I have tried am-gm inequality,i am getting that $xyz$ is greater than $36.9$.
I tried hit and trial,but it is of no use also.
Could anyone give a definite process?
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As pointed out by anant, please clarify if your domain is non negative integers, or non-negative reals. The approaches are different. – Calvin Lin Apr 28 '14 at 14:31
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In case you're interested, there's another way of solving this if the numbers are positive reals: just use the inequalities $$xyz\le \left(\frac{x+y+z}{3}\right)^3$$ $$xy+yz+zx\le \frac{(x+y+z)^2}{3}$$ Which, again, shows that the maximum is reached when $x=y=z$ given that they are positive reals. But this doesn't fully satisfy the conditions, just like Calvin's answer doesn't either. – user26486 Apr 28 '14 at 15:35
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Can u say how u get the second inequality. – user146181 Apr 28 '14 at 17:13
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@user146181 You should've written '@mathh' so that I can see that you're asking me a question (on my notifications). You're lucky I just went through this question again and saw your comment. First of all, talking about the inequality, prove that $a^2+b^2+c^2\ge ab+bc+ca$ by getting that $(a-b)^2+(b-c)^2+(c-a)^2\ge 0$ after multiplying both sides by $2$. After you've proved it, add $2(ab+bc+cd)$ to both sides of the inequality and use factorization... I'm sure you can finish the proof. – user26486 Apr 28 '14 at 17:17
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@user146181 Is it clear now? – user26486 Apr 28 '14 at 17:21
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@mathh i got that a2+b2+c2≥ab+bc+ca. – user146181 Apr 28 '14 at 17:53
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As I said in the comment, add $2(ab+bc+ca)$ to both sides of the inequality. You now have $$a^2+b^2+c^2+2(ab+bc+ca)=(a+b+c)^2\ge 3(ab+bc+ca)$$ – user26486 Apr 28 '14 at 18:01
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@mathh Thanks,for spending your precious time. – user146181 Apr 28 '14 at 18:06
2 Answers
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Hint: $x+1+y+1+z+1=13$
Bound $(x+1)(y+1)(z+1)$ from above using AM-GM.
Hence maximum is $(13/3)^3-11$ and occurs when $x=y=z$.
Calvin Lin
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@anant considering that OP used decimals in his answer, I believe that he wants a domain of real numbers. If it was integers, there are just a few cases to check – Calvin Lin Apr 28 '14 at 14:30
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Okay. For integers, how would we approach the problem if $x + y + z$ was large (say 2000)? – Anant Apr 28 '14 at 14:37
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@anant: For integers, the integers nearest to the value satisfying ${x+y+z=2000}$ when ${x=y=z}$ will yield the greatest value to the equation. For ${2000}$ it will be ${666,667,667}$ for ${x,y,z}$. :) – Abir Mukherjee Apr 28 '14 at 14:50
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Yes,i got it but can u inform me how u prepared exactly (x+1)(y+1)(z+1). – user146181 Apr 28 '14 at 17:06
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1I'm not sure why this answer has so many upvotes given that it doesn't take into account the fact that $x,y,z$ are non-negative integers. – user26486 Apr 28 '14 at 17:26
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@Calvin Lin Yes,i got it but can u inform me how u prepared exactly (x+1)(y+1)(z+1). – user146181 Apr 28 '14 at 18:18
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@mathh See my clarification attempt (first comment on the problem). I worked over the positive reals as the OP gave a decimal value, which indicated to me that he was thinking of it as real values. Had his answer been an integer, I would have done a case argument. – Calvin Lin Apr 29 '14 at 02:01
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@user146181 Experience. That looked like a good factorization to consider, mathh's solution in comments works too. – Calvin Lin Apr 29 '14 at 02:03
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$xyz+xy+yz+zx=xyz~\bigg(1+\dfrac1x+\dfrac1y+\dfrac1z\bigg)$, but the harmonic mean of the three variables is $H=\dfrac3{\dfrac1x+\dfrac1y+\dfrac1z}\iff\dfrac1x+\dfrac1y+\dfrac1z=\dfrac3H$ . At the same time, $A>G>H$, with $A=\dfrac{10}3$ and
$G=\sqrt[3]{xyz}~=>$ we are left with maximizing $G^3\bigg(1+\dfrac3H\bigg)\geqslant G^3\bigg(1+\dfrac3G\bigg)$, which lower bound peaks for $G_\text{max}=A=\dfrac{10}3$ , which only happens when $x=y=z$. Since the three variables are integers, the solutions obtained by rounding are $3,3$, and $4$, for which we have a maximum of $69$, which is just shy of the global maximum $\dfrac{1900}{3^3}\approx70.37$, obtained for $x=y=z=3\dfrac13\cdot$ QED.
Lucian
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