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Find the range of values for abc and a + b + c

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anemone

MHB POTW Director
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Feb 14, 2012
3,689
Let $a, b, c$ be positive real numbers satisfying \(\displaystyle \frac{1}{3}\le ab+bc+ca \le 3\).

Determine the range of values for

i) $abc$,

ii) $a+b+c$.
 

Opalg

MHB Oldtimer
Staff member
Feb 7, 2012
2,707
Anemone and kaliprasad have noticed that nobody ever replied to this challenge problem. Here is my attempt, using MarkFL's favourite method of Lagrange multipliers.

To find the extreme points of $abc$ subject to the restraint $bc+ca+ab = k$ (where $\frac13\leqslant k\leqslant 3$), put the partial derivatives of $abc - \lambda(bc+ca+ab - k)$ (with respect to $a$, $b$ and $c$) equal to $0$: $$bc - \lambda (b+c) = 0,\qquad ca - \lambda (c+a) = 0,\qquad ab - \lambda (a+b) = 0.$$ Write those equations as \(\displaystyle \frac1\lambda = \frac1b + \frac1c = \frac1c + \frac1a = \frac1a + \frac1b\) to see that $a=b=c$. That is the unique extremal point of $abc$. It must be a maximum because if we take $b=c=\varepsilon$ and $a = \dfrac{k-\varepsilon^2}{2\varepsilon}$ then $bc+ca+ab = k$ but $abc = \frac12\varepsilon(k-\varepsilon^2) \to0$ as $\varepsilon\to0$. So $abc\to0$ towards the boundary of the set $\{(a,b,c)\in \mathbb{R}^3:a>0,\,b>0,\,c>0\}.$ Thus the maximum possible value of $abc$ occurs when $k=3$ and $a=b=c= abc =1$. The range of values of $abc$ is therefore the half-open interval $(0,1]$.

An exactly similar calculation for the sum $a+b+c$ shows that it can take arbitrarily large values (when $b=c= \varepsilon$, $a = \dfrac{k-\varepsilon^2}{2\varepsilon}$ and $\varepsilon\to0$). There is again a unique extremal point when $a=b=c$, but this time it is a minimum, occurring when $a=b=c=\frac13$ and $a+b+c=1$. So the range of values of $a+b+c$ is the interval $[1,\infty).$
 

MarkFL

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Feb 24, 2012
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I think the reason I favor the method of Lagrange multipliers is because I am ignorant of the AM-GM method. (Wink) anemone has tried to teach me this, but I have been quite slow on the uptake. (Giggle)
 
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anemone

MHB POTW Director
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Feb 14, 2012
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I think the reason I favor the method of Lagrange multipliers is because I am ignorant of the AM-GM method. (Wink) anemone has tried to teach me this, but I have been quite slow on the uptake. (Giggle)
Hahaha...that isn't the case, Mark! That is because you don't like AM-GM for some reason, the same reason I have zero interest with the LM, I guess...:p
 

Opalg

MHB Oldtimer
Staff member
Feb 7, 2012
2,707
I think the reason I favor the method of Lagrange multipliers is because I am ignorant of the AM-GM method. (Wink) anemone has tried to teach me this, but I have been quite slow on the uptake. (Giggle)
Mention of the AM-GM method makes me see that this is the best way to approach this problem.15274680-light-bulb-icon.jpg

In fact, $\frac13(bc + ca + ab) \geqslant \sqrt[3]{a^2b^2c^2}.$ So if $bc+ca+ab \leqslant3$ it follows that $(abc)^{2 / 3} \leqslant1$ and so $abc\leqslant 1$.

For the other part of the problem, add the inequalities $b^2 + c^2 \geqslant 2bc$, $c^2+a^2 \geqslant 2ca$ and $a^2+b^2 \geqslant 2ab$ to get $2(a^2+b^2+c^2) \geqslant 2(bc+ca+ab)$ and hence $a^2+b^2+c^2 \geqslant bc+ca+ab.$ It follows that $(a+b+c)^2 = a^2+b^2+c^2 + 2(bc+ca+ab) \geqslant 3(bc+ca+ab) \geqslant1.$ Therefore $a+b+c\geqslant1.$