Bài178 choa,b,c>0 a2 +b2 +c2=3. Tìm min P
P=$\frac{a^3{}}{\sqrt[3]{a+3b}}+\frac{b^3{}}{\sqrt[3]{b+3c}}+\frac{c^3{}}{\sqrt[3]{c+3a}}$
Theo Cauchy - Schwarz, ta có:
$P=\frac{a^3}{\sqrt[3]{a+3b}}+\frac{b^3}{\sqrt[3]{b+3c}}+\frac{c^3}{\sqrt[3]{c+3a}}\geq \frac{(a^2+b^2+c^2)^2}{a\sqrt[3]{a+3b}+b\sqrt[3]{b+3c}+c\sqrt[3]{c+3a}}$
Cũng theo Cauchy - Schwarz, ta có:
$(a\sqrt[3]{a+3b}+b\sqrt[3]{b+3c}+c\sqrt[3]{c+3a})^2\leq (a^2+b^2+c^2)(\sqrt[3]{(a+3b)^2}+\sqrt[3]{(b+3c)^2}+\sqrt[3]{(c+3a)^2})=3(\sqrt[3]{(a+3b)^2}+\sqrt[3]{(b+3c)^2}+\sqrt[3]{(c+3a)^2})$
Theo AM - GM, ta có:
$\sqrt[3]{(a+3b)^2}=\frac{1}{\sqrt[3]{4}}\sqrt[3]{(a+3b)(a+3b).4}\leq \frac{1}{\sqrt[3]{4}}.\frac{2a+6b+4}{3} $
Tương tự với các biểu thức còn lại, cộng theo từng vế, ta được:
$\sqrt[3]{(a+3b)^2}+\sqrt[3]{(b+3c)^2}+\sqrt[3]{(c+3a)^2}\leq \frac{1}{\sqrt[3]{4}}.\frac{8(a+b+c)+12}{3}\leq \frac{12}{\sqrt[3]{4}}$
do $(a+b+c)^2\leq 3(a^2+b^2+c^2)\Rightarrow a+b+c\leq 3$
Từ đây, suy ra:
$(a\sqrt[3]{a+3b}+b\sqrt[3]{b+3c}+c\sqrt[3]{c+3a})^2\leq 3.\frac{12}{\sqrt[3]{4}}=\frac{36}{\sqrt[3]{4}}$
$\Rightarrow a\sqrt[3]{a+3b}+b\sqrt[3]{b+3c}+c\sqrt[3]{c+3a}\leq \frac{6}{\sqrt[6]{4}}$
$\Rightarrow P\geq \frac{9}{\frac{6}{\sqrt[6]{4}}}=\frac{3}{\sqrt[3]{4}}$
Vậy $MinP=\frac{3}{\sqrt[3]{4}} $ tại $a=b=c=1$.