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This is from Apostol's Calculus, Vol. II, Section 9.15 #11:

Find the maximum of $f(x,y,z)=\log x + \log y + 3 \log z$ on that portion of the sphere $x^2+y^2+z^2=5r^2$ where $x>0,\,y>0,\,z>0$. Use the result to prove that for real positive numbers $a,b,c$ we have $$abc^3\le 27\left(\frac{a+b+c} 5\right)^5$$

I had no trouble with the first part, using Lagrange's Multipliers. The maximum of $f$ subject to this constraint is $f(r,r,\sqrt 3 r )=5\log r + 3\log \sqrt 3$, and this answer matches the book's.

Now I see how we can take $f(a,b,c)=\log(abc^3)$. Then define $r>0$ by $a^2+b^2+c^2=5r^2\implies r=\sqrt {\frac{a^2+b^2+c^2} 5}$, so we can conclude that

$$abc^3\le3^{3/2}\left(\frac{a^2+b^2+c^2}5\right)^{5/2}$$

But this is a looser bound (for some numbers) than the one suggested by the text. In particular, if we consider $a=\frac 1 4,b= 1, c=1$, then $$27\left(\frac{a+b+c} 5\right)^5<3^{3/2}\left(\frac{a^2+b^2+c^2}5\right)^{5/2}$$ so I have the feeling that "I can't get there from here", at least not using the method suggested. Am I correct?

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I also thought that perhaps a simple proof sufficed for exceptional cases where one of the real numbers was small (I thought perhaps $<1$ would be the criteria), and for all other numbers the bound I found would be tighter, however Mathematica points out that $a=\frac{35}{32},b=\frac{13}8,c=4$ also makes the last inequality above true. –  process91 Aug 30 '12 at 3:49
    
A missing $\sqrt{\ }$ in the text perhaps? –  copper.hat Aug 30 '12 at 4:05
    
Are you sure your counterexample of $a=1/4,b=1,c=1$ invalidates the relation you wrote just above "so I have a feeling that..." –  Alex R. Aug 30 '12 at 4:13
    
@Sam On the contrary, for those values of $a,b,$ and $c$, the inequality is true, indicating the book's bound is, in some instances, tighter. –  process91 Aug 30 '12 at 4:31
    
@Michael Boratko, pardon I completely misread that. –  Alex R. Aug 30 '12 at 4:33
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1 Answer 1

up vote 0 down vote accepted

I think I'm going to delete this question - I realize no one (except maybe Apostol, and perhaps not even him) can really answer this question. Perhaps what was intended was that I should realize that a similar approach would yield the inequality desired, and in fact it does:

Maximize $f(x,y,z) = \log x + \log y + 3\log z$ subject to the constraint that $x+y+z=5r$, and then use this result to deduce that, for real numbers $a,b,c$ we have that $$abc^3\le27\left(\frac {a+b+c} 5 \right)^5$$

This actually works in a straightforward way. It's anyone's guess, I suppose, whether the author's "use this result" was meant in the broadest sense (i.e. generalize a strategy from the first result), or whether it was a typo in either the constraint or the inequality.

If no one has any complaints, I will delete this question tomorrow morning.

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Why not keep it, so if others hit the same problem they can avoid a fruitless direction? –  copper.hat Aug 30 '12 at 5:27
    
@copper.hat OK, will do. –  process91 Aug 30 '12 at 11:54
    
most probably it is a case of the author missing a square root and publishing a wrong answer in the text. –  ajinkya Apr 26 at 8:48
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