Question

Cho x,y > 0 va x + y = 1. CMR: 8\(\left(x^4+y^4\right)\) + \(\frac{1}{xy}\) ≥ 5

Answer 1

Với mọi x,y >0 có \(\left(x+y\right)^2\ge4xy\)

=> \(1\ge4xy\) (do x+y=1) <=> \(\frac{1}{xy}\ge4\)

​​Lại có \(x^2+y^2\ge2xy\)

<=> \(2\left(x^2+y^2\right)\ge\left(x+y\right)^2=1\)

<=> \(x^2+y^2\ge\frac{1}{2}\)

Có \(x^4+y^4\ge2x^2y^2\)

<=> \(2\left(x^4+y^4\right)\ge\left(x^2+y^2\right)^2\ge\left(\frac{1}{2}\right)^2\)

<=> \(8\left(x^4+y^4\right)\ge\frac{1}{4}.4=1\)

=> \(8\left(x^4+y^4\right)+\frac{1}{xy}\ge1+4=5\)

Dấu "=" xảy ra <=> x=y=\(\frac{1}{2}\)

\(2\left(x^2+y^2\right)\ge\left(x+y\right)^2=1\)​