Question

Answer 1

Câu 3 nhe

 

Answer 2

Câu 4:

Áp dụng BĐT cauchy: \(\dfrac{1}{a}+\dfrac{1}{b}\ge\dfrac{4}{a+b};ab\le\dfrac{\left(a+b\right)^2}{4}\)

\(S=\dfrac{1}{x^2+y^2}+\dfrac{3}{4xy}=\dfrac{1}{x^2+y^2}+\dfrac{1}{2xy}+\dfrac{1}{4xy}\\ \Leftrightarrow S\ge\dfrac{4}{x^2+y^2+2xy}+\dfrac{1}{4\cdot\dfrac{\left(x+y\right)^2}{4}}\\ \Leftrightarrow S\ge\dfrac{4}{\left(x+y\right)^2}+\dfrac{1}{\left(x+y\right)^2}=\dfrac{5}{\left(x+y\right)^2}=\dfrac{5}{1}=5\)

Dấu \("="\Leftrightarrow x=y=\dfrac{1}{2}\)