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acidnitric_hno3

vodinhtu1995@gmail.com said:
Cho a, b, c dương [tex]a^2 + b^2 + c^2 = 3 [/tex]. Tìm giá trị nhỏ nhất của biểu thức.
P=[tex] \frac{a^3}{\sqrt{b^2+3}} +\frac{b^3}{\sqrt{c^2+3}} + \frac{c^3}{\sqrt{a^2+3}}[/tex]

Em xin cảm ơn.
Bấm để xem đầy đủ nội dung ...

$P=\frac{a^3}{\sqrt{b^2+3}} +\frac{b^3}{\sqrt{c^2+3}} + \frac{c^3}{\sqrt{a^2+3}}$

Ta co: [TEX]\frac{a^3}{2\sqrt{b^2+3}}+\frac{a^3}{2\sqrt{b^2+3}}+\frac{b^2+3}{16}\geq \frac{3a^2}{4}[/TEX]

[TEX]\frac{b^3}{2\sqrt{c^2+3}}+\frac{b^3}{2\sqrt{c^2+3}}+\frac{c^2+3}{16}\geq \frac{3b^2}{4}[/TEX]

[TEX]\frac{c^3}{2\sqrt{a^2+3}}+\frac{c^3}{2\sqrt{a^2+3}}+\frac{a^2+3}{16}\geq \frac{3c^2}{4}[/TEX]

[TEX]P+ \frac{b^2+3+a^2+3+c^2+3}{16}\geq \frac{3}{4}(a^2+b^2+c^2)[/TEX]
[TEX]P\geq \frac{3}{2}[/TEX]
 
T

tranvanhung7997

Áp dụng BĐT Cauchy-Schwarz:
$P=\frac{a^3}{\sqrt[]{b^2+3}}+\frac{b^3}{\sqrt[]{c^2+3}}+\frac{c^3}{\sqrt[]{a^2+3}}=\frac{a^4}{a\sqrt[]{b^2+3}}+\frac{b^4}{b\sqrt[]{c^2+3}}+\frac{c^4}{c\sqrt[]{a^2+3}}$
$ \ge \frac{(a^2+b^2+c^2)^2}{a.\sqrt[]{b^2+3}+b.\sqrt[]{c^2+3}+c.\sqrt[]{a^2+3}}$
Ta có: $(a\sqrt[]{b^2+3}+b\sqrt[]{c^2+3}+c\sqrt[]{a^2+3})^2 \le (a^2+b^2+c^2)(a^2+b^2+c^2+9)=3.(3+9)=36$
\Rightarrow $a\sqrt[]{b^2+3}+b\sqrt[]{c^2+3}+c\sqrt[]{a^2+3} \le 6$
\Rightarrow $P \ge \frac{3^2}{6}=\frac{3}{2}$
Dấu "=" có \Leftrightarrow $a=b=c=1$
 
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