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If $$x_1+x_2+x_3+x_4+x_5=8$$ and $$x_1^2+x_2^2+x_3^2+x_4^2+x_5^2=16,$$ where $x_1,x_2,x_3,x_4,x_5\in\Bbb R$.

What is the largest possible value of $x_5$?

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closed as off-topic by John Doe, Saad, ahulpke, paul garrett, Namaste May 7 '18 at 15:10

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  • $\begingroup$ We're not here to do your homework for you. $\endgroup$ – Mattos May 7 '18 at 11:23
  • $\begingroup$ Welcome to Math.SE. Two things. First, make sure you show us any work you have done on the problem, we'll be more inclined to help you if we know you've shown an effort. Second, please learn use of Mathjax to formulate questions and answers correctly. Link to Mathjax help here: math.meta.stackexchange.com/questions/5020/… $\endgroup$ – Rhys Hughes May 7 '18 at 11:31
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As Logic_Problem_42 correctly pointed out, $x_5$ is at it's greatest when all the other values are equal. Hence, let $x_1=x_2=x_3=x_4=y$ and the resulting equations you will get are: $$4y^2+(x_5)^2=16$$ $$4y+x_5=8$$

We can rearrange the second into: $y=2-\frac{x_5}{4}$ and plug this into the first equation for:

$$4\bigg(2-\frac{x_5}{4}\bigg)^2+(x_5)^2=16$$ $$4\bigg(4+\frac{(x_5)^2}{16}-x_5\bigg)+(x_5)^2=16$$ $$16+\frac{(x_5)^2}{4}-4x_5+(x_5)^2=16$$ and solve from there. You should get $x_5=0$ and $x_5=\frac{16}{5}$

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$x_5$ cannot be larger than $4$. Can it be $4$?

Hint: for a given $a$ the minimum of $x_1^2+x_2^2+x_3^2+x_4^2$ under the condition $x_1+x_2+x_3+x_4=a$ is if $x_1=x_2=x_3=x_4$.

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  • $\begingroup$ X5 May be fraction also. Because it real number $\endgroup$ – user9640947 May 7 '18 at 11:25
  • $\begingroup$ @DietrichBurde if $x_5=4-\epsilon$, then via the first statement $x_1+x_2+x_3+x_4=4+\epsilon$, and that throws out the second statement. So $x_5$ has to be a fair bit lower than 4. $\endgroup$ – Rhys Hughes May 7 '18 at 11:35
  • $\begingroup$ (X1+x2+x3+x4+x5)<=x1^2+x2^2+x3^2+x4^2+x5^2 ........ $\endgroup$ – user9640947 May 7 '18 at 11:36
  • $\begingroup$ (X1+x2+x3+x4+x5)^2<=x1^2+x2^2+x3^2+x4^2+x5^2 ........k $\endgroup$ – user9640947 May 7 '18 at 11:37
  • $\begingroup$ @user9640947, please use Mathjax, otherwise what you're saying is hard to understand. math.meta.stackexchange.com/questions/5020/… $\endgroup$ – Rhys Hughes May 7 '18 at 11:38
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Using Inequality

$$\bigg(\frac{x^2_{1}+x^2_{2}+x^2_{3}+x^2_{4}}{4}\bigg)\geq \bigg(\frac{x_{1}+x_{2}+x_{3}+x_{4}}{4}\bigg)^2$$

$$\bigg(\frac{16-x^2_{5}}{4}\bigg)\geq \bigg(\frac{8-x_{5}}{4}\bigg)^2\Rightarrow x_{5}\in\bigg[0,\frac{16}{5}\bigg]$$

equality occus when $x_{1}=x_{2}=x_{3}=x_{4}$

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