Prove this indirect utility function is quasi-convex The indirect utility function is as follows:

$$ v(m,p) = \frac{m}{p_{1}^{1/2} p_{2}^{1/4} p_{3}^{1/4}} $$

I need to prove that it is quasi-convex. I tried both definition of a quasiconvex function and the property of a convex lower contour set but couldn't get it.

This is a question in my homework so you may just want to give hint.
If the other two methods did not work out for you, another possibility, albeit a more difficult one:

Set up the bordered Hessian:

$$\bar{H}=\begin{bmatrix}0 &amp; \frac{\partial v}{\partial p_1} &amp; \frac{\partial v}{\partial p_2} &amp; \frac{\partial v}{\partial p_3} &amp; \frac{\partial v}{\partial m}\\ 
\frac{\partial v}{\partial p_1} &amp; \frac{\partial^2 v}{\partial p_1^2} &amp; \frac{\partial^2 v}{\partial p_1\partial p_2} &amp; \frac{\partial^2 v}{\partial p_1\partial p_3} &amp; \frac{\partial^2 v}{\partial p_1\partial m}\\
\frac{\partial v}{\partial p_2} &amp; \frac{\partial^2 v}{\partial p_2 \partial p_1} &amp; \frac{\partial^2 v}{\partial p_2^2} &amp; \frac{\partial^2 v}{\partial p_2\partial p_3} &amp; \frac{\partial^2 v}{\partial p_2\partial m}\\
\frac{\partial v}{\partial p_3} &amp; \frac{\partial^2 v}{\partial p_3 \partial p_1} &amp; \frac{\partial^2 v}{\partial p_3 \partial p_2} &amp; \frac{\partial^2 v}{\partial p_3^2} &amp; \frac{\partial^2 v}{\partial p_3\partial m}\\
\frac{\partial v}{\partial m} &amp; \frac{\partial^2 v}{\partial m \partial p_1} &amp; \frac{\partial^2 v}{\partial m \partial p_2} &amp; \frac{\partial^2 v}{\partial m \partial p_3} &amp; \frac{\partial^2 v}{\partial m^2}\\
\end{bmatrix}$$

Define $\bar{H}_k$ as the $k_{th}$ order leading principal submatrix. We can apply the following theorem:

If the $\text{Det}\left(\bar{H}_k\right) &lt;0$ for all $k=2,3,4,5$, then $v$ is quasiconvex in $(\textbf{p},m)$. 

This is an application of the more general result:

Let $X$ be an open convex subset of $\mathbb{R}^n$ and let $f:X \rightarrow \mathbb{R}$ be a twice continuously differentiable function $\left(f\in C^2\right)$. Let $\bar{H}$ be the bordered Hessian and let $\bar{H}_k$ be the $k_{th}$ order leading principal submatrix. If the $\text{Det}\left(\bar{H}_k\right) &lt;0$ for all $k=2,3,\dots,n+1$, then $f$ is quasiconvex on $X$.