If-a-i-gt-0-b-i-gt-0-i-1-n-Prove-that-a-1-2-b-1-2-a-2-2-b-2-2-a-n-2-b-n-2-a-1-a-2-a-n-2-b-1-b-2-b-

Question Number 222582 by hardmath last updated on 30/Jun/25

If: a_i > 0 , b_i > 0 , i = 1,...,n^(−) Prove that: (√(a_1 ^2 + b_1 ^2 )) + (√(a_2 ^2 + b_2 ^2 )) +...+ (√(a_n ^2 + b_n ^2 )) ≥ ≥ (√((a_1 +a_2 +...+a_n )^2 + (b_1 +b_2 +...+b_n )^2 ))

$$\mathrm{If}:\:\:\:\mathrm{a}_{\boldsymbol{\mathrm{i}}} \:>\:\mathrm{0}\:\:\:,\:\:\:\mathrm{b}_{\boldsymbol{\mathrm{i}}} \:>\:\mathrm{0}\:\:\:,\:\:\:\mathrm{i}\:=\:\overline {\mathrm{1},…,\mathrm{n}} \\ $$$$\mathrm{Prove}\:\mathrm{that}: \\ $$$$\sqrt{\mathrm{a}_{\mathrm{1}} ^{\mathrm{2}} \:+\:\mathrm{b}_{\mathrm{1}} ^{\mathrm{2}} }\:+\:\sqrt{\mathrm{a}_{\mathrm{2}} ^{\mathrm{2}} \:+\:\mathrm{b}_{\mathrm{2}} ^{\mathrm{2}} }\:+…+\:\sqrt{\mathrm{a}_{\boldsymbol{\mathrm{n}}} ^{\mathrm{2}} \:+\:\mathrm{b}_{\boldsymbol{\mathrm{n}}} ^{\mathrm{2}} }\:\:\:\geqslant \\ $$$$\geqslant\:\sqrt{\left(\mathrm{a}_{\mathrm{1}} +\mathrm{a}_{\mathrm{2}} +…+\mathrm{a}_{\boldsymbol{\mathrm{n}}} \right)^{\mathrm{2}} \:+\:\left(\mathrm{b}_{\mathrm{1}} +\mathrm{b}_{\mathrm{2}} +…+\mathrm{b}_{\boldsymbol{\mathrm{n}}} \right)^{\mathrm{2}} } \\ $$

Answered by vnm last updated on 30/Jun/25

I think this is obvious if (a_i ,b_i ) are treated as vectors in the plane. The sum of lengths of vectors is not less than the length of their sum.

$$ \\ $$$$\mathrm{I}\:\mathrm{think}\:\mathrm{this}\:\mathrm{is}\:\mathrm{obvious}\:\mathrm{if}\:\left(\mathrm{a}_{{i}} ,\mathrm{b}_{{i}} \right) \\ $$$$\mathrm{are}\:\mathrm{treated}\:\mathrm{as}\:\mathrm{vectors}\:\mathrm{in}\:\mathrm{the} \\ $$$$\mathrm{plane}.\:\mathrm{The}\:\mathrm{sum}\:\mathrm{of}\:\mathrm{lengths}\:\mathrm{of} \\ $$$$\mathrm{vectors}\:\mathrm{is}\:\mathrm{not}\:\mathrm{less}\:\mathrm{than}\:\mathrm{the} \\ $$$$\mathrm{length}\:\mathrm{of}\:\mathrm{their}\:\mathrm{sum}. \\ $$

Commented by hardmath last updated on 30/Jun/25

$$\mathrm{solution}\:\mathrm{please} \\ $$

Answered by mr W last updated on 30/Jun/25

Commented by mr W last updated on 30/Jun/25

c_1 =(√(a_1 ^2 +b_1 ^2 )), ... AB=(√(AC^2 +CB^2 ))=(√((a_1 +a_2 +...+a_n )^2 +(b_1 +b_2 +...+b_n )^2 )) c_1 +c_2 +...+c_n ≥AB ⇒(√(a_1 ^2 +b_1 ^2 ))+(√(a_2 ^2 +b_2 ^2 ))+...+(√(a_n ^2 +b_n ^2 ))≥(√((a_1 +a_2 +...+a_n )^2 +(b_1 +b_2 +...+b_n )^2 ))

$${c}_{\mathrm{1}} =\sqrt{{a}_{\mathrm{1}} ^{\mathrm{2}} +{b}_{\mathrm{1}} ^{\mathrm{2}} },\:… \\ $$$${AB}=\sqrt{{AC}^{\mathrm{2}} +{CB}^{\mathrm{2}} }=\sqrt{\left({a}_{\mathrm{1}} +{a}_{\mathrm{2}} +…+{a}_{{n}} \right)^{\mathrm{2}} +\left({b}_{\mathrm{1}} +{b}_{\mathrm{2}} +…+{b}_{{n}} \right)^{\mathrm{2}} } \\ $$$${c}_{\mathrm{1}} +{c}_{\mathrm{2}} +…+{c}_{{n}} \geqslant{AB} \\ $$$$\Rightarrow\sqrt{{a}_{\mathrm{1}} ^{\mathrm{2}} +{b}_{\mathrm{1}} ^{\mathrm{2}} }+\sqrt{{a}_{\mathrm{2}} ^{\mathrm{2}} +{b}_{\mathrm{2}} ^{\mathrm{2}} }+…+\sqrt{{a}_{{n}} ^{\mathrm{2}} +{b}_{{n}} ^{\mathrm{2}} }\geqslant\sqrt{\left({a}_{\mathrm{1}} +{a}_{\mathrm{2}} +…+{a}_{{n}} \right)^{\mathrm{2}} +\left({b}_{\mathrm{1}} +{b}_{\mathrm{2}} +…+{b}_{{n}} \right)^{\mathrm{2}} } \\ $$

Commented by hardmath last updated on 03/Jul/25

$$\mathrm{thank}\:\mathrm{you}\:\mathrm{very}\:\mathrm{much}\:\mathrm{cool}\:\mathrm{dear}\:\mathrm{professor} \\ $$

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