\(ab+2bc+3ac\)
\(=ab+2bc+ac+2ac\)
\(=a\left(b+c\right)+2c\left(a+b\right)\)
\(=-a^2-2b^2\le0\) (đúng)
Dấu "=" khi \(x=y=z=0\)
Ta có:
a+b+c=0
=> a + b = -c
=> (a+b)2 = c2
=> a2 + 2ab + b2 = c2
=> ab = \(\dfrac{c^2-a^2-b^2}{2}\) (1)
Tương tự ta có: a2 + 2ac + c2 = b2
b2 + 2bc + c2 = a2
=> ac = \(\dfrac{b^2-a^2-c^2}{2}\) => 3ac = \(\dfrac{3b^2-3a^2-3c^2}{2}\) (2)
bc = \(\dfrac{a^2-b^2-c^2}{2}\) => 2bc = a2 - b2 - c2 (3)
Thay (1), (2), (3) vào bdt cần ch/m, ta có:
ab + 2bc + 3ac ≤ 0
<=> \(\dfrac{c^2-a^2-b^2}{2}\) + a2 - b2 - c2 + \(\dfrac{3b^2-3a^2-3c^2}{2}\)
<=> c2 - a2 - b2 + 2a2 - 2b2 - 2c2 + 3b2 - 3a2 - 3c2 ≤ 0
<=> -2a2 -4c2 ≤ 0
<=> -2(a2 + 2c2) ≤ 0 (Bdt đúng với mọi a, c)
Dau "=" xay ra khi a2 + 2c2 = 0
<=> a = c = b = 0.
