Câu hỏi của Vũ Phương Thảo

Question

Cho x,y >=0 và x+y=1

Tìm GTNN của A= $\left(\dfrac{x+1}{y}\right)^2+\left(\dfrac{y+1}{x}\right)^2$

Phạm Nguyễn Tất Đạt · Accepted Answer

$A=\left(\dfrac{x+1}{y}\right)^2+\left(\dfrac{y+1}{x}\right)^2$

$A=\left(\dfrac{x+x+y}{y}\right)^2+\left(\dfrac{y+x+y}{x}\right)^2$

$A=\left(\dfrac{2x}{y}+1\right)^2+\left(\dfrac{2y}{x}+1\right)^2$

$A=\dfrac{4x^2}{y^2}+\dfrac{4x}{y}+1+\dfrac{4y^2}{x^2}+\dfrac{4y}{x}+1$

$A\ge8+8+2=18$

$\Rightarrow MINA=18\Leftrightarrow x=y=\dfrac{1}{2}$Câu trả lời: $A=\left(\dfrac{x+1}{y}\right)^2+\left(\dfrac{y+1}{x}\right)^2$

$A=\left(\dfrac{x+x+y}{y}\right)^2+\left(\dfrac{y+x+y}{x}\right)^2$

$A=\left(\dfrac{2x}{y}+1\right)^2+\left(\dfrac{2y}{x}+1\right)^2$

$A=\dfrac{4x^2}{y^2}+\dfrac{4x}{y}+1+\dfrac{4y^2}{x^2}+\dfrac{4y}{x}+1$

$A\ge8+8+2=18$

$\Rightarrow MINA=18\Leftrightarrow x=y=\dfrac{1}{2}$

Vũ Phương Thảo · Answer

Akai HamuraCâu trả lời: Akai Hamura

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