bài này dễ ẹt ak 

nhưng giúp mình bài này đi 

chotam giac abc . co canh bc=12cm, duong cao ah=8cm

a> tinh s tam giac abc

b> tren canh bc lay diem e sao cho be=3/4bc. tinh s tam giac abe va s tam giac ace ( bằng nhiều cách )

c> lay diem chinh giua cua canh ac va m . tinh s tam giac ame

\(S=\dfrac{1}{a^3+b^3}+\dfrac{1}{a^2b}+\dfrac{1}{ab^2}\ge\dfrac{1}{a^3+b^3}+\dfrac{4}{a^2b+ab^2}\)

\(S\ge\left(\dfrac{1}{a^3+b^3}+\dfrac{1}{a^2b+ab^2}+\dfrac{1}{a^2b+ab^2}+\dfrac{1}{a^2b+ab^2}\right)+\dfrac{1}{ab\left(a+b\right)}\)

\(S\ge\dfrac{16}{a^3+b^3+3a^2b+3ab^2}+\dfrac{1}{\dfrac{\left(a+b\right)^2}{4}.\left(a+b\right)}=\dfrac{20}{\left(a+b\right)^3}\ge20\)

\(S_{min}=20\) khi \(a=b=\dfrac{1}{2}\)

ab+bc+ca = 4abc

<=> 1/a + 1/b + 1/c = 4

Áp dụng bđt : x^2+y^2+z^2 >= (x+y+z)^2/3 thì :

P >= 1/a^2+1/b^2+1/c^2)^2 /3

   >= [(1/a+1/b+1/c)^2/3]^2/3

     = [(4^2)/3^]2/3 = 256/27

Dấu "=" xảy ra <=> a=b=c=3/4

Vậy ........

Tk mk nha

\(\dfrac{4}{3}=a+2\sqrt{\dfrac{a}{4}.b}+\dfrac{1}{2}\sqrt[3]{\dfrac{a}{2}.2b.8c}\)

\(\dfrac{4}{3}\le a+\dfrac{a}{4}+b+\dfrac{1}{6}\left(\dfrac{a}{2}+2b+8c\right)=\dfrac{4}{3}\left(a+b+c\right)\)

\(\Rightarrow a+b+c\ge1\)

Dấu "=" xảy ra khi \(\left(a;b;c\right)=\left(\dfrac{16}{21};\dfrac{4}{21};\dfrac{1}{21}\right)\)

\(A=\frac{1}{a}\)\(+\frac{1}{a}\)\(+\frac{1}{a}\)\(+\frac{1}{a}\)\(+\frac{1}{ab}\)\(\ge\frac{25}{4a+ab}\)\(=\frac{25}{a\left(b+4\right)}\)\(\ge\frac{25}{\frac{1}{4}\left(a+b+4\right)^2}\)\(=1\)

\(A_{min=1}\)\(khi\){ a = 5 

                            b = 1

Lần đầu tiên làm toán lớp 8 , có gì sai sót mong bạn chỉ ra hộ mình

\(A=\frac{4}{a}+\frac{1}{ab}=\frac{1}{a}+\frac{1}{a}+\frac{1}{a}+\frac{1}{a}+\frac{1}{ab}\)

\(\Rightarrow A=\frac{1}{a}+\frac{1}{a}+\frac{1}{a}+\frac{1}{a}+\frac{1}{ab}\ge\frac{5^2}{4a+ab}=\frac{25}{a\left(4+b\right)}\)

\(=\frac{25}{\frac{1}{4}\left(a+b+4\right)^2}=\frac{25}{\frac{1}{4}.100}=1\)

Dấu "=" xảy ra \(\Leftrightarrow\frac{25}{a\left(b+4\right)}=1\Leftrightarrow a\left(b+4\right)=25\)

Mà a,b dương => MinA = 1 <=> a = 1 ; b = 5 ( tmđk )

\(P=\dfrac{a^2+b^2+c^2}{ab+bc+ca}\ge\dfrac{ab+bc+ca}{ab+bc+ca}=1\)

\(P_{min}=1\) khi \(a=b=c=1\)

\(P=\dfrac{\left(a+b+c\right)^2-2\left(ab+bc+ca\right)}{ab+bc+ca}=\dfrac{9}{ab+bc+ca}-2\)

Do \(a;b\ge1\Rightarrow\left(a-1\right)\left(b-1\right)\ge0\Rightarrow ab\ge a+b-1=2-c\)

\(\Rightarrow ab+c\left(a+b\right)\ge2-c+c\left(3-c\right)=-c^2+2c+2=c\left(2-c\right)+2\ge2\)

\(\Rightarrow P\le\dfrac{9}{2}-2=\dfrac{5}{2}\)

\(P_{max}=\dfrac{5}{2}\) khi \(\left(a;b;c\right)=\left(1;2;0\right);\left(2;1;0\right)\)

Ta có:\(A\ge\left(a+b+1\right)\frac{\left(a+b\right)^2}{2}+\frac{4}{a+b}\)

Đặt \(t=a+b\)thì \(t\ge2\) theo AM-GM

Ta có:\(A\ge\frac{t^3}{2}+\frac{t^2}{2}+\frac{4}{t}=\frac{t^3}{2}+\frac{t^2}{4}+\frac{t^2}{4}+\frac{2}{t}+\frac{2}{t}\ge4+1+3=8\)

Đẳng thức xảy ra khi \(a=b=1\)

Áp dụng bđt cosi ta dc

P>= (2canab+1)(2ab)+4/(2canab)

=8

dấu = xảy ra khi a=b