nice-calculus-Titu-s-lemma-for-any-positive-numbers-a-1-a-2-a-n-b-1-b-2-b-n-we-have-a-1-a-n-2-b-1-b-n-a-1-2-b-1-a-n-2-

Question Number 128023 by mnjuly1970 last updated on 04/Jan/21

\dots . n i c e c a l c u l u s \dots =

{T i t u}^{'} s l e m m a ::

f o r a n y p o s i t i v e n u m b e r s :

a_{1}, a_{2}, \dots, a_{n}, b_{1}, b_{2}, \dots, b_{n}

w e h a v e :

\frac{{(a_{1} + \dots + a_{n})}^{2}}{b_{1} + \dots + b_{n}} ⩽ \frac{a_{1}^{2}}{b_{1}} + \dots + \frac{a_{n}^{2}}{b_{n}}

p r o o f :

p u t : x = (x_{1}, \dots, x_{n}) \in R^{n}

: y = (y_{1}, \dots, y_{n}) \in R^{n}

{(x . y)}^{2} ⩽∣ x ∣^{2} ∣ y ∣^{2} (c a u c h y - s c h w a r z i n e q u a l i t y)

{(x_{1} y_{1} + \dots + x_{n} y_{n})}^{2} ⩽ (x_{1}^{2} + \dots + x_{n}^{2}) (y_{1}^{2} + \dots + y_{n}^{2})

b y a p p l y i n g s u b s i t u t i o n :

x_{i} = \frac{a_{i}}{\sqrt{b_{i}}}, y_{i} = \sqrt{b_{i}} (i = 1, 2, \dots, n)

\frac{a_{1}^{2} + \dots + a_{n}^{2}}{b_{2} + \dots + b_{n}} ⩽ \frac{a_{1}^{2}}{b_{1}} + \dots + \frac{a_{n}^{2}}{b_{n}} ✓ ✓