The answer is 3.2.

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In the question it is given real numbers not positive real.....then how can you apply Cauchy Schwartz.....

Ankur Verma
- 3 years ago

can u do it with TITU lemma

Anshaj Shukla
- 2 years, 10 months ago

This answer is really awesome ..nice

Magnas Bera
- 1 year, 11 months ago

I solved it is a completely different way, but I made a mistake in counting and didn't answer correctly. I just thought to post my solution anyway in a comment. The locus $x_1^2 + x_2^2 + x_3^2 + x_5^2 = 16$ is a 4th-dimensional sphere in $\mathbb R^5$ . The locus $x_1 + x_2 + x_3 + x_4 + x_5 = 8$ is an iper-plane and a very special one in fact each unit vector of the standard base (1,0,0,0,0), (0,1,0,0,0),(0,0,1,0,0) (0,0,0,1,0) and (0,0,0,0,1) forms with the normal vector of the plane (that is (1,1,1,1,1) btw) an angle of 45 degrees. The intersection of the plane and the sphere is a 3rd-dimensional sphere centered in $C = (8/5, 8/5, 8/5, 8/5, 8/5)$ and with radius $\dfrac{8}{\sqrt{5}}$ . This can be seen easily via basic geometrical analytics. (I used the formula of the distance between a point and an iperplane, and the fact that the point (1,1,1,2,3) belongs to the 3rd-dimensional sphere, and the Pythagorean theorem). The max value of $x_5$ is attained at the point $P$ of the 3rd-dimensional sphere that is as far as possible from the iperplane $x_5 = 8/5$ . We can discover it considering that from the center $C$ the point $P$ has distance $8/\sqrt{5}$ but since $P$ belongs to the iperplane $x_1+x_2+x_3+x_4+x_5 =8$ the length $8/\sqrt{5}$ must lie on this plane (we cannot travel in the (0,0,0,0,1) direction) so the max distance we can reach from the plane $x_5 = 8/5$ is the projection of $CP$ along the line $C + (0,0,0,0,1)t$ line. Due to the special position of the plane we can find this projection searching the side of a 5th-dimensional cube whose main diagonal is $r = 8/\sqrt{5}$ . This is in fact $\dfrac{r}{\sqrt{5}} = \dfrac{8}{5}$ . So have an estimate of the further distance we can gain (from the centre $C$ ) along the direction (0,0,0,0,1). Since the 5th coordinate of $C$ is $8/5$ we have the estimate that the 5th coordinate point of the 3rd dimensional sphere (intersection of the 4th dimensional sphere and the iperplane) is lower or equal that 8/5 + 8/5 = $\dfrac{16}{5}$ . We must still prove that this value is attained. We plug $x_5$ and due to the symmetry of problem $x=x_1=x_2=x_3=x_4$ and we find $4x + \dfrac{16}{5} = 8 \ \Rightarrow x = \dfrac{6}{5}$ and we check that this also satisfies the 4th dimensional sphere equation since $4 \cdot \dfrac{36}{25} + \dfrac{256}{25} = \dfrac{400}{25} = 16$

Andrea Palma
- 1 year, 5 months ago

Using Cauchy-Schwarz:

$\left(\frac{\sum_{i=1}^nx_{i}}{n}\right)^2\le\frac{\sum_{i=1}^n(x_{i})^2}{n}$ We have, $\left(\frac{x_1+x_2+…+x_4}{4}\right)^2\le\frac{x_1^2+x_2^2+..+x_4^2}{4}$ $\left(\frac{8-x_5}{4}\right)^2\le \frac{16-x_5^2}{4}$ On solving, $x_5(x_5-16/5)\le 0$ Hence maximum value of $x_5=\boxed{\frac{16}{5}}.$

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Could you please tell me what is the error in following solution.

as sum of square is 16, so highest possible value of x5^{2} will be 16 hence x5=4 =4/1

=> y1=4, y2=1 => so answer = 9

Chandan Baranwal
- 6 years, 9 months ago

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This implies x1=x2=x3=x4=0 from equation 2. Equation 1 will not be satisfied any more. Hence x5 can not be equal to 4.

mohit chawla
- 6 years, 9 months ago

We are given that:

$x_1+x_2+x_3+x_4+x_5=8 \Rightarrow x_1+x_2+x_3+x_4=8-x_5$ .

$(x_1+x_2+x_3+x_4)^{2}=(8-x_5)^{2}$ .

