\(a^2+b^2+c^2-3\left(a+b-c\right)=a^2-4a+4+b^2-4b+4+c^2-c+\dfrac{1}{4}+a+b+4c-\left(4+4+\dfrac{1}{4}\right)=\left(a-2\right)^2+\left(b-2\right)^2+\left(c-\dfrac{1}{2}\right)^2+a+b+4c-\dfrac{33}{4}\ge3\sqrt[3]{4abc}-\dfrac{33}{4}=3\sqrt[3]{4.2}-\dfrac{33}{4}=-\dfrac{9}{4}\left(đpcm\right)\)

\(dấu"="\Leftrightarrow a=b=2;c=\dfrac{1}{2}\)

\(\left|\overrightarrow{AB}+\overrightarrow{AC}\right|=\left|2\overrightarrow{AM}\right|=2AM\)\(\)(M là trung điểm BC)

\(=2AM=2\sqrt{AC^2-\left(\dfrac{BC}{2}\right)^2}=2\sqrt{a^2-\dfrac{a^2}{4}}=a\sqrt{3}\)

a; hạ các đường cao BK,CI vuông góc với AM

nên tổng các khoảng cách từ B và C đến AM là

\(d=BK+CI\)

xét tam giác BKM vuông tại K có: \(BK\le BM\)(bđt tam giác)

tương tự tam giác CMI vuông tại I : \(CI\le CM\)

\(\Rightarrow d=CI+BK\le BM+CM=BC\)(do \(BM=CM=\dfrac{1}{2}BC\))

b; để khoảng cách d lớn nhất

từ \(BK\le BM\) dấu"=" xảy ra khi K trùng M

tương tự \(CI\le CM\)dấu"=" xảy rakhi I trùng M

=>M trùng với hình chiếu B,C lên AM

\(\Sigma\dfrac{a}{\sqrt{2a+b}}=\Sigma\dfrac{a}{\sqrt{\left(a+b+c\right)\left(2a+b\right)}}\le\dfrac{\dfrac{a}{a+b+c}+\dfrac{a}{2a+b}+\dfrac{b}{a+b+c}+\dfrac{b}{2b+c}+\dfrac{c}{a+b+c}+\dfrac{c}{2c+a}}{2}=\dfrac{1+\dfrac{a}{2a+b}+\dfrac{b}{2b+c}+\dfrac{c}{2c+a}}{2}\)

\(ta\) \(đi\) \(cminh:\) \(\dfrac{a}{2a+b}+\dfrac{b}{2b+c}+\dfrac{c}{2c+a}\le1\left(1\right)\)

\(\left(1\right)\Leftrightarrow\left(2a+b\right)\left(2b+c\right)\left(2c+a\right)\ge a\left(2b+c\right)\left(2c+a\right)+b\left(2a+b\right)\left(2c+a\right)+c\left(2a+b\right)\left(2b+c\right)\)

\(\Leftrightarrow a^2c+b^2a+bc^2\ge3abc\)(đúng theo cô si 3 số)

do đó \(\left(1\right)\)đúng \(\Rightarrow\Sigma\dfrac{a}{\sqrt{2a+b}}\le\dfrac{1+1}{2}=1\)

\(đặt:a+1008=x;b+1008=y;c+1008=z\)

\(\Rightarrow A=\dfrac{x}{y+z}+\dfrac{y}{x+z}+\dfrac{z}{x+y}=\dfrac{x^2}{xy+xz}+\dfrac{y^2}{xy+yz}+\dfrac{z^2}{xz+yz}\ge\dfrac{\left(x+y+z\right)^2}{2\left(xy+yz+xz\right)}\ge\dfrac{\left(x+y+z\right)^2}{2.\dfrac{\left(x+y+z\right)^2}{3}}=\dfrac{3}{2}\)

\(dấu"="\Leftrightarrow x=y=z\Leftrightarrow a=b=c\)

\(chuẩn\) \(hóa\) \(a+b+c=3\) \(\left(do:f\left(ta;tb;tc\right)=f\left(a;b;c\right)\right)\left(0< a;b;c< 3\right)\)

\(\Rightarrow\dfrac{b+c}{-a+b+c}=\dfrac{3-a}{-a+3-a}=\dfrac{3-a}{3-2a}\ge3a-1\)\(\left(1\right)\)

\(thật\) \(vậy\) \(\left(1\right)\Leftrightarrow3-a\ge\left(3a-1\right)\left(3-2a\right)\Leftrightarrow3-a-\left(3-2a\right)\left(3a-1\right)\ge0\Leftrightarrow6\left(a-1\right)^2\ge0\left(đúng\right)\)

