Proving the inequality $2\mathsf E(Y^2)^2\le \mathsf E(Y^2)\mathsf E(Y)^2+\mathsf E(Y^4).$

Let $Y\in L^4$ be a non-negative real random variable. I want to prove that $$2\mathsf E(Y^2)^2\le \mathsf E(Y^2)\mathsf E(Y)^2+\mathsf E(Y^4).$$

My proof, based on the Hölder inequality, can be found below. Other approaches are welcome.

Solution 1:

Note: The other proof is very similar to this one, except that it is more elegant by only using Hölder once.

By Hölder's inequality, $$\mathsf E(Y^2)\mathsf E(Y)^2\ge \mathsf E(Y^{4/3})^3.$$

Therefore, $$\mathsf E(Y^2)\mathsf E(Y)^2+\mathsf E(Y^4)\ge\mathsf E(Y^{4/3})^3+\mathsf E(Y^4).$$

By the AM-GM inequality, $$\mathsf E(Y^{4/3})^3+\mathsf E(Y^4)\ge 2\sqrt{\mathsf E(Y^{4/3})^3 \mathsf E(Y^4)}.$$

Therefore it only remains to prove that $$\mathsf E(Y^2)^4\le \mathsf E(Y^{4/3})^3\mathsf E(Y^4),$$ which is once again true by Hölder.

Solution 2:

Another proof: By AM-GM, $$\mathsf E(Y^2)\mathsf E(Y)^2+\mathsf E(Y^4)\ge 2\sqrt{\mathsf E(Y^2)\mathsf E(Y)^2\mathsf E(Y^4)},$$ so we only need to prove $$\mathsf E(Y^2)^4\le\mathsf E(Y^2)\mathsf E(Y)^2\mathsf E(Y^4),$$ i.e. $$\mathsf E(Y^2)^3\le \mathsf E(Y)^2\mathsf E(Y^4),$$ which is true by Hölder's inequality.