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[daiduong1095]

Nguyên văn bởi daitoancvp

Cho $a,b,c,d $ dương. Chứng minh rằng:

$4.\sqrt[{16}]{{\dfrac{{32a(a + b)(a + b + c)}}{{3(a + b + c + d)^3 }}}} + \sqrt[4]{{\dfrac{{24bcd}}{{(a + b)(a + b + c)(a + b + c + d)}}}} \le 5 $

Dùng AM-GM

Khá vất vả

cho 16 số ta có:
$16.\sqrt[16]{{\dfrac{{32a(a + b)(a + b + c)}}{{3(a + b + c + d)^3 }}}}=16.\sqrt[16]{\frac{2a}{a+b}.\frac{3(a+b)}{2(a+b+c)}.\frac{3(a+ b)}{2(a+b+c)}.\frac{4(a+b+c)}{3(a+b+c+d)}.\frac{4( a+b+c)}{3(a+b+c+d)}.\frac{4(a+b+c)}{3(a+b+c+d)}} $
$\le \frac{2a}{a+b}+2.\frac{3(a+b)}{2(a+b+c)}+3.\frac{4 (a+b+c)}{3(a+b+c+d)}+10=\frac{2a}{a+b}+\frac{3(a+b )}{a+b+c}+\frac{4(a+b+c)}{a+b+c+d}+10 $ (1)

Lại có:$4.\sqrt[4]{{\dfrac{{24bcd}}{{(a + b)(a + b + c)(a + b + c + d)}}}}=4.\sqrt[4]{\frac{2b}{a+b}.\frac{3c}{a+b+c},\frac{4d}{a+b+c+d }} $
$\le \frac{2b}{a+b}+\frac{3c}{a+b+c}+\frac{4d}{a+b+c+d} +1 $ (2)

Cộng vế theo vế (1) và (2) suy ra đpcm.
[RIGHT][I][B]Nguồn: MathScope.ORG[/B][/I][/RIGHT]