Titu's lemma

For any real numbers ${\displaystyle a_{1},a_{2},a_{3},\ldots ,a_{n}}$  and positive reals ${\displaystyle b_{1},b_{2},b_{3},\ldots ,b_{n},}$  we have ${\displaystyle {\frac {a_{1}^{2}}{b_{1}}}+{\frac {a_{2}^{2}}{b_{2}}}+\cdots +{\frac {a_{n}^{2}}{b_{n}}}\geq {\frac {\left(a_{1}+a_{2}+\cdots +a_{n}\right)^{2}}{b_{1}+b_{2}+\cdots +b_{n}}}.}$  (Nairi Sedrakyan (1997), Arthur Engel (1998), Titu Andreescu (2003))

Similarly to the Cauchy–Schwarz inequality, one can generalize Sedrakyan's inequality to random variables. 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. Nesbitt's inequality.

For positive real numbers ${\displaystyle a,b,c:}$  ${\displaystyle {\frac {a}{b+c}}+{\frac {b}{a+c}}+{\frac {c}{a+b}}\geq {\frac {3}{2}}.}$ 

Example 2. International Mathematical Olympiad (IMO) 1995.

For positive real numbers ${\displaystyle a,b,c}$ , where ${\displaystyle abc=1}$  we have that ${\displaystyle {\frac {1}{a^{3}(b+c)}}+{\frac {1}{b^{3}(a+c)}}+{\frac {1}{c^{3}(a+b)}}\geq {\frac {3}{2}}.}$ 

Example 3.

For positive real numbers ${\displaystyle a,b}$  we have that ${\displaystyle 8(a^{4}+b^{4})\geq (a+b)^{4}.}$ 

Example 4.

For positive real numbers ${\displaystyle a,b,c}$  we have that ${\displaystyle {\frac {1}{a+b}}+{\frac {1}{b+c}}+{\frac {1}{a+c}}\geq {\frac {9}{2(a+b+c)}}.}$ 

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