4
$\begingroup$

Let $a_1,a_2,. . . ,a_n \in \mathbb{R^+} $. Prove that : $$\frac{ a_{1} }{ a_{2} ^{2}+ a_{3} ^{2}+\cdots+ a_{n} ^{2}} +\frac{ a_{2} }{ a_{1} ^{2}+ a_{3} ^{2}+\cdots+ a_{n} ^{2}} +\cdots+\frac{ a_{n} }{ a_{1} ^{2}+ a_{2} ^{2}+\cdots+ a_{n-1} ^{2}} \geq \frac{ 4}{ a_{1}+ a_{2} + a_{3}+\cdots+ a_{n}}$$


I do not know where to start. please help me .

$\endgroup$
  • 4
    $\begingroup$ Is the last fraction on the left-hand side supposed to be $\frac{ a_{n} }{ a_{1} ^{2}+ a_{2} ^{2}+\cdots+ a_{n-1} ^{2}}$? $\endgroup$ – Arthur Jul 17 '17 at 11:04
  • $\begingroup$ prove this in the case $n=2$ $\endgroup$ – Dr. Sonnhard Graubner Jul 17 '17 at 11:15
  • $\begingroup$ Induction on $n=2$ base case should work. $\endgroup$ – 高田航 Jul 17 '17 at 11:18
  • $\begingroup$ I have a proof for $n=3$ and $n=4$ $\endgroup$ – Michael Rozenberg Jul 17 '17 at 12:23
  • $\begingroup$ @MichaelRozenberg . please write . $\endgroup$ – Almot1960 Jul 17 '17 at 12:29
2
$\begingroup$

I think we can start by C-S: $$\sum_{k=1}^n\frac{a_k}{\sum\limits_{i\neq k}a_i^2}=\sum_{k=1}^n\frac{a_k^2}{a_k\sum\limits_{i\neq k}a_i^2}\geq\frac{\left(\sum\limits_{k=1}a_k\right)^2}{\sum\limits_{k=1}^na_k\sum\limits_{i\neq k}a_i^2}.$$ Thus, it remains to prove that $$\left(\sum\limits_{k=1}a_k\right)^3\geq4\sum\limits_{k=1}^n\left(a_k\sum\limits_{i\neq k}a_i^2\right)$$ or $$\left(\sum\limits_{k=1}a_k\right)^3\geq4\sum\limits_{k=1}^n\left(a_k^2\sum\limits_{i\neq k}a_i\right).$$ Now, let $\sum\limits_{k=1}^na_k=1$.

Hence, we need to prove that $$\sum\limits_{k=1}^nf(a_k)\leq\frac{1}{4},$$ where $f(x)=x^2-x^3$.

But $f''(x)=2-6x$, which says that $f$ has an unique inflection point on $(0,1)$.

Thus, by Vasc's HCF Theorem it's enough to prove our inequality for

$a_1=a_2=...=a_{n-1}=x$ and $a_n=1-(n-1)x$, where $0<x<\frac{1}{n-1}$.

Thus, it's enough to prove that $g(x)\geq0,$ where $$g(x)=\frac{1}{4}-(n-1)(x^2-x^3)-(1-(n-1)x)^2+(1-(n-1)x)^3.$$ We have $$g'(x)=(n-1)(1-nx)(3(n-2)x-1).$$ We see that $0<\frac{1}{3(n-2)}<\frac{1}{n}<\frac{1}{n-1}$, $x_{min}=\frac{1}{3(n-2)}$ and $x_{max}=\frac{1}{n}$, which says $$g(x)\geq\min\left\{g\left(\frac{1}{3(n-2)}\right),g\left(\frac{1}{n-1}\right)\right\}=\min\left\{\frac{11n^2-56n+72}{108(n-2)^2},\frac{(n-3)^2}{4(n-1)^2}\right\}\geq0.$$ Done!

For the proof we can use also the Vasc's EV Method.

$\endgroup$
2
$\begingroup$

Notice that $$\left( \frac{a_1}{a_2^2 + a_3^2 + ... + a_n^2} + \frac{a_2}{a_1^2 + a_3^2 + ... + a_n^2} + \ ... \ \frac{a_n}{a_1^2 + a_2^2+... + a_{n-1}^2}\right)(a_1 + a_2 + ... + a_n) \ge \left( \sqrt{\frac{a_1^2}{a_2^2 + a_3^2 + ... + a_n^2}} + \sqrt{\frac{a_2^2}{a_1^2 + a_3^2 + ... + a_n^2}} + \ ... \ \sqrt{\frac{a_n^2}{a_1^2 + a_2^2... + a_{n-1}^2}}\right)^2$$ by Cauchy-Schwarz inequality. WLOG, take $a_1^2 + a_2^2 + ... + a_n^2 = 1$ (because of homogenity). Thus it remains to prove that $$\frac{a_1}{\sqrt{1-a_1^2}} + \frac{a_2}{\sqrt{1-a_2^2}}+ \ ... \ +\frac{a_n}{\sqrt{1-a_n^2}} \ge 2$$Now take $x_m = \frac{a_m}{\sqrt{1-a_m^2}} \ \forall \ 1 \le m \le n$. Then it remains to prove that $$x_1 + x_2 + ... + x_n \ge 2$$ Now since $a_m^2 =\frac{x_m^2}{1+x_m^2}$, we have $$\frac{x_1^2}{1+x_1^2} + \ ... \ + \frac{x_n^2}{1+x_n^2} = 1$$ But $\frac{x_m^2}{1+x_m^2} = x_m \left( \frac{x_m}{1+x_m^2} \right) \le x_m \left( \frac 12\right)$ since $x_m \ge 0$ and it follows that $$1 = x_1 \left( \frac{x_1}{1+x_1^2} \right) + ... + x_n \left( \frac{x_n}{1+x_n^2} \right) \le \frac{x_1}2 + ... + \frac{x_n}2 = \frac 12 \left(x_1 + ... + x_n \right)$$ i.e. $$x_1 + x_2 + ... + x_n \ge 2$$ Done.

Note that $2$ is the minimum attainable value of $$\frac{a_1}{\sqrt{1-a_1^2}} + \frac{a_2}{\sqrt{1-a_2^2}}+ \ ... \ +\frac{a_n}{\sqrt{1-a_n^2}} \ge 2$$ (take $a_1 = a_2 =\frac 1{\sqrt 2}$ and all other $a_i $s equal to $0$).

$\endgroup$

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.