Tìm GTLN $T=\frac{4}{a+b}+\frac{4}{b+c}+\frac{4}{c+a}-\frac{1}{a}-\frac{1}{b}-\frac{1}{c}$

c ko pít nữa, đề cô cho là tìm min, còn min ko ra thì tìm max thui =.= –  @_@ *Mèo9119* @_@ 06-06-16 09:31 AM
ừ đến đoạn vẽ bbt Max =9 tại c=1/3 –  ๖ۣۜQueenღ 06-06-16 08:17 AM
max chắc rồi –  tran85295 05-06-16 11:44 PM
c xem lại chắc là max r –  ๖ۣۜQueenღ 05-06-16 11:29 PM
hình như bài này k có min @_@ –  tran85295 20-04-16 07:26 PM
Click  or Click 
!! ~~ Ngại làm ~~!! 

Note : Nghe nhạc : Click

a,b,c có cho là độ dài 3 cạnh tam giác đâu :)) –  tran85295 06-06-16 03:50 PM
??? sai gì –  ☼SunShine❤️ 06-06-16 03:44 PM
sai rồi :))) –  tran85295 06-06-16 01:26 PM
hhehe free nhạc :)) ngại cái công latex quá –  ☼SunShine❤️ 06-06-16 12:56 PM
câu trả lời độc đáo vậy? –  @_@ *Mèo9119* @_@ 06-06-16 12:55 PM
hờ hờ... –  @_@ *Mèo9119* @_@ 06-06-16 12:55 PM
Ta sẽ chứng minh $T \le9$
$\Leftrightarrow \sum\frac{4}{a+b} \le9+\sum\frac{1}{a}$
$\Leftrightarrow \frac{4\biggl[(a+b)(b+c)+(b+c)(c+a)+(c+a)(a+b) \biggr]}{(a+b)(b+c)(c+a)} \le 9+\frac{ab+bc+ca}{abc}$
$\Leftrightarrow \frac{4\bigg[a^2+b^2+c^2+3(ab+bc+ca) \bigg]}{(a+b+c)(ab+bc+ca)-abc} \le \frac{9abc+ab+bc+ca}{abc}$
$\Leftrightarrow \frac{1+ab+bc+ca}{ab+bc+ca-abc} \le \frac 14 \Bigg[ \frac{9abc+ab+bc+ca}{abc} \Biggr]$
­
$\Leftrightarrow 4abc(ab+bc+ca) +(ab+bc+ca)^2 \ge 4abc+9a^2b^2c^2$
$\bullet$Ta có $abc(ab+bc+ca)=abc(ab+bc+ca)(a+b+c) \ge abc.3\sqrt{(abc)^2}.3\sqrt{abc}=9(abc)^2$
Nên chỉ cần cm 
$3abc(ab+bc+ca)+(ab+bc+ca)^2 \ge 4abc$
$\Leftrightarrow (ab+bc+ca)^2-3abc(a+b+c) \ge abc\big[(a+b+c)^2-3(ab+bc+ca) \big]$
$\Leftrightarrow c(a-b)^2+a(b-c)^2+b(c-a)^2 \ge \frac 12abc \left[(a-b)^2+(b-c)^2+(c-a)^2 \right]$
$\Leftrightarrow \sum \Bigg[ c(a-b)^2 \left( 1-\frac{ab}2\right) \Bigg] \ge0$
Dễ thấy bdt cuối luôn đúng
$\Rightarrow \max T=9\Leftrightarrow a=b=c=\frac 13$
nhìn mà nản quá :3 –  @_@ *Mèo9119* @_@ 07-06-16 08:27 PM
~~~~~~~~~~
$T=\frac{4}{1-a}-\frac{1}{a}+\frac{4}{1-b}-\frac{1}{b}+\frac{4}{1-c}-\frac{1}{c}$
$=\frac{5a-1}{a-a^2}+\frac{5b-1}{b-b^2}+\frac{5c-1}{c-c^2}$
Dự đoán dấu = xảy ra khi $a=b=c=\frac{1}{3}$
$\frac{5a-1}{a-a^2} \le ma+n$
<=> $5a-1 \le ma^2 +na-ma^3-na^2$
<=> $ma^3+a^2(n-m)+a(5-n)-1 \le 0$
Xét $f(a)=ma^3+a^2(n-m)+a(5-n)-1$
$f'(a)=3a^2m+2a(n-m+5-n$
Dấu = xảy ra khi $a=\frac{1}{3}$
=> $\begin{cases}f(\frac{1}{3})=0 \\ f'(\frac{1}{3})=0 \end{cases}$
=> $\begin{cases}\frac{2}{27}m+\frac{2}{9}n=\frac{2}{3} \\ \frac{1}{3}m+\frac{1}{3}n=5 \end{cases}$
=> $\begin{cases}m=18 \\ n=-3 \end{cases}$
~~~~~~~~~~
Ta chứng minh: $\frac{5a-1}{a-a^2}\le 18a-3$
thiêu sđiều kiện nên c ko CM đc kaj dưới cùng kia –  @_@ *Mèo9119* @_@ 07-06-16 07:47 PM
Với $a=\frac 34$ thì $\frac{5a-1}{a-a^2}>18a-3$????? –  tran85295 07-06-16 01:46 PM
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