1,đặt $\frac{1}{a}$ =x² ,$\frac{1}{b}$ =y² ,$\frac{1}{c}$ =z²
bdt⇔x²+y²+z²≥xy+yz+zx
2(x²+y²+z²)≥2(xy+yz+zx)
(x²+y²)+(y²+z²)+(z²+x²)≥ 2xy + 2yz+2zx
(x²-2xy+y²)+(y²-2yz+z²)+(z²-2zx+x²) ≥ 0
(x-y)²+(y-z)²+(z-x)²≥0 với mọi x,y,z
2,$\sqrt{(a+c)(b+d}$ $\geq$ $\sqrt{ab}$ +$\sqrt{cd}$
($\sqrt{(a+c)(b+d}$ )²≥($\sqrt{ab}$ +$\sqrt{cd}$ )²
(a+c)(b+d)≥ab+cd 2.$\sqrt{ab.cd}$
ab+cb+ad+cd≥ab+cd 2.$\sqrt{ab.cd}$
cb+ad≥2.$\sqrt{ab.cd}$
cb-2.$\sqrt{(cb).(ad)}$+ad≥0
($\sqrt{cd}$-$\sqrt{ad}$)²≥0 với mọi a,b,c,d>=0
