Terms overload

We wish to maximise $f(\mathbf{x}) = x_1$ subject to the relation $g(\mathbf{x}) \; = \; (x_1 + x_2 + \cdots + x_n)^2 + x_1^2 + x_2^2 + \cdots + x_n^2 \; = \; n$ By Lagrange Multipliers, we need to find $\lambda$ such that $\nabla f - \lambda \nabla g \,=\, \mathbf{0}$ . For $2 \le i \le n$ , the $i$ th component of $\nabla f - \lambda \nabla g$ is $-\lambda\big[2(x_1+x_2 +\cdots + x_n) + 2x_i\big] \; = \; 0$ so we deduce that $x_2=x_3=\cdots=x_n = -y$ where $x_1=ny$ . Considering the first component of $\nabla f - \lambda \nabla g$ would determine $\lambda$ , but we are not interested in that. We have $\begin{array}{rcl} n \; = \; g(\mathbf{x}) & = & (x_1+x_2+\cdots+x_n)^2 + x_1^2 + x_2^2 + \cdots + x_n^2 \\ & = & y^2 + (ny)^2 + (n-1)y^2 \; = \; n(n+1)y^2 \end{array}$ and hence $y = \frac{1}{\sqrt{n+1}}$ . Thus $x_1 = \frac{n}{\sqrt{n+1}}$ .

Thus it follows that $|x_i| \le \frac{n}{\sqrt{n+1}}$ for all $1 \le i \le n$ whenever $g(\mathbf{x}) = n$ .

Now $80089 = 283^2$ and it is easy to show that $m \; < \; \frac{m^2+1}{\sqrt{m^2+2}} \; < \; m+1$ and hence the smallest integer value of $N$ that we want to answer the question is $\left\lfloor \frac{80090}{\sqrt{80091}}\right\rfloor+1 \; = \; 284$ The answer is not $283$ , but calculators will not spot this, since $\frac{80090}{\sqrt{80091}}=283.000000022$ .

This follows from two claims:

Step 1 : If

$(a_1 + a_2 + \ldots + a_{80090})^2 + a_1^2 + a_2^2 + \ldots + a_{80090}^2 \le 80090 (*),$

then $|a_i| \le 283.9$ for all $i$ .

Proof : If $|a_i| > 283.9$ , then $a_i^2 > 80090$ so the inequality (*) does not hold. (QED)

Step 2 : There exist real numbers $a_1, \ldots, a_{80090}$ such that (*) holds and $a_1 = 283$ .

Proof : Indeed, let $a_1 = 283$ and $a_2 = a_3 = \ldots = a_{80090} = -\frac 1 {283}$ . Then the expression in (*) is exactly 80090. (QED)

We need to find the largest $a_{i}$ ever possible to find the smallest N. We can accomplish this by making all but one $a_{i}$ 0. This simplifies the inequality to $a^{2}≤80090$ .

From this we know that the highest possible $a$ is $\sqrt{80090} ≈ 283.0018$ . The integer immediately above is 284.

N must be 284 .