While searching for a proof of the algebra property of Sobolev spaces ($H^s(\mathbb R^n)$ is an algebra when $s > n/2$), I found these notes. On page two the author states that if $t \ge 1$ then Hölder's inequality implies

$$((1+|x|)+(1+|y|))^t\le 2^t((1+|x|)^t + (1+|y|)^t).$$

I don't see why Hölder's inequality implies that the above inequality is true. Perhaps the author meant that for $t\ge 1$ and $a=(1+|x|),$ $b=(1+|y|)$ we obtain

$$(a+b)^t \le 2^t(a^t+b^t).$$

Something similar to Cauchy/Young's inequality might work here since

$$t = 1 \implies (a+b)\le 2(a+b),$$ $$t = 2 \implies (a+b)^2\le 4(a^2+b^2),$$

and the inequality covers the general case for $t\ge 1$. Trying

$$(a+b)^t=(a+b)(a+b)^{t-1}\le \frac{1}{2}\left(\frac{2(a+b)^2 + (a+b)^{2t}}{(a+b)^2}\right)$$

doesn't look like the correct approach. Does the inequality follow from either Young's or Hölder's inequality?