Convexity over the specified domain

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

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

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 $$x$$

• Character limit wouldn't let me, but can we say as for 1, Hessian is symmetric if and on ly if $a_{21}$ is -2(happens if and only if $x1x2$ goes to infinity), which I guess is not the case here, I think. For 6, I think we need the diagonal values to be greater than or equal to the rest for it to be positive semidefinite. Correction: $a_{12}$ -> $a_{21}$ Commented Jan 21, 2021 at 22:22
• For 1, the Hessian should always be symmetric so check your calculations. For the second part of 5, try the Hessian. For 6, what is the Hessian? The eigenvalues need to be nonnegative, just looking at the diagonal is not enough. Commented Jan 21, 2021 at 22:26
• Oh I am sorry, the second Hessian given above is for the 6. My bad, it's a little bit late here. I'll try to do your suggestions tomorrow. Thanks again for the help. Commented Jan 21, 2021 at 22:30
• check the $(1/3)x_2^2$ in the Hessian for 6 too Commented Jan 21, 2021 at 22:35

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.