Question

A) (a+1)(a-3)(a-4)(a-6)+10>0 B) (ab+bc+ca)²>=3abc(a+b+c)

Answer 1

a) \(VT=\left(a-1\right)\left(a-3\right)\left(a-4\right)\left(a-6\right)+10=\left(a^2-7a+6\right)\left(a^2-7a+12\right)+10\)

Đặt : \(a^2-7a+9=t\) , ta có :

\(VT=\left(t-3\right)\left(t+3\right)+10=t^2-9+10=t^2+1>0\)

⇒ đpcm

b) \(\left(ab+bc+ca\right)^2\) ≥ \(3abc\left(a+b+c\right)\)

⇔ \(\left(ab+bc+ca\right)^2-3ab.ac-3ab.bc-3ac.bc\) ≥ 0

Đặt : x = ab ; y = bc ; z = ac . Ta có :

\(\left(x+y+z\right)^2-3xz-3xy-3yz\) ≥ 0

⇔ \(x^2+y^2+z^2-xy-yz-xz\) ≥ 0

⇔ \(2\left(x^2+y^2+z^2-xy-yz-xz\right)\) ≥ 0

⇔ \(x^2-2xy+y^2+y^2-2yz+z^2+x^2-2xz+z^2\) ≥ 0

⇔ \(\left(x-y\right)^2+\left(y-z\right)^2+\left(x-z\right)^2\) ≥ 0 ( Luôn đúng )

⇒ đpcm