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In general how do you prove inequalities using calculus, I believe it is using maxima or minima right? For example

$$a^2b+b^2c+c^2a \le 3, \qquad a,b,c \ge 0,\quad a+b+c=3.$$

How would you use calculus, just a sketch, I am interested in the method itself.

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  • $\begingroup$ Lagrange multipliers $\endgroup$
    – Blex
    Jun 30, 2015 at 19:52
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    $\begingroup$ I also think that if you eliminate one variable c=3-a-b and then consider as a function of "a", then you can try to take the derivative and so on. $\endgroup$
    – user72012
    Jun 30, 2015 at 20:00
  • $\begingroup$ Honestly, the notation used here is just awful. I'd much prefer: $$a^2b+b^2c+c^2a \le 3 \quad \dashv\quad a,b,c \ge 0,\quad a+b+c=3$$ or similar. $\endgroup$
    – goblin GONE
    Jul 1, 2015 at 3:30
  • $\begingroup$ @user72012, what is easier, lagrange multipliers or partial derivative method? $\endgroup$
    – Amad27
    Jul 1, 2015 at 11:49
  • $\begingroup$ @Amad27 I think Lagrange multipliers method is more advanced... $\endgroup$
    – user72012
    Jul 1, 2015 at 12:51

3 Answers 3

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For this I'd suggest Lagrange multipliers. Using the usual terminology, your objective function is $f(a,b,c)=a^2b+b^2c+c^2a$ and your constraint function is $g(a,b,c)=a+b+c=3$. So you introduce the Lagrange multiplier variable $\lambda$ and solve the system of four equations

$$\nabla f = \lambda \nabla g \\ g=3.$$

This expands to

$$2ab+c^2=\lambda \\ 2bc+a^2=\lambda \\ 2ac+b^2=\lambda \\ a+b+c=3.$$

The solution(s) to this system are critical points of the objective function in the regime of the equality constraint. To solve your problem you need to find the ones of these which have $a,b,c \geq 0$. Then any global maximum will either be one of these, or will be on the boundary, i.e. will have at least one of the variables equal to zero. To find an extremum on the boundary, you can just replace the corresponding variable(s) with zero in the objective and constraint functions and then do the same kind of procedure.

The algebra here is not that easy, unfortunately, but the actual calculus part is done.

It is also possible to convert this to a two dimensional problem without an equality constraint, by writing $c=3-a-b$ and replacing $c$ with this expression in the objective function. So you wind up wanting to maximize

$$f(a,b)=a^2b+b^2(3-a-b)+(3-a-b)^2a$$

over the triangle $a \geq 0,b \geq 0,a+b \leq 3$. I suspect this is harder than the Lagrange multiplier approach.

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  • $\begingroup$ @BrandonHumpert As far as I can tell Amad27 wanted to know how to do the calculus part of the problem, not the algebra. If I'm wrong, they can comment for themselves. $\endgroup$
    – Ian
    Jun 30, 2015 at 20:06
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    $\begingroup$ You may also want to mention how you know this is in fact a maximum, and not minimum, or an inflection point. $\endgroup$
    – Macavity
    Jul 1, 2015 at 1:22
  • $\begingroup$ @Ian, don't worry about the algebra, great answer (+1), but which variable are we solving the system of 4 equations for? Is it for $\lambda$? $\endgroup$
    – Amad27
    Jul 1, 2015 at 11:44
  • $\begingroup$ @Amad27 You're solving for $a,b,c$. You sometimes also solve for $\lambda$, although in this case I probably wouldn't. Instead you can just use substitution in the first three equations to arrive at $2ab+c^2=2bc+a^2$,$2bc+a^2=2ac+b^2$. $\endgroup$
    – Ian
    Jul 1, 2015 at 14:09
  • $\begingroup$ mmm... this system is very tough to solve actually. $\endgroup$
    – Amad27
    Jul 1, 2015 at 22:20
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$f(a,b) = a^2b + b^2(3-a-b) + (3-a-b)^2a = a^2b + 3b^2-ab^2-b^3+9a+a^3+ab^2-6a^2-6ab+2a^2b=3a^2b+3b^2-b^3+9a+a^3-6a^2-6ab$. Thus: $f_a = 6ab+9+3a^2-12a-6b = 0 = f_b = 3a^2+6b-3b^2-6a\Rightarrow 3(a^2-b^2) = 6(a-b)\Rightarrow (a-b)(a+b-2)=0$. There are $2$ cases:

Case $1$: $a = b \Rightarrow 6a^2+9+3a^2-12a-6a = 0\Rightarrow 9a^2-18a+9 = 0\rightarrow 9(a-1)^2 = 0\to a = 1=b, c = 1$, since $a+b+c=3$.

Case $2$: $a+b = 2 \Rightarrow 6a(2-a)+9+3a^2-12a-6(2-a) = 0\Rightarrow 12a-6a^2+9+3a^2 - 12a-12+6a=0\to -3a^2-3+6a=0\to -3(a-1)^2=0 \to a = 1, b = 1, c = 1$.

Either case gives $f_{max} = f(1,1,1) = 3\to a^2b+b^2c+c^2a \leq 3$

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    $\begingroup$ There is also the step of showing you have found the maximum, not a minima or a point of inflection. $\endgroup$
    – Macavity
    Jul 1, 2015 at 1:16
  • $\begingroup$ (+1) Great answer, why did you set $f_a = f_b = 0$? though? How do you find crtical points in a 3d system? $\endgroup$
    – Amad27
    Jul 1, 2015 at 11:47
  • $\begingroup$ @Amad27 In more than one dimension, the first order optimality condition is that the gradient is the zero vector, or in other words that all the partial derivatives are zero. $\endgroup$
    – Ian
    Jul 1, 2015 at 19:58
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this inequality is not true $$a^2b+b^2c+c^2a\le4$$ with equality at $$(a,b,c)=(2,1,0)$$ also for the same condition we have $$a^2b+b^2c+c^2a+abc\le 4$$

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