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

             ....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_n )   ✓✓

$$\:\:\:\:\:\:\:\:\:\:\:\:\:....{nice}\:\:{calculus}...= \\ $$$$\:\:{Titu}'{s}\:{lemma}:: \\ $$$$\:{for}\:{any}\:{positive}\:{numbers}\:: \\ $$$${a}_{\mathrm{1}} ,{a}_{\mathrm{2}} ,...,{a}_{{n}} \:,\:{b}_{\mathrm{1}} ,{b}_{\mathrm{2}} ,...,{b}_{{n}} \\ $$$$\:{we}\:{have}: \\ $$$$\:\frac{\left({a}_{\mathrm{1}} +...+{a}_{{n}} \right)^{\mathrm{2}} }{{b}_{\mathrm{1}} +...+{b}_{{n}} }\leqslant\frac{{a}_{\mathrm{1}} ^{\mathrm{2}} }{{b}_{\mathrm{1}} }\:+...+\frac{{a}_{{n}} ^{\mathrm{2}} }{{b}_{{n}} }\: \\ $$$${proof}\:: \\ $$$${put}\::\:{x}=\left({x}_{\mathrm{1}} ,...,{x}_{{n}} \right)\in\mathbb{R}^{{n}} \\ $$$$\:\:\:\:\:\:\:\:\::{y}=\left({y}_{\mathrm{1}} ,...,{y}_{{n}} \right)\in\mathbb{R}^{{n}} \\ $$$$\left({x}.{y}\right)^{\mathrm{2}} \leqslant\mid{x}\mid^{\mathrm{2}} \mid{y}\mid^{\mathrm{2}} \left({cauchy}−{schwarz}\:\:{inequality}\right) \\ $$$$\left({x}_{\mathrm{1}} {y}_{\mathrm{1}} +...+{x}_{{n}} {y}_{{n}} \right)^{\mathrm{2}} \leqslant\left({x}_{\mathrm{1}\:\:\:} ^{\mathrm{2}} +...+{x}_{{n}} ^{\mathrm{2}} \right)\left({y}_{\mathrm{1}} ^{\mathrm{2}} +...+{y}_{{n}\:} ^{\mathrm{2}} \right) \\ $$$$\:{by}\:{applying}\:{subsitution}\:: \\ $$$$\:{x}_{{i}} =\frac{{a}_{{i}} }{\:\sqrt{{b}_{{i}} }}\:\:\:,\:\:{y}_{{i}} =\sqrt{{b}_{{i}} }\:\left({i}=\mathrm{1},\mathrm{2}\:,...,{n}\right) \\ $$$$\:\frac{{a}_{\mathrm{1}\:} ^{\mathrm{2}} +...+{a}_{{n}\:} ^{\mathrm{2}} }{{b}_{\mathrm{2}} +...+{b}_{{n}} }\leqslant\frac{{a}_{\mathrm{1}} ^{\mathrm{2}} }{{b}_{\mathrm{1}} }+...+\frac{{a}_{{n}} ^{\mathrm{2}} }{{b}_{{n}} }\:\:\:\checkmark\checkmark \\ $$$$ \\ $$

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