Question

cho x>o, y>o thoa man dieu kien 1/x+1/y=1/2. tìm GTNN của biểu thức A=\(\sqrt{x}+\sqrt{y}\).giúp mình nhé!!

Answer 1

1/x+1/y=1/2 <=> (x+y)/xy=1/2 <=>[(\(\sqrt{x}+\sqrt{y}\))²-2\(\sqrt{xy}\)]/xy=1/2 <=>(\(\sqrt{x}+\sqrt{y}\))²=xy/2+2\(\sqrt{xy}\)=A²

1/2=1/x+1/y\(\ge\)2/\(\sqrt{xy}\)(bdt cosi cho 1/x và 1/y) <=>1/2 \(\ge\frac{2}{\sqrt{xy}}\)<=> \(\sqrt{xy}\ge\)4

Vậy A²\(\ge\)4²/2+2.4=16 <=> A\(\ge\)4( vì A >0)

Dấu = xảy ra khi 1/x=1/y và \(\sqrt{xy}=4\)=> x=y=4

Answer 2

\(\frac{1}{2}=\frac{1}{x}+\frac{1}{y}=\left(\frac{1}{\sqrt{x}}\right)^2+\left(\frac{1}{\sqrt{y}}\right)^2\ge\frac{1}{2}\left(\frac{1}{\sqrt{x}}+\frac{1}{\sqrt{y}}\right)^2\)

=> \(\left(\frac{1}{\sqrt{x}}+\frac{1}{\sqrt{y}}\right)^2\le1\)

=> \(\frac{1}{\sqrt{x}}+\frac{1}{\sqrt{y}}\le1\)

=> \(1\ge\frac{1^2}{\sqrt{x}}+\frac{1^2}{\sqrt{y}}\ge\frac{\left(1+1\right)^2}{\sqrt{x}+\sqrt{y}}=\frac{4}{\sqrt{x}+\sqrt{y}}\)

=> \(\sqrt{x}+\sqrt{y}\ge4\)

Dấu " = " xảy ra <=> \(\hept{\begin{cases}\frac{1}{\sqrt{x}}=\frac{1}{\sqrt{y}}\\\frac{1}{x}+\frac{1}{y}=\frac{1}{2}\end{cases}}\Leftrightarrow x=y=4\)

Vậy min A = 4 đạt tại x = y= 4.