+, Ta có: \(a+b+c+ab+bc+ac=6abc\) nên \(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ac}+\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=6\)

+, \(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\)

\(=\frac13\left(\frac{3}{a^2}+\frac{3}{b^2}+\frac{3}{c^2}\right)\)

\(=\frac13\left\lbrack\left(\frac{1}{a^2}+1\right)+\left(\frac{1}{b^2}+1\right)+\left(\frac{1}{c^2}+1\right)-3+\left(\frac{1}{a^2}+\frac{1}{b^2}\right)+\left(\frac{1}{b^2}+\frac{1}{c^2}\right)+\left(\frac{1}{a^2}+\frac{1}{c^2}\right)\right\rbrack\)

\(\ge\frac13\left(2\sqrt{\frac{1}{a^2}.1}+2\sqrt{\frac{1}{b^2}.1}+2\sqrt{\frac{1}{c^2}+1}-3+2\sqrt{\frac{1}{a^2b^2}}+2\sqrt{\frac{1}{b^2c^2}}+2\sqrt{\frac{1}{a^2c^2}}\right)\)

\(=\frac13\left(\frac{2}{a}+\frac{2}{b}+\frac{2}{c}-3+\frac{2}{ab}+\frac{2}{bc}+\frac{2}{ac}\right)\)

\(=\frac23\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ac}\right)-1\)

\(=\frac{2.6}{3}-1\)

\(=3\) ( đpcm )

( Dấu "=" xảy ra khi \(a=b=c=1\) )

\(x^2+y^2+xy-3x-3y+3\)

\(=\left(x^2-2x+1\right)+\left(y^2-2y+1\right)+\left(xy-x-y+1\right)\)

\(=\left(x-1\right)^2+\left(y-1\right)^2+\left\lbrack x\left(y-1\right)-\left(y-1\right)\right\rbrack\)

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

\(=\left\lbrack\left(x-1\right)^2+\left(x-1\right)\left(y-1\right)+\frac{\left(y-1\right)^2}{4}\right\rbrack+\frac{3\left(y-1\right)^2}{4}\)

\(=\left\lbrack\left(x-1\right)+\frac{y-1}{4}\right\rbrack^2+\frac{3\left(y-1\right)^2}{4}\)

\(\ge0\) ( đpcm )

( Dấu "=" xảy ra khi \(x=y=1\) )

1)

\(a^2-ab+b^2\)

\(=a^2-ab+\frac{b^2}{4}+\frac{3b^2}{4}\)

\(=\left(a-\frac{b}{2}\right)^2+\frac{3b^2}{4}\)

\(\ge0\) ( đpcm )

( Dấu "=" xảy ra khi \(a=b=0\) )

2)

\(a^2-ab+b^2\)

\(=\left(\frac{3a^2}{4}-\frac{3ab}{2}+\frac{3b^2}{4}\right)+\left(\frac{a^2}{4}+\frac{ab}{2}+\frac{b^2}{4}\right)\)

\(=\frac{3\left(a-b\right)^2}{4}+\frac{\left(a+b\right)^2}{4}\)

\(\ge\frac14\left(a+b\right)^2\) ( đpcm )

( Dấu "=" xảy ra khi \(a=b\) )

\(\sqrt{a^2-ab+b^2}+\sqrt{b^2-bc+c^2}+\sqrt{a^2-ac+c^2}\)

\(=\sum_{cyc}^{a,b,c}\sqrt{a^2-ab+b^2}\)

\(=\sum_{cyc}^{a,b,c}\sqrt{\left(\frac{3a^2}{4}+\frac{3b^2}{4}-\frac{3ab}{2}\right)+\left(\frac{a^2}{4}+\frac{b^2}{4}+\frac{ab}{2}\right)}\)

\(=\sum_{cyc}^{a,b,c}\sqrt{\frac34\left(a-b\right)^2+\frac14\left(a+b\right)^2}\)

\(\ge\sum_{cyc}^{a,b,c}\sqrt{\frac14\left(a+b\right)^2}\)

\(=\sum_{cyc}^{a,b,c}\frac{a+b}{2}\)

\(=\frac{a+b}{2}+\frac{b+c}{2}+\frac{a+c}{2}\)

\(=a+b+c\)

\(=3\) ( đpcm ).

( Dấu "=" xảy ra khi \(a=b=c=1\) )

+, Ta có:

\(x\left(x-y\right)\left(x-z\right)+y\left(y-x\right)\left(y-z\right)+z\left(z-x\right)\left(z-y\right)\)

\(=x\left(x-y\right)\left(x-z\right)-y\left(y-x\right)\left(z-y\right)+z\left(z-x\right)\left(z-y\right)\)

\(=x\left(x-y\right)\left(x-z\right)+\left(z-y\right)\left\lbrack z\left(z-x\right)-y\left(y-x\right)\right\rbrack\)

\(=x\left(x-y\right)\left(x-z\right)+\left(z-y\right)\left(z^2-y^2-xz+xy\right)\)

\(=x\left(x-y\right)\left(x-z\right)+\left(z-y\right)\left\lbrack\left(z-y\right)\left(y+z\right)-x\left(z-y\right)\right\rbrack\)

\(=x\left(y-x\right)\left(z-x\right)+\left(z-y\right)^2\left(y+z-x\right)\)

+, Do \(z\ge y\ge x\ge0\) nên \(\begin{cases}x\ge0\\ y-x\ge0\\ z-x\ge0\\ y+z-x\ge0\end{cases}\) thì \(\begin{cases}x\left(y-x\right)\left(z-x\right)\ge0\\ \left(z-y\right)^2\left(y+z-x\right)\ge0\end{cases}\)

( do \(\left(z-y^{}\right)^2\ge0\) \(\forall y,z\) )

hay \(x\left(y-x\right)\left(z-x\right)+\left(y+z-x\right)\left(z-y\right)^2\ge0\)

cho nên \(x\left(x-y\right)\left(x-z\right)+y\left(y-x\right)\left(y-z\right)+z\left(z-x\right)\left(z-y\right)\ge0\) ( đpcm ).

( Dấu "=" xảy ra khi \(x=y=z\) ).

\(2\left(\frac23\right)^2.2x^2.10^2+\frac14.45^1:4\)

= \(\frac49.4.100x^2+\frac{45}{16}\)

=\(\frac{1600x^2}{9}+\frac{45}{16}\)

\(\sqrt{4^2+2^2}=2\sqrt5\) ❏

△CAH ∼ △CBA ( g.g ) nên CH / AC = AH / AB ( cặp cạnh tương ứng tỉ lệ ) thì AC / AB = CH / AH.

△ABE vuông tại A có D là trung điểm BE nên AD là tia phân giác góc BAE ( tính chất ) thì AG là tia phân giác góc BAC hay CG / BG = AC / AB ( tính chất ) mà AC / AB = CH / AH ( cmt ) nên CG / BG = CH / AH thì ( CG / BG ) + 1 = ( CH / AH ) + 1 = BC / BG = ( AH + CH ) / AH mà AH = HI nên BC / BG = ( HI + CH ) / AH thì BG / BC = AH / ( HI + CH ) ❏

Đặt các điểm như hình.

EF // AB // CD nên EF / AB = CF / BC và EF / CD = BF / BC ( Định lí Thales )

mà ( CF / BC ) + ( BF / BC ) = BC / BC = 1 nên ( EF / AB ) + ( EF / CD ) = ( EF / 3 ) + ( EF / 2 ) = 1 thì EF = 6 / 5 ( m )