[Toán 10] $abc\le 50$

N

nttthn_97

Bài 2: x thuộc [0;1]
$S^2=x^2(13\sqrt{1-x^2}+9\sqrt{1+x^2})^2$

$(13\sqrt{1-x^2}+9\sqrt{1+x^2})^2=(\sqrt{13}.\sqrt{13}.\sqrt{1-x^2}+3\sqrt{3}.\sqrt{3}.\sqrt{1+x^2})^2$[TEX]\leq[/TEX]$(13+27)(13-13x^3+3+3x^2)=40(16-10x^2)$(BĐT BUnhiacopxki)

$10x^2(16-10x^2)$[TEX]\leq[/TEX]$(\frac{10x^2+16-10x^2}{b})^2=64$

[TEX]\Rightarrow[/TEX]$S^2$[TEX]\leq[/TEX]256

[TEX]\Rightarrow[/TEX]$S$[TEX]\leq[/TEX]16

Dấu bằng xảy ra [TEX]\Leftrightarrow[/TEX].....
 
M

minhtuyb

Bài 1: Chú ý điểm rơi ở $a=b=\dfrac{5}{2};c=8$
$$abc=\dfrac{25}{256}.\dfrac{16a}{5}\dfrac{16a}{5}c\\ \le \dfrac{(\dfrac{16a}{5}+\dfrac{16a}{5}+c)^3}{27}.25/256\\ \le \dfrac{(13+\dfrac{11}{5}.5)^3}{27}.\dfrac{25}{256}\\ =50\ \square$$
 
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