z^2}) \leq xyz \\ \Leftrightarrow \frac{4}{y+z+\sqrt{y^2+z^2}} \leq x \\ \Leftrightarrow 4 \leq x(y+z+\sqrt{y^2+z^2}$$ The last inequality follows from $$x(y+z+\sqrt{y^2+z^2}) \ge 2x\sqrt{y^2+z^2} = 2 x \sqrt{9-x^2} \ge 4$$ and the condition on $x$.

Thus we may assume $z = 0$. Then $f(x,y,z) = 2(x+y) \leq 2\sqrt{2(x^2+y^2)} = 2\sqrt{18} < 10$.

Case 2b: $x^2 \ge \frac{9 + \sqrt{65}}{2}$. ($x$ is "large")

We get a trivial bound on $y+z$ by Cauchy-Schwarz: $$y + z \leq \sqrt{2(y^2+z^2)} = \sqrt{2(9-x^2)} \leq \sqrt{9 - \sqrt{65}} < 1$$ Then $$f(x,y,z) = 2(x+y+z) - xyz < 2(x+y+z) \leq 2 (3+1) < 10.$$

All in all, it's possible to do this without calculus, although I can't understand why one would not use a tool like Lagrange multiplier that is especially well suited for this situation. It's exactly the power of those tools that render brute-force solutions like this one unnecessary.

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without loss of generality we can assume that $|x|\le |y|\le |z|$,and Note that $3z^2\ge x^2+y^2+z^2$, which implies that $z^2\ge 3$

we have \begin{align} [2(x+y+z)-xyz]^2&=[2(x+y)+z(2-xy)]^2\\ &\le [(x+y)^2+z^2][4+(2-xy)^2]\\ &=(9+2xy)(8-4xy+(xy)^2))\\ &=2(xy)^3+(xy)^2-20(xy)+72\\ &=(xy+2)^2(2xy-7)+100 \end{align} from $z^2\ge 3$ it follows that $2xy\le x^2+y^2=9-z^2\le 6$,then $$[2(x+y+z)-xyz]^2\le 100$$ or equivalently $$2(x+y+z)-xyz\le 10$$

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Here is an attempt. I believe it has to be reconsidered for negative numbers, though.

$0\leq (x+y+z-1)^2$

$0\leq x^2+y^2+z^2+1-2(x+y+z)+2xyz(\frac{1}{x}+\frac{1}{y}+\frac{1}{z})$

$2(x+y+z)-2xyz(\frac{1}{x}+\frac{1}{y}+\frac{1}{z})\leq x^2+y^2+z^2+1$

EDIT: As pointed out my bound is the wrong way :( But I hope the first steps gives others some ideas.

That's almost what we want. We only need to bound the harmonic mean by the quadratic mean now.

$\frac{3}{\frac{1}{x}+\frac{1}{y}+\frac{1}{z}}\leq \sqrt{\frac{x^2+y^2+z^2}{3}}$

In our case: $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\geq \sqrt{3}$

which is enough.

Hope there is no mistake :)

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You need an upper bound for $1/x + 1/y + 1/z$ to get conclusion, not the lower bound. – user27126 Dec 8 '12 at 9:37
Oh true. And that means this way cannot work? :( – Gerenuk Dec 8 '12 at 9:45