Guaranteed Real Roots

For the equation to have real roots, it's discriminant $\Delta \geq 0$ .

$\Delta = 4(a_1^2 + a_2^2 + ... + a_{n+1}^2 - 4a_{n+1}(a_1+a_2+...+a_n)$

$= 4(a_1^2 + a_2^2 + ... + a_{n+1}^2) - 4a_{n+1}(a_1+a_2+...+a_n) + na_{n+1}^2 - na_{n+1}^2$

$= (2a_1 - a_{n+1})^2 + ... + (2a_n - a_{n+1})^2 + a_{n+1}^2(4 - n)$

This is clearly non-negative for $n \leq 4$ with maximum $n = 4$ .

For $n > 4$ the last square will be greater than all preceding squares & will make $\Delta$ negative. Like put all $a_i = m$ & $a_{n+1} = m+1$ for some constant m. This gives us negative discriminant & hence non-real roots.

For the equation to have real roots, we must have the discriminant at least 0, that is $4\left((a^2_1+a^2_2+\cdots+a^2_n+a^2_{n+1})-a_{n+1}\left(a_1+a_2+\cdots+a_n\right)\right) \ge 0 \implies$ $a^2_{n+1}-a_{n+1}\left(a_1+a_2+\cdots + a_n\right)+\left(a^2_1+a^2_2+\cdots +a^2_n\right) \ge 0$ . Viewing this as a quadratic in $a_{n+1}$ , its always non-negative iff the discriminant is always non-positive. So we should have the discriminant $\le 0$ , so $(a_1+a_2+\cdots +a_n)^2-4(a^2_1+a^2_2+\cdots+a^2_n) \le 0$ . For $n \ge 5$ , this is not always true (just plug in everything = 1). For $n = 4$ though, the claim follows from Cauchy-Schwarz: $(a^2_1+a^2_2+a^2_3+a^2_4)(1+1+1+1) \ge (a_1+a_2+a_3+a_4)^2$ .

In order for this to always have real roots, the discriminant must always be nonnegative, which means that $4(a_1^2+a_2^2+ \cdots a_{n+1}^2)- 4a_{n+1}(a_1+\cdots + a_n)$ . When $n = 4$ , this is true, because if we rearrange, this is equivalent to $(2a_1-a_{5+1})^2+(2a_2-a_{5})^2+(2a_3-a_{5})^2+(2a_4-a_{5})^2 \ge 0$ , which is true by the trivial inequality.

The statement will not hold when $n \ge 5$ . This can be shown by taking $a_1=a_2= \cdots = a_n = 1, a_{n+1} = 2$ . Then the left hand side of the discriminant evaluates to $4n+16$ , while the right hand side evaluates to $8n$ . When $n > 4$ , this does not hold.

Thus the largest value for which the statement holds is $n=4$

The discriminant of the L.H.S is $\Delta = \left( 2\sqrt{a_1^2+a_2^2+\cdots+a_{n+1}^2} \right)^2 - 4a_{n+1}(a_1+a_2+\cdots+a_n) = 4(a_1^2+a_2^2+\cdots+a_{n+1}^2 - a_{n+1}(a_1+a_2+\cdots+a_n)$ . We need $\Delta \ge 0$ for all $a_1,a_2,\ldots,a_{n+1}$ . By setting $a_1=a_2=\cdots=a_n=1, a_{n+1}=2$ , we obtain $\Delta = 4(n+4-2n) \ge 0$ , or equivalently, $4-n \ge 0$ and $n \le 4$ .

Conversely, for $n=4$ , we will prove that $a_1^2+a_2^2+a_3^2+a_4^2+a_5^2 \ge a_5(a_1+a_2+a_3+a_4)$ for all real numbers $a_1,\ldots,a_5$ . Indeed, by using AM-GM inequality and Cauchy-Schwarz inequality, we have $\begin{aligned} a_1^2+a_2^2+a_3^2+a_4^2+a_5^2 & \ge 2a_5\sqrt{a_1^2+a_2^2+a_3^2+a_4^2} \\ &= 2a_5\sqrt{4(a_1^2+a_2^2+a_3^2+a_4^2)} \\ &\ge 2a_5(a_1+\cdots+a_4). \end{aligned}$

Therefore, the largest possible value of $n$ is 4.

