Với \(a>b>c:\hept{\begin{cases}\frac{2a^2}{a-b}\ge\frac{2a^2-2b^2}{a-b}=\frac{2\left(a-b\right)\left(a+b\right)}{a-b}=2a-2b\\\frac{b^2}{b-c}\ge\frac{b^2-c^2}{b-c}=\frac{\left(b-c\right)\left(b+c\right)}{b-c}=b+c\end{cases}}\)

\(\Rightarrow\frac{2a^2}{a-b}+\frac{b^2}{b-c}\ge2a+3b+c\)

Dấu đẳng thức xảy ra \(\Leftrightarrow b=c=0\)(Vô lí với \(b>c\))

Vậy \(\frac{2a^2}{a-b}+\frac{b^2}{b-c}>2a+3b+c\)

\(\Leftrightarrow\frac{\left(2a^2+3b^2\right)\left(a+b\right)}{2a^3+3b^3}+\frac{\left(2b^2+3a^2\right)\left(a+b\right)}{2b^3+3a^3}\le4\)

\(\Leftrightarrow\frac{2a^3+3b^3+2a^2b+3ab^2}{2a^3+3b^3}+\frac{2b^3+3a^3+2ab^2+3ab^2}{2b^3+3a^3}\le4\)

\(\Leftrightarrow\frac{2a^2b+3ab^2}{2a^3+3b^3}+\frac{2ab^2+3ab^2}{2b^3+3a^3}\le2\)

\(\Leftrightarrow\frac{2\left(\frac{a}{b}\right)^2+3\left(\frac{a}{b}\right)}{2\left(\frac{a}{b}\right)^3+3}+\frac{2\left(\frac{a}{b}\right)+3\left(\frac{a}{b}\right)^2}{3\left(\frac{a}{b}\right)^3+2}\le2\)

Đặt \(\frac{a}{b}=x>0\Rightarrow\frac{2x^2+3x}{2x^3+3}+\frac{3x^2+2x}{3x^3+2}\le2\)

\(\Leftrightarrow\left(x-1\right)^2\left(12x^4+12x^3-x^2+12x+12\right)\ge0\) (luôn đúng)

Dấu "=" xảy ra khi \(x=1\) hay \(a=b\)

Hơi trâu bò :D

a: 3a+2b>=3b+2a

=>3a-2a>=3b-2b

=>a>=b(đúng)

b: =>a^2-2ab+b^2<=2a^2+2b^2

=>2a^2+2b^2-a^2+2ab-b^2>=0

=>(a+b)^2>=0(luôn đúng)

c: =>5a^2+5b^2>=4a^2-4ab+b^2

=>a^2+4ab+4b^2>=0

=>(a+2b)^2>=0(luôn đúng)

a) đặt \(\frac{a}{b}=\frac{c}{d}=k\)

=>a=bk

c=dk

ta có \(\frac{2a}{+3b2a-3b}=\frac{2bk+3b}{2bk-3b}=\frac{b\left(2k+3\right)}{b\left(2k-3\right)}=\frac{2k+3}{2k-3}\left(1\right)\)

\(\frac{2c+3d}{2c-3d}=\frac{2dk+3d}{2dk-3d}=\frac{d\left(2k+3\right)}{d\left(2k-3\right)}=\frac{2k+3}{2k-3}\left(2\right)\)

từ (1) và(2) ta có\(\frac{2a+3b}{2a-3b}=\frac{2c+3d}{2c-3d}\)

b)

đặt\(\frac{a}{b}=\frac{c}{d}=k\)

ta có\(\frac{ab}{ad}=\frac{bk.b}{dk.d}=\frac{kb^2}{kd^2}=\frac{b^2}{d^2}\left(1\right)\)

\(\frac{\left(a+b\right)^2}{\left(c+d\right)^2}=\frac{b^2k^2+b^2}{d^2k^2+d^2}=\frac{b^2\left(k^2+1\right)}{d^2\left(k^2+1\right)}=\frac{b^2}{d^2}\left(2\right)\)

từ (1) và(2) \(\Rightarrow\frac{ab}{cd}=\frac{\left(a+b\right)^2}{\left(c+d^2\right)}\)

ban kia lam dung roi do 

k tui nha

thanks