Proving inequalities using Calculus 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. 
 A: 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.
A: $f(a,b) = a^2b + b^2(3-a-b) + (3-a-b)^2a = a^2b + 3b^2-ab^2-b^3+9a+a^3+ab^2-6a^2-6ab+2a^2b=3a^2b+3b^2-b^3+9a+a^3-6a^2-6ab$. 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 $f_{max} = f(1,1,1) = 3\to a^2b+b^2c+c^2a \leq 3$
A: 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 $$a^2b+b^2c+c^2a+abc\le 4$$