The criterion for a quadratic expression to have only real roots, is simply that the discriminant $\Delta$ must be nonnegative (i.e. positive or zero). In this case:

$\Delta = 4(a_1^2 + \ldots + a_{n+1}^2) - 4 a_{n+1} (a_1 + \ldots + a_n)$

[ In principle, for the proof it is enough to skip to the end, starting at "More specifically...", but I thought I'd share how I got there. ]

We thus want to establish whether this can be negative. Let us look at this expression as if it were a function of $a_{n+1}$ . For $a_{n+1} \rightarrow \pm \infty$ , $\Delta$ is obviously positive, regardless of the values of the other "parameters". Therefore, in order for the parabolic function $\Delta(a_{n+1})$ to be negative for some value(s) of $a_{n+1}$ , it must be zero at two points. In order for that to be the case, the discriminant of this new parabolic function $\Delta(a_{n+1})$ should be positive. The discriminant is

$\Delta' = 16(a_1 + \ldots + a_n)^2 - 64(a_1^2 + \ldots + a_n^2)$

This is positive iff

$\Delta' > 0 \iff \frac{(a_1 + \ldots + a_n)^2}{16} > \frac{a_1^2 + \ldots + a_n^2}{4} \\ \quad \iff \frac{|a_1 + \ldots + a_n|}{4} > \sqrt{\frac{a_1^2 + \ldots + a_n^2}{4}}$

In the last line, for $n=4$ , we recognize the arithmetic mean and the quadratic mean of the numbers $a_1$ through $a_4$ on each side of the inequality. However, the quadratic mean is always at least as large as the arithmetic mean (this is a specific result from the generalized mean inequality), therefore, $\Delta'$ is never positive for $n=4$ . The same holds for $n<4$ , since we can artificially add $a_4=0$ (and $a_3=0$ , $a_2=0$ if applicable) and we can use the same argument.

On the other hand, for $n>4$ , the above inequality can easily hold, for instance in the case $a_1 = \ldots = a_n = 1$ , guaranteeing that a solution must exist with $\Delta \leq 0$ .

More specifically, for $n=4$ we can rewrite:

$\Delta = 4(a_1 + a_2 + a_3 + a_4 + a_5)^2 - 4 a_5 (a_1^2 + a_2^2 + a_3^2 + a_4^2) \\ \quad = (a_5 - 2 a_1)^2 + (a_5 - 2 a_2)^2 + (a_5 - 2 a_3)^2 + (a_5 - 2 a_4)^2$

which clearly demonstrates that $\Delta$ cannot be negative for $n=4$ . The cases with $n<4$ are similar, since you can just add $a_i$ 's that are zero.

For $n>4$ , we can for instance take $a_1 = \ldots = a_5 = 1$ , $a_6 = \ldots = a_n = 0$ (for $n>5$ ) and $a_{n+1} = 2$ , which results in the equation $2x^2 - 6x + 5 = 0$ , which has $\Delta = -4$ and hence no real roots, as a counterexample.

Therefore, the largest positive integer $n$ for which the given equation necessarily has real roots, is $\boxed{4}$

Since we are only interested in the nature of the roots, we can assume without loss of generality that $a_{n+1} =1$

Next we can rearrange the equation and complete the square to give

$(x-\sqrt{1+a_1+a_2+...+a_n})^2=1+(a_1^2-a_1) + (a_2^2-a_2) + ...+ (a_n^2-a_n)$

Sketching the quadratic $y=x^2-x$ you can see that each of the terms inside the bracket is $a_i^2-a_i \le -1/4$ . So each term can take away at most $\frac{1}{4}$ and you can have at most 4 terms to guarantee real roots.

This equation has real roofs if and only if its discriminant is non-negative. n=1,2 it is an easy problem. n=3,4 We see the quadratic is positively definite. n=5. the quadratic is not positively definite. Thus n=4

It is apparent that for the equation to possess real roots, the discriminant $\geq 0$ . Therefore, we obtain the inequality

$4(a_1^{2}+\cdots+a_{n+1}^{2})-4(a_{n+1})(a_1+ \cdots a_n) \geq 0$

Division of $4$ on both sides yield

$(a_1^{2}+\cdots+a_{n+1}^{2})-(a_{n+1})(a_1+ \cdots a_n) \geq 0$

Now, if we view the function on $L.H.S.$ as a function of $a_1$ , we obtain the inequality

$a_1^{2}-a_{n+1}a_1+C \geq 0$ , where $C$ is some arbitrary constant.

