....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 /b_n ) proof : put : x=(x_1 ,...,x_n )∈R^n :y=(y_1 ,...,y_n )∈R^n (x.y)^2 ≤∣x∣^2 ∣y∣^2 (cauchy−schwarz inequality) (x_1 y_1 +...+x_n y_n )^2 ≤(x_(1 ) ^2 +...+x_n ^2 )(y_1 ^2 +...+y_(n ) ^2 ) by applying subsitution : x_i =(a_i /( (√b_i ))) , y_i =(√b_i ) (i=1,2 ,...,n) ((a_(1 ) ^2 +...+a_(n ) ^2 )/(b_2 +...+b_n ))≤(a_1 ^2 /b_1 )+...+(a_n ^2 /b


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

$. . . . n i c e c a l c u l u s . . . =$ ${T i t u}^{'} s l e m m a ::$ $f o r a n y p o s i t i v e n u m b e r s :$ $a_{1}, a_{2}, . . ., a_{n}, b_{1}, b_{2}, . . ., b_{n}$ $w e h a v e :$ $\frac{{(a_{1} + . . . + a_{n})}^{2}}{b_{1} + . . . + b_{n}} ⩽ \frac{a_{1}^{2}}{b_{1}} + . . . + \frac{a_{n}^{2}}{b_{n}}$ $p r o o f :$ $p u t : x = (x_{1}, . . ., x_{n}) \in R^{n}$ $: y = (y_{1}, . . ., y_{n}) \in R^{n}$ ${(x . y)}^{2} ⩽∣ x ∣^{2} ∣ y ∣^{2} (c a u c h y - s c h w a r z i n e q u a l i t y)$ ${(x_{1} y_{1} + . . . + x_{n} y_{n})}^{2} ⩽ (x_{1}^{2} + . . . + x_{n}^{2}) (y_{1}^{2} + . . . + y_{n}^{2})$ $b y a p p l y i n g s u b s i t u t i o n :$ $x_{i} = \frac{a_{i}}{\sqrt{b_{i}}}, y_{i} = \sqrt{b_{i}} (i = 1, 2, . . ., n)$ $\frac{a_{1}^{2} + . . . + a_{n}^{2}}{b_{2} + . . . + b_{n}} ⩽ \frac{a_{1}^{2}}{b_{1}} + . . . + \frac{a_{n}^{2}}{b_{n}} ✓ ✓$
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