\(\Rightarrow A\ge3a-1+3b-1+3c-1=3\left(a+b+c\right)-3=3.3-3=6\)\(\)

kẻ thêm \(AI\perp AP\left(AI\cap CD\left(I\right)\right)\)

\(\Rightarrow AD=AB\left(ABCDvuoong\right)\);góc ABM=góc ADI=90

ta có \(AE//FP\Rightarrow\) góc BAM=góc API(so le trong)

lại có: trong tam giác AIP vuông tại A có

góc AIP+ góc API=90

=>góc BAM+góc AIP=90(1)

lại có trong tam giác AID vuông tại D có góc DAI+góc AIP=90(2)

từ(1)(2)=>góc DAI=góc BAM

=>tam giác ABM=tam giác ADI(g.c.g)

=>BM=DI và AI=AM

do AK là phân giác DAM=>góc DAK=góc KAP

lại có: góc IAK=góc IAD+góc DAK

=>góc IAK=góc KAP+góc BAM=góc KAB

có góc KAB=góc AKI(so le trong)

=>góc IAK=góc AKI =>tam giác AIK cân tại I=>AI=IK

lại có AE\(//\)IF

\(AI\perp AP,EF\perp AP\Rightarrow AI//EF\Rightarrow\) AEFI là hình bình hành

\(\Rightarrow\)\(EF=AI=IK=ID+DK=BM+DK\left(đpcm\right)\)

\(b;\) xét tam giác AIP vuông tại A đường cao AD

\(\Rightarrow\dfrac{1}{AD^2}=\dfrac{1}{AI^2}+\dfrac{1}{AP^2}\Rightarrow\dfrac{1}{AB^2}=\dfrac{1}{AM^2}+\dfrac{1}{AP^2}\)(do AD=AB;AI=AM)

\(\sqrt[3]{9}T=\sqrt[3]{3}\Sigma\sqrt{a+2b}=\Sigma\sqrt[3]{3.3\left(a+2b\right)}\le\dfrac{a+2b+3+3+b+2c+3+3+c+2a+3+3}{3}=\dfrac{3\left(a+b+c\right)+18}{3}=9\Leftrightarrow T\le\dfrac{9}{\sqrt[3]{9}}\)

\(2,,,,K=\Sigma\sqrt{x^2+\dfrac{1}{x^2}}\ge\sqrt{\left(x+y+z\right)^2+\left(\dfrac{1}{x^{ }}+\dfrac{1}{y^{ }}+\dfrac{1}{z}\right)^2}\ge\sqrt{\left(x+y+z\right)^2+\left(\dfrac{9}{x+y+z}\right)^2}=\sqrt{\left(x+y+z\right)^2+\dfrac{81}{\left(x+y+z\right)^2}}=\sqrt{\left(x+y+z\right)^2+\dfrac{1}{\left(x+y+z\right)^2}+\dfrac{80}{\left(x+y+z\right)^2}}\ge\sqrt{2+80}=\sqrt{82}\)

\(\dfrac{6}{a}+\dfrac{3}{b^2}+\dfrac{2}{c^3}=\dfrac{6}{a}+6a+\dfrac{3}{b^2}+3b+3b+\dfrac{2}{c^3}+2c+2c+2c-6\left(a+b+c\right)\ge2\sqrt{6.6}+3\sqrt[3]{3.3.3}+4\sqrt[4]{2.2.2.2}-6.3=11\)

\(dấu"="\Leftrightarrow a=b=c=1\)

\(\sqrt{a+bc}=\sqrt{a\left(a+b+c\right)+bc}=\sqrt{\left(a+b\right)\left(a+c\right)}\Rightarrow\dfrac{a}{\sqrt{a+bc}}=\dfrac{a}{\sqrt{\left(a+b\right)\left(a+c\right)}}\le\dfrac{\dfrac{a}{a+b}+\dfrac{a}{a+c}}{2}\)

\(\Rightarrow\Sigma\dfrac{a}{\sqrt{a+bc}}=\Sigma\dfrac{a}{\sqrt{\left(a+c\right)\left(a+b\right)}}\le\dfrac{\dfrac{a}{a+c}+\dfrac{a}{a+b}+\dfrac{b}{b+c}+\dfrac{b}{a+b}+\dfrac{c}{c+a}+\dfrac{c}{c+b}}{2}=\dfrac{3}{2}\)

\(dấu"="\Leftrightarrow a=b=c=\dfrac{1}{3}\)