Convexity over the specified domain

Question

I have the following problem:

My attempts:

for i: Hessian matrix I found for 1 is given below.

$$ \begin{bmatrix} \frac {x_2^2}{4(x_1x_2)^\frac 32}+4& \frac {8(x1x2)^{\frac 32}-x1x2}{4( {x1x2)^{\frac 32}}}& 0 \\ \frac {8(x1x2)^{\frac 32}-x1x2}{4( {x1x2)^{\frac 32}}} & \frac {x_1^2}{4(x_1x_2)^\frac 32}+4 & -2 \\ 0 & -2 & 6 \\ \end{bmatrix} $$ for iii: I could not figure out how to approach to this as I know ln function is convex.

for v: Hessian matrix I found for second part of 5 is given below. $$ \begin{bmatrix} 12x_1^2+12x_1+4x_2^2+4x_2+10 & 8x_1x_2+4x_1+4x_2+2 \\ 8x_1x_2+4x_1+4x_2+2 & 12x_2^2+12x_2+4x_1^2+4x_1+10 \\ \end{bmatrix} $$

for vi: Hessian matrix I found for 6 is given a

$$ \begin{bmatrix} 12x_1^2+\frac {13}3x_2^2 & \frac {26}3x_1x_2 \\ \frac {26}3x_1x_2 & 12x_2^2+\frac {13}3x_1^2 \\ \end{bmatrix} $$

Answer 1

For 1, you cannot separate the function. Try the Hessian.

For 3, try the perspective of the convex function $f(x) = \sum_i x_i \log x_i$

For 4: try relating the function to a norm, writing $Q=UU^T$

For 5: the square root of a nondecreasing and convex function is not necessarily convex, think $\sqrt{x}$. Can you relate this to the result of 4?

For 6: the product of convex functions is not always convex, e.g., multiply $x^2$ with $x$

Answer 2

I found it. I showed each Hessian is positive semidefinite on the mentioned domain, no matter which values $x_1, x_2$ and $x_3 $ take. I did that by finding determinants of sub square matrices and the Hessians itself. That way, I was able to prove that.