Obviously, the minimum of $L.H.S.$ of the inequality is obtained when $a_1=\frac{a_{n+1}}{2}$ .

Similarly, viewing the function in terms of $a_2, \cdots a_n$ , we know that the minimum of the inequality is obtained when $a_2, \cdots ,a_n= \frac{a_{n+1}}{2}$ .

Now, we view the function in terms of $a_{n+1}$ . Note that we get

$a_{n+1}^{2}- (a_1+a_2+\cdots a_n)(a_{n+1})+C \geq 0$ , where $C$ is some arbitrary constant.

Obviously, the minimum of $L.H.S.$ of the inequality is obtained when $a_{n+1}=\frac{a_1+ \cdots a_n}{2}$ .

Note that, all along, we are assuming that we can indeed find $a_1,a_2,a_3...a_{n+1}$ which satisfies $a_1, a_2, \cdots ,a_n= \frac{a_{n+1}}{2}$ and $a_{n+1}=\frac{a_1+ \cdots a_n}{2}$ for all values of n. This might not be the case!

For $n=4$ , we can indeed achieve such minimum. The minimum, $0$ is achieved when we have $(a_1,a_2, a_3, a_4,a_5)=(\frac{x}{2}, \frac{x}{2}, \frac{x}{2}, \frac{x}{2}, x)$ for all values of $x$ . Since the minimum of the inequality is $0$ , the discriminant is always $\geq 0$ . Thus, there exists real roots to the original equation.

For $n=5$ , don't bother about achieving the minimum. Just pick $(a_1,a_2, a_3, a_4,a_5,a_6)$ = $(1,1,1,1,1,\sqrt{3})$ and note that the inequality yields a negative value $8-5\sqrt{3}$ , which is less than $0$ and we are done.

x^2/4+y^2>=xy

A 2nd degree polynomial is guaranteed to have real roots iff its discriminant is greater than or equal to $0$ . In this case, this condition is written as

$4(a_1^2 + \ldots + a_{n+1}^2) - 4a_{n+1}(a_1+ \ldots + a_n) \geq 0$

or, equivalently,

$a_{n+1}^2 - a_{n+1}(a_1 + \ldots + a_n) + (a_1^2 + \ldots + a_n^2) \geq 0$ .

Again, we are faced with a 2nd degree polynomial, this time in $a_{n+1}$ , and we wish to determine when it is guaranteed to assume non-negative values. This happens iff its discriminant is non-positive, in other words, when

$(a_1 + \ldots + a_n)^2 \leq 4(a_1^2 + \ldots + a_n^2)$ . (1)

As a side note, if one rewrites this inequality as $|a_1 + \ldots + a_n| \leq 2 \sqrt{a_1^2 + \ldots + a_n^2}$ (2), it is interesting to observe that the question we are trying to answer can be paraphrased as follows:

What is the largest $n$ such that for any vector of $R^n$ , the absolute value of sum of its coordinates is not greater than twice its norm?

Coming back to (1), we now note that the RHS of the inequality is minimal when $a_1 = \ldots = a_n = \alpha$ (*) , so we can reduce the problem to finding the largest $n$ such that for any $\alpha$ , we have

$(n \alpha)^2 \leq 4 n \alpha^2$ , which yields

$n(n-4) \leq 0$ , and it is now easy to see that the largest $n$ for which this condition holds is $4$ .

(*) To see why this is true, consider inequality (2) and visualize of $(a_1,\ldots,a_n)$ as a vector in $R^n$ . The set of vectors whose sum of coordinates is equal to a constant is an $n$ -plane which is normal to the vector $(1,\ldots,1)$ , so the vector in that $n$ -plane that lies closest to the origin, i.e., has the smallest norm, is a multiple of $(1,\ldots,1)$ , hence $a_1 = \ldots = a_n$ .