Show that $-2abX \le a^2Y +b^2Z\implies 4X^2 \le 4 Y Z $, $a,b \in \mathbb R $, $X;Y;Z \ge 0 $

Question

Show that $-2abX \le a^2Y +b^2Z\implies 4X^2 \le 4YZ $

where $X,Y,Z $ are nonnegative and $a,b \in \mathbb R $.

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This looks almost as I could use Young's inequality, but not quiet.

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The above comes from a proof that claims that for $X,Y \in L ^2(P) $ ,

$(2E[|XY |])^2\le 4 E [X ^2]E [Y^2 ] $

assuming $X,Y $ to be nonnegative and bounded. Then I can prove Schwartz inequality using monotone convergence theorem.

Answer 1

It seems you need to clarify the question. I suppose you are intend to ask:

If for all $a, b\in \mathbb{R}$, we have $-2ab X\le a^2Y+b^2 Z$, then $X^2\le YZ$.

In this case, $$ 2YZ=\min_{a,b} \{\frac{a^2Y+b^2 Z}{|ab|} \} \ge 2|X|, $$ proving the claim.