Find largest possible value

Question

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 $x_5$?

Answer 1

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 $x_5=\frac{16}{5}$

Answer 2

$x_5$ 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 $x_1=x_2=x_3=x_4$.

Answer 3

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 $x_{1}=x_{2}=x_{3}=x_{4}$