Question

cho x,y,z dương thỏa \(x^2+y^2+z^2=3\)

C/M \(\dfrac{x}{3-yz}+\dfrac{y}{3-zx}+\dfrac{z}{3-xy}\le\dfrac{3}{2}\)

Answer 1

\(A=\sum\dfrac{x}{3-yz}\le\dfrac{x}{x^2+y^2+z^2-\dfrac{y^2+z^2}{2}}=\dfrac{2x}{x^2+3}\le\dfrac{x^2+1}{x^2+3}=1-\dfrac{2}{x^2+3}.\)

Ta co \(\dfrac{1}{x^2+3}+\dfrac{1}{y^2+3}+\dfrac{1}{z^2+3}\ge\dfrac{9}{3+9}=\dfrac{3}{4}.\)

=>\(A\le3-2.\dfrac{3}{4}=\dfrac{3}{2}\)

A max = 3/ 2 khi x =y =z =1

Answer 2

nguồn:Chuyên toán HN năm nay @Ace Legona xem thử