n thuộc nha bạn mình viết nhầm

Một hình vuông có cạnh là 8 cm. Vậy diện tích hình vuông là:

A.  32 cm                                        

B.     64 c m 2                                                       

C.  64 cm

áp dụng bdt côsi \(\dfrac{a^2}{b^3}+\dfrac{1}{a}+\dfrac{1}{a}\ge\dfrac{3}{b}\)

tuông tu \(\dfrac{b^2}{c^3}+\dfrac{1}{b}+\dfrac{1}{b}\ge\dfrac{3}{c}\)

\(\dfrac{c^2}{a^3}+\dfrac{1}{c}+\dfrac{1}{c}\ge\dfrac{3}{a}\)

suy ra vt +\(2\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\ge3\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\)

suy ra dpcm

dau = xay ra khi a=b=c

ta có \(\left(x+y+z\right)^2=x^2+y^2+z^2+2\left(xy+yz+xz\right)\)

\(\Rightarrow x^2+y^2+z^2=9\)

áp dung bu nhi a \(2\left(y^2+z^2\right)\ge\left(y+z\right)^2\)

\(\Leftrightarrow2\left(9-x^2\right)\ge\left(5-x\right)^2\)

\(\Leftrightarrow18-2x^2\ge25-10x+x^2\)

\(\Leftrightarrow0< =3x^2-10x+7\)

suy ra 1<=x<=7/3

theo de bai ta co \(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=1\) suy ra ab+bc+ac=abc

\(\dfrac{a^2}{a+bc}=\dfrac{a^3}{a^2+abc}=\dfrac{a^3}{a^2+ab+bc+ac}=\dfrac{a^3}{\left(a+b\right)\left(a+c\right)}\)

nên vt =\(\dfrac{a^3}{\left(a+b\right)\left(a+c\right)}+\dfrac{b^3}{\left(b+a\right)\left(b+c\right)}+\dfrac{c^3}{\left(a+c\right)\left(c+b\right)}\)

nx \(\dfrac{a^3}{\left(a+b\right)\left(a+c\right)}+\dfrac{a+b}{8}+\dfrac{a+c}{8}\) >= \(\dfrac{3a}{4}\)

ttu vt>= \(\dfrac{3\left(a+b+c\right)}{4}-\left(\dfrac{a+b}{8}+\dfrac{a+c}{8}+\dfrac{a+b}{8}+\dfrac{b+c}{8}+\dfrac{a+c}{8}+\dfrac{b+c}{8}\right)\) =\(\dfrac{a+b+c}{4}\)

dau = say ra a=b=c=3