4
$\begingroup$

let $a_{1},a_{2},\cdots,a_{n}\ge 0,n\ge 3$,and such $$a^2_{1}+a^2_{2}+\cdots+a^2_{n}=1$$ show that $$a_{1}+a_{2}+\cdots+a_{n}\ge\sqrt{3}(a_{1}a_{2}+a_{2}a_{3}+\cdots+a_{n}a_{1})$$

I can prove when $n=3$, it need to prove $$a_{1}+a_{2}+a_{3}\ge\sqrt{3}(a_{1}a_{2}+a_{2}a_{3}+a_{3}a_{1})$$ where $a^2_{1}+a^2_{2}+a^2_{3}=1$.

since $$ (a_{1}+a_{2}+a_{3})^2\ge 3(a_{1}a_{2}+a_{2}a_{3}+a_{3}a_{1})\tag{1}$$ and $$1=a^2_{1}+a^2_{2}+a^2_{3}\ge a_{1}a_{2}+a_{2}a_{3}+a_{3}a_{1}\tag{2}$$ $(1)\times (2)$ we have $$(a_{1}+a_{2}+a_{3})^2\ge 3(a_{1}a_{2}+a_{2}a_{3}+a_{3}a_{1})^2$$ so we have$$a_{1}+a_{2}+a_{3}\ge\sqrt{3}(a_{1}a_{2}+a_{2}a_{3}+a_{3}a_{1})$$

But for $n\ge 4$ ,I want to show $$f(a_{1},a_{2},\cdots,a_{n})-f(a_{1},a_{2},a_{3},\cdots,a_{n-2},\sqrt{a^2_{n-1}+a^2_{n}},0)\ge 0$$,where $$a_{1}=\max(a_{1},a_{2},\cdots,a_{n}).f(a_{1},a_{2},\cdots,a_{n})=\dfrac{1}{\sqrt{3}}(a_{1}+\cdots+a_{n})-(a_{1}a_{2}+\cdots+a_{n}a_{1})$$ and $$f(a_{1},a_{2},\cdots,a_{n})-f(a_{1},a_{2},a_{3},\cdots,a_{n-2},\sqrt{a^2_{n-1}+a^2_{n}},0)$$$$=\dfrac{1}{\sqrt{3}}(a_{n-1}+a_{n}-\sqrt{a^2_{n-1}+a^2_{n}})+a_{n-2}\sqrt{a^2_{n}+a^2_{n-1}}-a_{n-2}a_{n-1}-a_{n-1}a_{n}-a_{n}a_{1}$$I can't prove it

$\endgroup$
  • $\begingroup$ Now I found this inequality when $(a_{1},a_{2},a_{3},\cdots,a_{n})=(a.a,a,0,0,\cdots,0)$ equality holds $\endgroup$ – inequality Feb 11 at 13:35
  • $\begingroup$ Just a note that after calculation I found out that actually the equality does not hold when $(a_1,...,a_n) = ({\sqrt{3}\over 3}, {\sqrt{3}\over 3}, {\sqrt{3}\over 3}, 0,...,0)$. The left side is $\sqrt{3}$ and the right side is ${2\sqrt{3}\over 3}$. $\endgroup$ – cr001 Feb 11 at 18:20
3
$\begingroup$

Observe that $f(a_1,a_2,...,a_n)=a^2_{1}+a^2_{2}+\cdots+a^2_{n}$ is a symmetric polynomial, so you may write it in terms of the elementary symmetric polynomials.

As a matter of fact, you may write $a^2_{1}+a^2_{2}+\cdots+a^2_{n}=(a_1+a_2+...+a_n)^2-2(a_1a_2+a_1a_3+...+a_{n-1}a_n)$

Therefore we have: $(a_1+a_2+...+a_n)^2-2(a_1a_2+a_1a_3+...+a_{n-1}a_n)=1$, which you may rearrange to get:

$(a_1+a_2+...+a_n)^2 = 1+2(a_1a_2+a_1a_3+...+a_{n-1}a_n)$.

You may now use the rearrangement inequality : (look at this : https://brilliant.org/wiki/rearrangement-inequality/) to conclude that $1=a^2_{1}+a^2_{2}+\cdots+a^2_{n} \geq a_1a_2+a_2a_3+...+a_{n-1}a_{n}$

Putting everything together we get: $(a_1+a_2+...+a_n)^2 = 1+2(a_1a_2+a_1a_3+...+a_{n-1}a_n) $

$\geq (a_1a_2+a_2a_3+...+a_{n-1}a_{n})+2(a_1a_2+a_1a_3+...a_{n-1}a_{n} )$

$\geq(a_1a_2+a_2a_3+...+a_{n-1}a_{n})+2(a_1a_2+a_2a_3+...+a_{n-1}a_{n}) $

=$3(a_1a_2+a_2a_3+...+a_{n-1}a_{n}) $

(on a sidenote, I'm not 100% sure what $(a_{1}a_{2}+a_{2}a_{3}+\cdots+a_{n}a_{1})$ means in your question, as in I'm not sure whether this is $\sum\prod_{1 \leq i <j \leq n} a_ia_j$, or simply $a_1a_2+a_2a_3+a_3a_4+...a_{n-1}a_{n}$ )

Edit by cr001:

$$a_1a_2+a_2a_3+...+a_na_1\leq {a_1}^2+{a_2}^2+...+{a_n}^2\leq 1\implies$$

$$a_1a_2+a_2a_3+...+a_na_1 \geq (a_1a_2+a_2a_3+...+a_na_1)^2$$

which concludes the proof.

$\endgroup$
  • $\begingroup$ Why the downvote? $\endgroup$ – Dr. Mathva Feb 11 at 15:17
  • $\begingroup$ Added the last step for the proof, which is basically $x\leq 1 \implies x\geq x^2$ but this might not seem obvious to many. $\endgroup$ – cr001 Feb 11 at 16:24
3
$\begingroup$

For $n=3$ you have a proof.

We'll prove that for all $n\geq4$ the following stronger inequality is true: $$\sum_{k=1}^na_k\geq2\sum_{k=1}^na_ka_{k+1},$$ where $a_{n+1}=a_1$.

Indeed, we need to prove that $$\sum_{k=1}^na_k^2\left(\sum_{k=1}^na_k\right)^2\geq4\left(\sum_{k=1}^na_ka_{k+1}\right)^2,$$ which is true because $$\sum_{k=1}^na_k^2-\sum_{k=1}^na_ka_{k+1}=\frac{1}{2}\sum_{k=1}^n\left(a_k-a_{k+1}\right)^2\geq0$$ and $$\left(\sum_{k=1}^na_k\right)^2\geq4\sum_{k=1}^na_ka_{k+1}$$ is true by AM-GM: $$\sum_{k=1}^na_ka_{k+1}\leq(a_1+a_3+...)(a_2+a_4+...)\leq\left(\frac{\sum\limits_{k=1}^na_k}{2}\right)^2$$ because for odd $n$ we can assume that $a_1=\min\limits_{i}\{a_i\}$.

For example, for $n=5$ it works so: $$a_1a_2+a_2a_3+a_3a_4+a_4a_5+a_5a_1\leq $$ $$\leq a_1a_2+a_2a_3+a_3a_4+a_4a_5+a_5a_2+a_4a_1=$$ $$=(a_1+a_3+a_5)(a_2+a_4)\leq\left(\frac{a_1+a_3+a_5+a_2+a_4}{2}\right)^2.$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.