Is my solution to finding the Hicksian demand correct? Maximize $x_1^{1/2} + x_2^{1/2}$ subejct to the budget constraint Maximize $x_1^{\frac{1}{2}} + x_2^{\frac{1}{2}}$ subejct to the budget constraint $p_1x_1+p_2x_2=m$
Setting up the Lagrange and finding the first-order conditions:
$L(x_1, x_2, \lambda)=x_1^{\frac{1}{2}} + x_2^{\frac{1}{2}}+\lambda(p_1x_1+p_2x_2-m)$
$\frac{\partial L}{\partial x_1}=\frac{1}{2}x_1^\frac{-1}{2}+\lambda p_1=0$
$\frac{\partial L}{\partial x_2}=\frac{1}{2}x_2^\frac{-1}{2}+\lambda p_2=0$
Equating these,
$2p_2x_1^\frac{-1}{2}=2p_1x_2^\frac{-1}{2}$
Solving for $x_1$ and $x_2$, we get
$x_1=\frac{p_2^2x_2}{p_1^2}$
$x_2=\frac{p_1^2x_1}{p_2^2}$
Substituting these into the budget constraint and solving to find the Hicksian demand ($x^*(p,m)$),
$p_1x_1+p_2x_2=m$
$\frac{p_2^2x_2}{p_1}+p_2x_2=m$
Solving for $x_2$,
$x_2(\frac{p_2^2}{p_1}+p_2)=m$
$x_2^*=\frac{m}{p_2}.\frac{p_2}{p_1+p_2}=\frac{m}{p_1+p_2}$
We do the same thing for $x_1$ and we get
$x_1^*=\frac{m}{p_1+p_2}$
We now have the Hickdian demands, substitute these into the objective function to get the indirect utility function and by duality, we know, $V(p,E(p,u))=u$,
$V(p,m)=(\frac{m}{p_1+p_2})^\frac{1}{2}+(\frac{m}{p_1+p_2})^\frac{1}{2}=u$
Solving this to find m,
$\frac{2m^\frac{1}{2}}{p_1^\frac{1}{2}+p_2^\frac12}=u$
$m^\frac{1}{2}=\frac{u}{2}(p_1^\frac{1}{2}+p_2^\frac{1}{2})$
$m=[\frac{u}{2}(p_1^\frac{1}{2}+p_2^\frac{1}{2})]^2$
Finally we also know by duality that $E(p,V(p,m))=m$, therefore
$E=[\frac{u}{2}(p_1^\frac{1}{2}+p_2^\frac{1}{2})]^2$
To find the Hicksian demand we use Shephards Lemma (take the partial derivative of the expenditure function)
$\frac{\partial E}{\partial p_i}=2.\frac{u}{2}(p_1^\frac{1}{2}+p_2^\frac{1}{2}).\frac{1}{2}.\frac{u}{2}.p_1^\frac{-1}{2}$
Simplifying this
$h_1^*=\frac{u^2}{4}.p_1^\frac{-1}{2}(p_1^\frac{1}{2}+p_2^\frac{1}{2})=\frac{u^2}{4}(p_1^\frac{-1}{4}+p_2^\frac{1}{2})$
Is this correct? (I've tried to make each of my steps really clear so this is easy to follow)
Here is what I am thinking (and i could have made a mistake since CES algebra is a bit tedious)
First you have the MRS condition from Lagrange which boils down to
$$p_1 x_2^{\rho-1} = p_2 x_1^{\rho-1}$$
from here I always aim to reconstruct the utility function (because it is a constant equal to the max utility or in this instance you know it will satisfy constraint and be u). So to get the utility function to appear I do this
$$p_1^{\frac{\rho}{\rho-1}} x_2^{\rho} = p_2^{\frac{\rho}{\rho-1}} x_1^{\rho}$$
I then add $p_1^{\frac{\rho}{\rho-1}} x_1^{\rho}$ to both sides
$$p_1^{\frac{\rho}{\rho-1}} x_2^{\rho} + p_1^{\frac{\rho}{\rho-1}} x_1^{\rho}= p_1^{\frac{\rho}{\rho-1}} x_1^{\rho}+p_2^{\frac{\rho}{\rho-1}} x_1^{\rho}$$
and isolate
$$p_1^{\frac{\rho}{\rho-1}}( x_2^{\rho} +  x_1^{\rho})= (p_1^{\frac{\rho}{\rho-1}}+p_2^{\frac{\rho}{\rho-1}} )x_1^{\rho}$$
utility function appears on LHS and I impose constraint
$$p_1^{\frac{\rho}{\rho-1}}u= (p_1^{\frac{\rho}{\rho-1}}+p_2^{\frac{\rho}{\rho-1}} )x_1^{\rho}$$
finding
$$\frac{p_1^{\frac{\rho}{\rho-1}}}{(p_1^{\frac{\rho}{\rho-1}}+p_2^{\frac{\rho}{\rho-1}} )}u=x_1^{\rho}$$
which reduces to
$$\frac{p_1^{\frac{1}{\rho-1}}}{(p_1^{\frac{\rho}{\rho-1}}+p_2^{\frac{\rho}{\rho-1}} )^\frac{1}{\rho}}u^\frac{1}{\rho}=x_1$$
Inserting value of $\rho = 1/2$ I get
$$\frac{p_1^{-2}}{(p_1^{-1}+p_2^{-1} )^2}u^2=x_1$$