Question

Answer 1

Ta có x³ + y³ \(\le8-6xy\)

<=> (x + y)³ - 8 - 3xy(x + y) + 6xy \(\le0\)

<=> (x + y - 2)[(x + y)² + 2(x + y) + 4)] - 3xy(x + y - 2) \(\le0\)

<=> (x + y - 2)(x² - xy + y² + 2x + 2y + 4) \(\le0\)

<=> x + y - 2 \(\le\) 0 (Vì x² - xy + y² + 2x + 2y + 4 > 0 \(\forall xy\)) 

<=> x + y \(\le2\)

Khi đó 

P = \(\dfrac{1}{x^2+y^2}+\dfrac{2}{xy}=\dfrac{1}{x^2+y^2}+\dfrac{1}{2xy}+\dfrac{3}{2xy}\ge\dfrac{4}{x^2+y^2+2xy}+\dfrac{3}{2xy}\)

\(\dfrac{4}{\left(x+y\right)^2}+\dfrac{3}{2xy}\ge\dfrac{4}{\left(x+y\right)^2}+\dfrac{6}{\left(x+y\right)^2}=\dfrac{10}{\left(x+y\right)^2}\ge\dfrac{10}{2^2}=\dfrac{5}{2}\)

Dấu "=" xảy ra <=> x = y = 1