C/m BĐT phụ:   \(\frac{1}{x}+\frac{1}{y}\ge\frac{4}{x+y}\)  (*)      (x,y dương)

Ta có:   \(\left(x-y\right)^2\ge0\)       

\(\Leftrightarrow\)\(x^2-2xy+y^2\ge0\)

\(\Leftrightarrow\)\(x^2+y^2\ge2xy\)

\(\Leftrightarrow\)\(x^2+2xy+y^2\ge4xy\)

\(\Leftrightarrow\)\(\left(x+y\right)^2\ge4xy\)

\(\Leftrightarrow\)\(\frac{x+y}{xy}\ge\frac{4}{x+y}\)

\(\Leftrightarrow\)\(\frac{1}{x}+\frac{1}{y}\ge\frac{4}{x+y}\)   (BĐT đã đc chứng minh)

Dấu "=" xảy ra  \(\Leftrightarrow\)\(x=y\)

ÁP dụng BĐT (*) ta có:

\(\frac{1}{p-a}+\frac{1}{p-b}\ge\frac{4}{p-a+p-b}=\frac{4}{2p-\left(a+b\right)}=\frac{4}{c}\)  (1)

\(\frac{1}{p-b}+\frac{1}{p-c}\ge\frac{4}{p-b+p-c}=\frac{4}{2p-\left(b+c\right)}=\frac{4}{a}\)  (2)

\(\frac{1}{p-c}+\frac{1}{p-a}\ge\frac{4}{p-c+p-a}=\frac{4}{2p-\left(c+a\right)}=\frac{4}{b}\) (3)

Lấy (1); (2); (3) cộng theo vế ta được:

          \(2\left(\frac{1}{p-a}+\frac{1}{p-b}+\frac{1}{p-c}\right)\ge4\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\)

\(\Leftrightarrow\)\(\frac{1}{p-a}+\frac{1}{p-b}+\frac{1}{p-c}\ge2\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\)  (đpcm)

Dấu "=" xảy ra  \(\Leftrightarrow\)\(a=b=c\)

Khi đó  \(\Delta ABC\)là tam giác đều

A=(x+2y)^2 + (y-1)^2 - 4 => min A = -4 khi y=1, x=-2y=-2

\(VT=\left(x-1\right)\left(x-3\right)\left(x-4\right)\left(x-6\right)+9\)

\(=\left(x^2-7x+6\right)\left(x^2-7x+12\right)+9\)

Đặt   \(x^2-7x+9=a\)  ta có:

\(VT=\left(a-3\right)\left(a+3\right)+9\)\(=a^2-9+9=a^2\)\(\ge0\)  (đpcm)

hinh bn tu ve nhe

\(\infty:\)dong dang

\(\Delta ABD\infty\Delta ACE\)(g.g)     \(\Rightarrow\frac{AE}{AD}=\frac{AC}{AB}\)

\(\Rightarrow AE.AB=AD.AC\)  (1)

\(\Delta AMB\infty\Delta AEM\)(g.g)    \(\Rightarrow\frac{AM}{AE}=\frac{AB}{AM}\Rightarrow AM^2=AE.AB\)(2)

\(\Delta ANC\infty\Delta ADN\)(g.g)      \(\Rightarrow\frac{AN}{AD}=\frac{AC}{AN}\Rightarrow AN^2=AD.AC\)(3)

Tu (1), (2), (3) \(\Rightarrow AM^2=AN^2\Rightarrow AM=AN\)

\(\Rightarrow\)\(\Delta AMN\)can tai A

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