Xét \(F+1=ab+bc+2ac+a^2+b^2+c^2\)
\(\Leftrightarrow F+1=\left(a+c\right)^2+b\left(a+c\right)+b^2\)
\(\Leftrightarrow\left(a+c\right)^2+b\left(a+c\right)+b^2-F-1=0\left(6\right)\)
Ta coi (6) là pt bậc 2 ẩn \(t=\left(a+c\right)\)
Để (6) có nghiệm thì
\(\Delta=b^2-4.1.\left(b^2-F-1\right)\ge0\)
\(\Rightarrow F\ge-1+\frac{3}{4}b^2\ge-1\)
Dấu = khi b=0 và \(a=-c=\pm\frac{\sqrt{2}}{2}\)
nãy là cách 1:
Cách 2: Từ \(a^2+b^2+c^2=1\)
\(\Rightarrow F=ab+bc+2ac=a^2+b^2+c^2+ab+bc+2ca-1\)
\(=\left[\left(a^2+2ac+c^2\right)+b\left(a+c\right)+\frac{b^2}{4}\right]+3\cdot\frac{b^2}{4}-1\)
\(=\left[\left(a+c\right)^2+b\left(a+c\right)+\frac{b^2}{4}\right]+3\cdot\frac{b^2}{4}-1\)
\(=\left[a+c+\frac{b}{2}\right]^2+3\cdot\frac{b^2}{4}-1\ge-1\)
Cách 3:Ta có: \(\left(a+b+c\right)^2\ge0\left(1\right)\)
\(\left(a+c\right)^2\ge0\left(2\right)\) và \(b^2\ge0\left(3\right)\)
Cộng theo vế (1),(2),(3) có:
\(\left(a+b+c\right)^2+\left(a+c\right)^2+b^2\ge0\)
\(\Leftrightarrow a^2+b^2+c^2+2\left(ab+bc+ca\right)+a^2+2ac+c^2+b^2\ge0\)
\(\Leftrightarrow2\left(a^2+b^2+c^2\right)+2\left(ab+bc+2ca\right)\ge0\)
\(\Leftrightarrow\left(a^2+b^2+c^2\right)+\left(ab+bc+2ca\right)\ge0\)
\(\Leftrightarrow\left(ab+bc+2ca\right)\ge-\left(a^2+b^2+c^2\right)=-1\)
\(\Rightarrow F\ge-1\)
Cách 4: Có:
\(\left(a+b+c\right)^2\ge0\)
\(\Rightarrow ab+bc+ca\ge-\frac{a^2+b^2+c^2}{2}=-\frac{1}{2}\left(1\right)\)
\(\left(a+c\right)^2\ge0\Rightarrow ca\ge-\frac{a^2+c^2}{2}=\frac{b^2-1}{2}\ge-\frac{1}{2}\left(2\right)\)
Cộng theo vế (1) và (2) có \(F\ge-1\)
Cách 5:Có \(a^2+b^2+c^2=1\)
\(\Leftrightarrow2\left(a^2+b^2+c^2\right)=2\)
\(\Rightarrow2F+2=2\left(ab+bc+2ca\right)+2\left(a^2+b^2+c^2\right)\)
\(=\left(a+b+c\right)^2+\left(a+c\right)^2+b^2\ge0\)
\(\Rightarrow F\ge-1\)
Cách 6:
Có: \(F=ab+bc+2ac=a^2+b^2+c^2+ab+bc+2ca-1\)
\(=\left(a^2+2ac+c^2\right)+b\left(a+c\right)+b^2-1\)
\(=\left(a+c\right)^2+b\left(a+c\right)+b^2-1\)
\(=\frac{3}{4}\left(a+b+c\right)^2+\frac{1}{4}\left(a-b+c\right)^2-1\ge-1\)
\(\Rightarrow F\ge-1\)
Xét \(F+1=ab+bc+2ca+a^2+b^2+c^2\)
\(\Leftrightarrow F+1=\left(a+c\right)^2+b\left(a+c\right)+b^2\)
\(\Leftrightarrow\left(a+c\right)^2+b\left(a+c\right)+b^2-F-1=0\left(6\right)\)
Ta coi ( 6 ) là phương trình bậc 2 ẩn \(t=\left(a+c\right)\)
Để ( 6 ) có nghiệm thì :
\(\Delta=b^2-4.1.\left(b^2-F-1\right)\ge0\)
\(\Rightarrow F\ge-1+\frac{3}{4}b^2\ge-1\)
Dấu = khi b = 0 và \(a=-c=\sqrt{\frac{2}{2}}\)và \(\sqrt{-\frac{2}{2}}\)
ủng hộ mik nhé .
đúng vậy toán này khó thiệt
toi cung chua biet cau tra loi du la xem qua nhung toi van ko hieu
