$3+\frac{a}{2b}+\frac{2b}{3c}+\frac{3c}{a}\geq a+2b+3c+\frac{1}{a}+\frac{1}{2b}+\frac{1}{3c}$

C

congchuaanhsang

Nhận thấy a.2b.3c=1 nên ta đặt $a=\dfrac{y}{x}$; $2b=\dfrac{z}{y}$; $3c=\dfrac{x}{z}$

Khi đó VT\geqVP\Leftrightarrow$\dfrac{3xyz+x^3+y^3+z^3}{xyz}$\geq$\dfrac{x^2y+y^2x+y^2z+z^2y+x^2z+z^2x}{xyz}$

\Leftrightarrow$3xyz+x^3+y^3+z^3-x^2y-y^2x-y^2z-z^2y-z^2x-x^2z$\geq0

\Leftrightarrow$x(x-y)(x-z)+y(y-z)(y-x)+z(z-x)(z-y)$\geq0(đúng theo bđt Schur)

Vậy bđt được cm
 
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