Titu's Lemma

Similarly to the Cauchy–Schwarz inequality, one can generalize Sedrakyan's inequality to random variable. In this formulation let ${\displaystyle X}$ be a real random variable, and let ${\displaystyle Y}$ be a positive random variable. X and Y need not be independent, but we assume ${\displaystyle E[|X|]}$ and ${\displaystyle E[Y]}$ are both defined. Then

$${\displaystyle \operatorname {E} [X^{2}/Y]\geq \operatorname {E} [|X|]^{2}/\operatorname {E} [Y]\geq \operatorname {E} [X]^{2}/\operatorname {E} [Y].}$$

Example 1.

Proof: Use ${\displaystyle n=3,}$ ${\displaystyle \left(a_{1},a_{2},a_{3}\right):=(a,b,c),}$ and ${\displaystyle \left(b_{1},b_{2},b_{3}\right):=(a(b+c),b(c+a),c(a+b))}$ to conclude:

$${\displaystyle {\frac {a^{2}}{a(b+c)}}+{\frac {b^{2}}{b(c+a)}}+{\frac {c^{2}}{c(a+b)}}\geq {\frac {(a+b+c)^{2}}{a(b+c)+b(c+a)+c(a+b)}}={\frac {a^{2}+b^{2}+c^{2}+2(ab+bc+ca)}{2(ab+bc+ca)}}={\frac {a^{2}+b^{2}+c^{2}}{2(ab+bc+ca)}}+1\geq {\frac {1}{2}}(1)+1={\frac {3}{2}}.\blacksquare }$$

Example 2.

We have that ${\displaystyle {\frac {{\Big (}{\frac {1}{a}}{\Big )}^{2}}{a(b+c)}}+{\frac {{\Big (}{\frac {1}{b}}{\Big )}^{2}}{b(a+c)}}+{\frac {{\Big (}{\frac {1}{c}}{\Big )}^{2}}{c(a+b)}}\geq {\frac {{\Big (}{\frac {1}{a}}+{\frac {1}{b}}+{\frac {1}{c}}{\Big )}^{2}}{2(ab+bc+ac)}}={\frac {ab+bc+ac}{2a^{2}b^{2}c^{2}}}\geq {\frac {3{\sqrt[{3}]{a^{2}b^{2}c^{2}}}}{2a^{2}b^{2}c^{2}}}={\frac {3}{2}}.}$

Example 3.

We have ${\displaystyle {\frac {a^{2}}{1}}+{\frac {b^{2}}{1}}\geq {\frac {(a+b)^{2}}{2}}}$ so that ${\displaystyle a^{4}+b^{4}={\frac {\left(a^{2}\right)^{2}}{1}}+{\frac {\left(b^{2}\right)^{2}}{1}}\geq {\frac {\left(a^{2}+b^{2}\right)^{2}}{2}}\geq {\frac {\left({\frac {(a+b)^{2}}{2}}\right)^{2}}{2}}={\frac {(a+b)^{4}}{8}}.}$

Example 4.

We have that ${\displaystyle {\frac {1}{a+b}}+{\frac {1}{b+c}}+{\frac {1}{a+c}}\geq {\frac {(1+1+1)^{2}}{2(a+b+c)}}={\frac {9}{2(a+b+c)}}.}$