Question

cho x>0;y>0;\(x+y\le1\) chứng minh \(\dfrac{1}{x^2+xy}+\dfrac{1}{y^2+xy}\ge4\)

Answer 1

Áp dụng BĐT Cô si cho 2 số dương a,b ta có \(\dfrac{a+b}{2}\ge\sqrt{ab}\)

\(\dfrac{1}{a}+\dfrac{1}{b}\ge2.\sqrt{\dfrac{1}{a}.\dfrac{1}{b}}=>\left(a+b\right)\left(\dfrac{1}{a}+\dfrac{1}{b}\right)\ge2\sqrt{ab}.2\sqrt{\dfrac{1}{a}.\dfrac{1}{b}}\)

suy ra \(\left(a+b\right)\left(\dfrac{1}{a}+\dfrac{1}{b}\right)\ge4\Rightarrow\dfrac{1}{a}+\dfrac{1}{b}\ge\dfrac{4}{a+b}\).Áp dụng vào bài toán ta có :\(\dfrac{1}{x^2+xy}+\dfrac{1}{y^2+xy}\ge\dfrac{4}{x^2+xy+y^2+xy}=\dfrac{4}{\left(x+y\right)^2}\ge4\) (Do \(x+y\le1\))

Answer 2

Áp dụng bất đẳng thức Cauchy-Schwarz:

\(\dfrac{1}{x^2+xy}+\dfrac{1}{y^2+xy}\ge\dfrac{\left(1+1\right)^2}{x^2+2xy+y^2}=\dfrac{4}{\left(x+y\right)^2}\ge\dfrac{4}{1}=4\)