Câu hỏi của Annie Scarlet

Question

Cho x,y>0 thỏa mãn $x^2+y^2=2$. Tìm GTNN của $P=\frac{x^2}{\sqrt{y}}+\frac{y^2}{\sqrt{x}}$

Nguyễn Việt Lâm · Accepted Answer

Đặt $\left(\sqrt{x};\sqrt{y}\right)=\left(a;b\right)\Rightarrow a^4+b^4=2$

$\Rightarrow2=a^4+b^4\ge2a^2b^2\Rightarrow ab\le1$

Lại có: $a^4+a^4+a^4+1\ge4\sqrt[4]{a^{12}}=4a^3$ ; $3b^4+1\ge4b^3$

$\Rightarrow3\left(a^4+b^4\right)+2\ge4\left(a^3+b^3\right)\Rightarrow a^3+b^3\le2$

$P=\frac{a^4}{b}+\frac{b^4}{a}=\frac{a^8}{a^4b}+\frac{b^8}{ab^4}\ge\frac{\left(a^4+b^4\right)^2}{ab\left(a^3+b^3\right)}\ge\frac{2^2}{1.2}=2$

$P_{min}=2$ khi $a=b=1$Câu trả lời: Đặt $\left(\sqrt{x};\sqrt{y}\right)=\left(a;b\right)\Rightarrow a^4+b^4=2$

$\Rightarrow2=a^4+b^4\ge2a^2b^2\Rightarrow ab\le1$

Lại có: $a^4+a^4+a^4+1\ge4\sqrt[4]{a^{12}}=4a^3$ ; $3b^4+1\ge4b^3$

$\Rightarrow3\left(a^4+b^4\right)+2\ge4\left(a^3+b^3\right)\Rightarrow a^3+b^3\le2$

$P=\frac{a^4}{b}+\frac{b^4}{a}=\frac{a^8}{a^4b}+\frac{b^8}{ab^4}\ge\frac{\left(a^4+b^4\right)^2}{ab\left(a^3+b^3\right)}\ge\frac{2^2}{1.2}=2$

$P_{min}=2$ khi $a=b=1$

Violympic toán 9