# Prove that there are $1 \leq i, j, k \leq n$

Given $$a_1, a_2, ..., a_n> 0$$ knowing that $$a_1 + ... + a_n = 3$$, and that $$a_1^2 + ... + a_n^2 \ge 1$$ prove that there are $$1 \leq i, j, k \leq n$$ such that: $$a_i + a_j + a_k \geq 1$$.

I think it has a very good geometric output, but I think it would be hard work for inequality or algebra ...

Cause Cauchy I have no idea ....

I thought of $$a_2 ^ 2, ..., a_n ^ 2,$$ as square areas of sides $$a_1, a_2, ... a_n$$ (Is there any failure to think about it?) So just prove that when you make ak the way it was made, you exceed the side of the square ...

I also accept new solutions

Would it have a more interesting way to solve?

• Should it be $a_1^2+\ldots+a_n^2=1$? – Milten Nov 1 '19 at 22:38
• @Milten I didn't understand what you meant – Behemooth Nov 1 '19 at 22:40
• @Milten Between $a^ 2_n$ and $1$ was to have what? – Behemooth Nov 1 '19 at 22:44
• It was supposed to be $\geq$, does that make much difference? – Behemooth Nov 1 '19 at 22:48
• So anyway, if $a_i+a_j+a_k<1$ for all $i,j,k$, then at most two $a$'s can be bigger than or equal to $\frac13$, and $n$ must be at least $10$. That may be useful? – Milten Nov 1 '19 at 22:54

## 1 Answer

From $$a^2+b^2=\frac 12[(a+b)^2+(a-b)^2]$$, if we keep $$a+b$$ fixed and increase their difference $$|a-b|$$, then $$a^2+b^2$$ is getting strictly bigger. This is the fundamental operation here - we will modify'' pairs $$a_i, a_j$$, i.e. preserving their sum but increase their difference, to increase $$a_i^2+a_j^2$$.

Now assume $$a_1\geq a_2\geq a_3\geq ...\geq a_n$$. We first modify $$a_1, a_2$$: change $$a_1$$ to $$a_1+a_2-a_3$$ and change $$a_2$$ into $$a_3$$, by doing this we increase $$\sum a_i^2$$ while preserving $$\sum a_i$$ and $$a_1+a_2+a_3$$. So we can assume $$a_1\geq a_2=a_3\geq a_4...$$. Next, we modify $$a_4$$ and $$a_n$$: increase $$a_4$$ and decrease $$a_n$$ until we see either $$a_4=a_3$$ or $$a_n=0$$. Next modify $$a_5$$ (or $$a_4$$ if we still have $$a_4) and the last positive term $$a_j$$... So by doing this, we can assume $$a_1\geq a_2=a_3=...=a_{k}\geq a_{k+1}$$ while $$a_{k+2}=....=0$$, and $$\sum a_i=3$$, and $$a_1+a_2+a_3$$ remains the same, and $$\sum a_i^2$$ is larger than the beginning.

Now if $$a_1+a_2+a_3$$ (that is the biggest $$3$$-sum) is less than $$1$$, assume $$a_1=1-2\epsilon$$, $$a_2=...=a_k=\delta$$, we get $$\delta<\epsilon$$. On the other hand $$a_1\geq a_2$$ implies $$1-2\epsilon\geq \delta$$. Now $$\sum a_j=1-2\epsilon+(k-1)\delta+a_{k+1}=3$$ with $$a_{k+1}\leq \delta$$. So $$1-2\epsilon+k\delta\geq 3\geq 1-2\epsilon+(k-1)\delta.$$ So $$k-1\leq \frac{2+2\epsilon}{\delta}\leq k.$$ Then compute $$\sum a_i^2=(1-2\epsilon)^2+(k-1)\delta^2+a_{k+1}^2\leq (1-2\epsilon)^2+\frac{2+2\epsilon}{\delta}\delta^2+\delta^2.$$ So $$\delta<\epsilon$$ implies $$\sum a_i^2\leq 1-4\epsilon+4\epsilon^2+2\delta+2\epsilon\delta+\delta^2<1-2\epsilon+7\epsilon^2.$$ So if $$\epsilon< \frac 27$$ the above is $$<1$$ and we get the contradiction. On the other hand, if $$\epsilon>\frac 13$$, we see $$a_1=1-2\epsilon <\frac 13$$ and so all $$a_i<\frac 13$$, so $$\sum a_i^2< \frac 13 \sum a_i=1$$, again, contradiction.