$x_1^{2}+2x_1x_2+x_2^{2}+x_3^{2}+2x_3x_4+x_4^{2}+2(x_1+x_2) (x_3+x_4)=64-16x_5+x_5^{2}$ $(1)$

We are also given that:

$x_1^{2}+x_2^{2}+x_3^{2}+x_4^{2}=16-x_5^{2}$ . $(2)$ .

Notice that:

$(1)-(2) \Rightarrow x_5^{2}-8x_5+24=x_1x_2+x_3x_4+(x_1+x_2)(x_3+x_4)$ . $(3)$

To find the maximum of $x_5$ we must first find the maximum of the $RHS$ ,or:

$x_1x_2+x_3x_4+(x_1+x_2)(x_3+x_4)= x_1x_2+x_3x_4+x_1x_3+x_1x_4+x_2x_3+x_2x_4$ .

We know that:

$x_1x_2 \leq \frac{x_1^{2}+x_2^{2}}{2}$

$x_3x_4 \leq \frac{x_3^{2}+x_4^{2}}{2}$ .

$x_1x_3 \leq \frac{x_1^{2}+x_3^{2}}{2}$

$x_1x_4 \leq \frac{x_1^{2}+x_3^{2}}{2}$

$x_2x_3 \leq \frac{x_2^{2}+x_3^{2}}{2}$

$x_2x_4 \leq \frac{x_2^{2}+x_4^{2}}{2}$

Summing these up we get:

$x_1x_2+x_3x_4+x_1x_3+x_1x_4+x_2x_3+x_2x_4 \leq \frac{3} {2}*(x_1^{2}+x_2^{2}+x_3^{2}+x_4^{2})=24 -\frac{3x_5^{2}}{2}$ .

Substituting these final value on the $RHS$ of $(3)$ ,we get:

$x_5^{2}-8x_5+24=24-\frac{3x_5^{2}}{2}$ .

Multiply both sides by $2$ and after making some arrangements we get:

$5x_5^{2}-16x_5=0 \Rightarrow \boxed{x_5=\frac{16}{5}}$ .

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Nice solution! I actually thought of doing the same thing, but decided to go ahead with the Lagrange multipliers method. You just have to remember that you need to check that the set of equations you get from all the inequalities is consistent. In your case it was because what you end up with reduces to $x_1=x_2=x_3=x_4$ . If you don't know what I'm talking about, try this problem: https://brilliant.org/practice/classical-inequalities-applications-level-4-5/?p=4

James Wilson
- 3 years, 5 months ago

Could have been shorter.

Rudrayan Kundu
- 2 years, 7 months ago

This problem is solvable because in reality is a system of two equations and two variables. This can be seen by noticing the fact that $x_1=x_2=x_3=x_4\neq0$ . There are two conditions implicit in the latter equation: the first, all four variables being different from zero ensures that both equation are satisfied; the second, if all variables are equal $x_5$ can have a maximum value (give it a thought).

Given the latter, we have now the following:

$4x_1+x_5=8$

$4x_1^2+x_5^2=16$

Solving for $x_5$ one gets that it must be either zero or $\frac{16}{5}$ , and the final answer follows.

Moreover, if one writes the original equations as

$\sum_{i=1}^n x_i=8$

$\sum_{i=1}^n x_i^2=16$ ,

the problem is solvable if $n\geq4$ ; nevertheless, if $n=4$ , $x_1=x_2=x_3=x_4=2$ - in other words, there is no "maximum variable". For $n\geq5$ we will have a maximum value for $x_5$ (give it another thought!)

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Exactly the same solution I have.

Yes, it is quite obvious that $x_1=x_2=x_3=x_4$ for $x_5$ to be at its maximum. Any increase in any of those would mean a decrease in $x_5$ . Any decrease would mean relatively increasing the others resulting to a decreased $x_5$ .

Roman Frago
- 6 years, 3 months ago

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Solved also by finding cross section between 5-dimensional plane and 5-dimensional sphere.

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From the given equations, $x_1+x_2+x_3+x_4=8-x_5$ and $x_1^2 + x_2^2 + x_3^2 + x_4^2=16-x_5^2$ .

Apply Cauchy-Schwarz to get:

$(x_1^2+x_2^2+x_3^2+x_4^2)(1+1+1+1) \geq (x_1+x_2+x_3+x_4)^2$

So, $4(16-x_5^2) \geq (8-x_5)^2$ , which reduces to $\frac{16}{5} \geq x_5 \geq 0$ .

The equality case occurs where $x_1=x_2=x_3=x_4=1.2$ and $x_5=\boxed{3.2}.$

This is a familiar AIME , RMO, and Brilliant problem.