We know that sum of $x$ , $y$ , $\dots$ , $l$ is constant and they are positive. ($x+y+z + \dots +l = c)$ Prove that $P = x^k y^m \dots l^t$ is maximum if $\frac{x}{k} = \frac{y}{m} = \dots = \frac{l}{t}$
My try : I tried to use inequalities like AM-GM but didn't get any result.
Using the method of Lagrange's Multipliers:
Let our $n $ variables $x,y,\cdots,l $. Let $$f (x,y,z,\cdots, l) =x^my^n\cdots l^t $$ subject to the condition $\phi (x,y,\cdots l)= x+y+\cdots + l=c $. Consider the auxiliary function $$F (x,y,\cdots,L,\lambda) = x^my^n\cdots l^t +\lambda (x+y+\cdots +l-c)=0$$ For extreme values $$\frac {\partial F}{\partial x} =mx^{m-1}y^n\cdots l^t +\lambda =0 \tag {1}$$ $$\frac {\partial F}{\partial y} =ny^{n-1 }x^m\cdots l^t +\lambda =0 \tag {2}$$ $$\vdots $$ $$\frac {\partial F}{\partial l} =tl^{t-1} \cdots x^my^n +\lambda =0 \tag {n}$$ We compare $(1) $ with $x (1)+y (2)+\cdots l (n) $ to get the value of $\lambda $. Then it becomes clear that $$\frac {x}{m}=\frac {y}{n}=\cdots =\frac {l}{t} $$ is fulfilled by these extreme values.
Hope it helps.
AM-GM is right. But you need to apply it to a set of terms with constant sum (so that you know what the AM is), and you need to have $k$ terms for $x$, etc., to get an $x^k$ in the GM. These sound like contradictory aims, but in fact they aren't. Try applying AM-GM to $\frac xk,\ldots,\frac xk,\frac ym,\ldots,\frac ym,\ldots$, where you have $k$ terms of the first type, $m$ of the second, and so on.
You can use the method of Lagrange multipliers:
$$\mathcal L(x,y,..,l;\lambda)=P-\lambda(x+y+z+...+l-c)$$
\begin{cases}{\partial\mathcal L\over \partial x}=k{P\over x}-\lambda=0\\{\partial\mathcal L\over \partial y}=m{P\over y}-\lambda=0\\ {\partial\mathcal L\over \partial l}=t{P\over l}-\lambda=0\\x+y+z+...+l-c=0\end{cases}
and you get: $k{P\over x}=m{P\over y}=...=t{P\over l}\longrightarrow{k\over x}={m\over y}=...={t\over l}$