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Five real numbers $a_1, a_2, a_3, a_4\;\text{and}\; a_5\;$ are such that

$$\sqrt{a_1- 1} + 2\sqrt{a_2- 4}+3\sqrt{a_3- 9} +4\sqrt{a_4- 16} + 5 \sqrt{a_4- 25} =\frac{a_1+a_2+a_3+a_4+a_5}{2}.$$

Find $a_1+a_2+a_3+a_4+a_5.$

Thanks for checking this out!

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For real $\sqrt{a-b^2},$ we need $a-b^2\ge0$

and for $b>0,$ and as $\sqrt{a-b^2}\ge0$ by AM-GM inequality,

$$\dfrac{(\sqrt{a-b^2})^2+(b)^2}2\ge b\sqrt{a-b^2}$$

the equality will occur if $\sqrt{a-b^2}=b$

$$\implies\dfrac{(\sqrt{a_1-1})^2+1^2+\cdots+(\sqrt{a_5-5^2})^2+5^2}2=\dfrac{a_1+a_2+a_3+a_4+a_5}2 \ge \sqrt{a_1-1}+\cdots$$

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Define$$b_1=\sqrt{a_1-1}\\b_2=\sqrt{a_2-4}\\b_3=\sqrt{a_3-9}\\b_4=\sqrt{a_4-16}\\b_5=\sqrt{a_5-25}\\$$therefore by substitution we have $$b_1+2b_2+3b_3+4b_4+5b_5={b_1^2+b_2^2+b_3^2+b_4^2+b_5^2\over 2}+{55\over 2}$$or$$2b_1+4b_2+6b_3+8b_4+10b_5={b_1^2+b_2^2+b_3^2+b_4^2+b_5^2}+{55}$$by rearranging we obtain$$(b_1-1)^2+(b_2-2)^2+(b_3-3)^2+(b_4-4)^2+(b_5-5)^2=0$$therefore $$b_i=i\quad ,\quad i=1,2,3,4,5$$and we have $$a_i=2i^2$$which yields to $$\sum a_i=110$$

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  • $\begingroup$ excellent solution!!! $\endgroup$ – user376343 Dec 16 '18 at 10:48

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