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Math Help - Differential calculus of higher variables

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    Differential calculus of higher variables

    if x1.x2.x3.....xn=1 where xi is positive then show that x1+x2+x3+.....+xn >=n (greater than or equal to)

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    This was given as a problem in the chapter of Differential calculus of functions of several variables. How to solve it
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  2. #2
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    Re: Differential calculus of higher variables

    Quote Originally Posted by vinmech32 View Post
    if x1.x2.x3.....xn=1 where xi is positive then show that x1+x2+x3+.....+xn >=n (greater than or equal to)

    Guys need your help

    This was given as a problem in the chapter of Differential calculus of functions of several variables. How to solve it
    This result is an easy application of the arithmetic-geometric mean inequality. But if you want to use it as an exercise in several-variable calculus, then you can regard it as asking you to find the minimum value of the function f(x_1,x_2,\ldots,x_n) = x_1+x_2+\ldots +x_n in the region \{x_i>0: 1\leqslant i\leqslant n\}, subject to the constraint x_1x_2\cdots x_n=1. It then becomes an exercise in the use of Lagrange multipliers, and you can find that the minimum value of f is n, occurring at the point (x_1,x_2,\ldots,x_n) = (1,1,\ldots,1).
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    Re: Differential calculus of higher variables

    Quote Originally Posted by Opalg View Post
    It then becomes an exercise in the use of Lagrange multipliers
    ... unless you wrote x_n as \frac{1}{x_1 x_2 \ldots x_{n-1}},

    and minimised

    y\ =\ x_1 + x_2 + \ldots + x_{n-1} + \frac{1}{x_1 x_2 \ldots x_{n-1}}

    Setting all the first derivatives equal to zero makes all the x's equal to one, and then you can show the Hessian is positive definite at that point.
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    Re: Differential calculus of higher variables

    Quote Originally Posted by Opalg View Post
    This result is an easy application of the arithmetic-geometric mean inequality. But if you want to use it as an exercise in several-variable calculus, then you can regard it as asking you to find the minimum value of the function f(x_1,x_2,\ldots,x_n) = x_1+x_2+\ldots +x_n in the region \{x_i>0: 1\leqslant i\leqslant n\}, subject to the constraint x_1x_2\cdots x_n=1. It then becomes an exercise in the use of Lagrange multipliers, and you can find that the minimum value of f is n, occurring at the point (x_1,x_2,\ldots,x_n) = (1,1,\ldots,1).

    Thanks for the reply.. Can u pls brief me the process of evaluating it by the use of lagrange multipliers...(any examples)
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    Re: Differential calculus of higher variables

    Quote Originally Posted by vinmech32 View Post
    Can u pls brief me the process of evaluating it by the use of lagrange multipliers...(any examples)
    Let
    \begin{aligned}F(x_1,x_2,\ldots,x_n,\lambda) &= f(x_1,x_2,\ldots,x_n) + \lambda(x_1x_2\cdots x_n-1) \\ &= x_1+x_2+\ldots +x_n + \lambda(x_1x_2\cdots x_n-1).\end{aligned}

    Put the partial derivative with respect to x_i equal to 0:

    \frac{\partial F}{\partial x_i} = 1 + \lambda x_1x_2\cdots x_{i-1}x_{i+1}\cdots x_n = 0.

    Multiply by x_i to see that x_i = -\lambda (because x_1x_2\cdots x_n=1). Thus all the x_i's are equal, and since their product is 1 they must all be equal to 1. Therefore the only critical point is (x_1,x_2,\ldots,x_n) = (1,1,\ldots,1).

    You can find other worked examples here.
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    Re: Differential calculus of higher variables

    thanks guys for your assistance.....
    u r doing a great job...
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