Elementary proof that $|x|^p$ is convex.

Update

I let the solution stand as is, even though it uses non-allowed tools. Maybe someone in the future will find it useful.

Old solution

Here is a solution using the Hölder inequality. If that one is not allowed I suggest that you specify more clearly what is and what is not allowed to use among the "standard tools".

We want to show that (here $w_1$ and $w_2$ are non-negative numbers with $w_1+w_2=1$ and $x$ and $y$ are real numbers) $$ |w_1x+w_2y|^p\leq w_1|x|^p+w_2|y|^p, $$ that is $$ |w_1x+w_2y|\leq \bigl(w_1|x|^p+w_2|y|^p\bigr)^{1/p} $$ (If any of the involved quantities is $0$ we just note that the inequality is true).

We let $q$ be the dual exponent, i.e. defined such that $1/p+1/q=1$. Then, we have, using first the triangle inequality and then the Hölder inequality, $$ \begin{aligned} |w_1x+w_2y|&\leq w_1|x|+w_2|y|\\ &=\bigl(w_1^{1/p}|x|\bigr)(w_1)^{1/q}+\bigl(w_2^{1/p}|y|\bigr)(w_2)^{1/q}\\ &\leq\bigl(w_1|x|^p+w_2|y|^p\bigr)^{1/p}\bigl(w_1+w_2\bigr)^{1/q}\\ &=\bigl(w_1|x|^p+w_2|y|^p\bigr)^{1/p}. \end{aligned} $$ Since requested, by Hölder inequality, I mean $$ a_1b_1+a_2b_2\leq(a_1^p+a_2^p)^{1/p}(b_1^q+b_2^q)^{1/q}, $$ where $$ a_1=w_1^{1/p}|x|,\quad a_2=w_2^{1/p}|y|,\quad b_1=w_1^{1/q},\quad\text{and}\quad b_2=w_2^{1/q}. $$

Another for a request about obtaining strict convexity.

If $x$ and $y$ have different signs, then we have strict inequality in the triangle inequality. Thus, we can assume that $x$ and $y$ have the same sign.

Further, one has equality in the Hölder inequality precisely when there exists a constant $c$ such that $$ |b_1|=c|a_1|^{p-1},\quad\text{and}\quad |b_2|=c|a_2|^{p-1}. $$ In our case this simplifies to $$ w_1^{1/q}=cw_1^{(p-1)/p}|x|^{p-1},\quad\text{and}\quad w_2^{1/q}=cw_2^{(p-1)/p}|y|^{p-1}. $$ Since $1/q=(p-1)/p$ this simplifies to $$ 1=c|x|^{p-1},\quad\text{and}\quad 1=c|y|^{p-1}, $$ and hence to $|x|=|y|$. Since we assumed that $x$ and $y$ had the same sign, we conclude that we have equality precisely when $x=y$. All this implies that we have strict convexity.

I think yes. The composition of a convex function with an increasing convex one is still convex. More precisely, if $f$ is convex and $g$ is convex and increasing, then $(g\circ f)(x)=g\bigl(f(x)\bigr)$ is convex. Now $f(x)=|x|$ is convex by a triangle inequality and $g(x)=x^p$ is increasing and convex. Here we could use derivatives.