So now we can assume $$\frac 27\leq \epsilon\leq \frac 13$$. Now if $$\delta\leq 0.27$$, we see $$\sum a_i^2\leq (1-2\epsilon)^2+\frac{2+2\epsilon}{\delta}\delta^2+\delta^2 \leq (1-\frac 47)^2+(2+\frac 23)0.27+0.27^2<1,$$ contradiction. So $$0.27<\delta<\epsilon\leq \frac 13$$. So $$8=\frac{2}{\frac 13}+2< \frac{2}{\delta}+2 <\frac{2+2\epsilon}{\delta}< \frac{2+\frac 23}{0.27}<10.$$ So $$8< k$$ and $$k-1 <10$$, i.e. $$k=9, 10$$.

The $$k=9$$ case: do a further modification'' to increase $$a_2, .., a_9$$ at the cost of $$a_{k+1}=a_{10}$$, until $$a_2=a_3=...=a_9=\epsilon$$. If $$a_{10}$$ is truned into $$0$$ while $$a_2, ...,a_9$$ increases only to $$\delta'<\epsilon$$, then $$3=a_1+...+a_9=1-2\epsilon+8\delta'< 1-2\epsilon+8\epsilon,$$ we get $$\epsilon>\frac 13$$, that is not allowed. So EITHER (I) $$a_{10}=0$$ and $$a_2=...=a_9=\epsilon$$, then the above argument becomes $$3=1+8\epsilon$$, which means $$\epsilon=\frac 13$$, so all $$a_j=\frac 13$$. We calculate $$\sum a_i^2=1$$; since now $$a_1+a_2+a_3=1$$ we do need to do some modifications to turn the original $$a_1, a_2, ...$$ to this, and during the process the sum of square increased. So for the original $$a_1, ...$$ we have $$\sum a_j^2<1$$, contradiction. OR (II) $$a_2=...=a_9=\epsilon$$ and $$a_{10}>0$$; in this case
$$a_1+...+a_9+a_{10}=(1-2\epsilon)+8\epsilon+(2-6\epsilon)=3,$$ and $$\sum a_i^2=(1-2\epsilon)^2+8\epsilon^2+(2-6\epsilon)^2= 5-28\epsilon+48\epsilon^2.$$ Find maximum of this for $$\frac 27\leq\epsilon\leq \frac 13$$: we see this equals $$1$$ at $$\epsilon=\frac 13$$, equals $$\frac{45}{49}$$ when $$\epsilon=\frac 27$$. Only when $$\epsilon=\frac 13$$ we have $$\sum_i a_i^2=1$$, again we need to do some modification to turn the original $$a_1, a_2, ...$$ to this (i.e. all $$a_i=\frac 13$$) so for the original $$a_1, a_2, ...$$ we have $$\sum a_j^2<1$$, contradiction.

The $$k=10$$ case: do a further modification'' to increase $$a_2, .., a_{10}$$ at the cost of $$a_{k+1}=a_{11}$$, until $$a_2=a_3=...=a_{10}=\epsilon$$. If $$a_{11}$$ turns into $$0$$ before $$a_2=...=\epsilon$$, this just becomes case $$k=9$$. Otherwise, $$a_1+...+a_{11}=(1-2\epsilon)+9\epsilon+(2-7\epsilon)=3, \ \ \ \$$ now $$a_{11}\geq 0$$ requires $$\epsilon\leq \frac 27$$. If $$\epsilon<\frac 27$$, we have already seen $$\sum a_i^2<1$$. If $$\epsilon=\frac 27$$, we see $$a_{11}=0$$ and this is reduced to the $$k=9$$ case.

• Small typo: In the case $k=9$, case (I), you forgot to square the terms in the two sums. – Milten Nov 3 '19 at 11:45
• Thanks, corrected. – Yuval Nov 4 '19 at 18:01