Differential calculus of higher variables

**vinmech32** · Dec 1st 2011, 11:52 PM

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

**Opalg** · Dec 2nd 2011, 03:55 AM

Originally Posted by vinmech32

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 $(x_1,x_2,\ldots,x_n) = (1,1,\ldots,1).$

**tom@ballooncalculus** · Dec 2nd 2011, 05:11 AM

Originally Posted by Opalg

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.

**vinmech32** · Dec 3rd 2011, 04:07 PM

Originally Posted by Opalg

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)

**Opalg** · Dec 4th 2011, 12:01 AM

Originally Posted by vinmech32

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.

**vinmech32** · Dec 5th 2011, 12:58 AM

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

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