PROBLEM 4574 - CRUX MATHEMATICORUM vol. 46, no 8

En, p 415........................................................................................................................

 

 Fr, p 418.....................................................................................................................

 

ANSWER CiP

Equality $\Leftrightarrow x_{1}=\cdots=x_{n}=16$

 SOLUTION CiP

           LEMMA     If   $0\leqslant x <1$  then

(1)                                       $\frac{1}{1-x}\geqslant 1+4x^{2}$.

                               Equality if and only if  $x=0$ or $x=\frac{1}{2}$.

     Proof of Lemma

We perform simple calculation

$\frac{1}{1-x}-1-4x^{2}=\frac{x(1-2x)^{2}}{1-x} \geqslant 0$.

The conclusions result.

$\square$(LEMMA)

           Let $f(x)=\frac{1}{8-\sqrt{x}}$  where $0<x<64$. If we write

$f(x)=\frac{1}{8} \cdot \frac{1}{1-\frac{\sqrt{x}}{8}} \geqslant \frac{1}{8} \cdot(1+4(\frac{\sqrt{x}}{8})^{2})$

 where I applied the Lemma for $\frac {\sqrt{x}}{8}$ instead of $x$, then it turns out 

(2)                                $f(x) \geqslant \frac{1}{8}+\frac{x}{128}$.

      We put in the formula (2), one by one $x=x_{1}, x_{2}, \cdots x_{n}$ and add the obtained results.

So we have

$\sum_{i=1}^{n}f(x_{i})\geqslant \sum_{i=1}^{n}\left ( \frac{1}{8}+\frac{x_{i}}{128} \right )=\frac{n}{8}+ \frac{\sum_{i=1}^{n}x_{i}}{128}=\frac{n}{8}+\frac{16 \cdot n}{128}=\frac{n}{8}+\frac{n}{8}=\frac{n}{4}$

 
what needed to be shown.

$\blacksquare$

The equal sign occur when in (2) eqal signs occur, i.e. $\frac{\sqrt{x_{i}}}{8}=\frac{1}{2}$, so all $x_{i}=16.$

 

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Tracking Number: 10788
Received:        Mon Oct 12 16:50:40 2020

From:  Ciobanu Petre
       Scoala Gimnaziala "Samuil Micu" SADU
       Sibiu, Romania
Email: ptr.ciobanu@gmail.com

Type:  Solve a Crux (Numbered) Problem
       (problem 4574)

Files:
  yynwlsnl.pdf


Comments:
See my Blog
https://ogeometrie-cip.blogspot.com/2020/10/problem-4574-crux-mathematicorum-vol-46.html

Crux Mathematicorum <crux@cms.math.ca>	lun., 12 oct., 19:52
către eu

 =============

         Added April 7, 2021

 

           See their solution, V47n03, pages 159-160 (they give two solutions)

          The first solution starts from identity

$\frac{1}{1-t}=1+t^2+t+\frac{t^3}{1-t}$

 and for the interval $t\in ( 0,\;1 )$ we have 

$t+\frac{t^3}{1-t}\geq 3t^2\;(\Leftrightarrow t(2t-1)^2 \geq 0)$.

 N.CiP In fact, it is equivalent to inequality (1).

          

          The second solution uses that for $x \in (0, \; 64)$ we have

$\frac{1}{8-\sqrt{x}} \geq \frac {x+16}{128}\; \Leftrightarrow \sqrt{x}(\sqrt{x}-4)^2 \geq 0.$

N. CiP It is essentially the inequality (2).

= end added=