Question

Cho a , b , c ,d > 0 và abcd =1

CM : a² +b²+c²+d² +a(b+c)+ b(c+d) + d(c+a) \(\ge\) 10

Answer 1

Ta có : a² + b² \(\ge2ab\)

\(c^2+d^2\ge2cd\)

Do abcd = 1 nên cd =\(\dfrac{1}{ab}\)( dùng \(x+\dfrac{1}{x}\ge\dfrac{1}{2}\))

Ta có :\(a^2+b^2+c^2\ge2\left(ab+cd\right)=2\left(ab+\dfrac{1}{ab}\right)\ge4\)(1)

Mặt khác : a(b+c) +b(c+d)+d(c+a)

=(ab+cd)+(ac+bd)+(bc+ad)

=\(\left(ab+\dfrac{1}{ab}\right)+\left(ac+\dfrac{1}{ac}\right)+\left(bc+\dfrac{1}{bc}\right)\ge2+2+2\)

Vậy \(a^2+b^2+c^2+d^2+a\left(b+c\right)+b\left(c+d\right)+d\left(c+a\right)\ge10\)