cho a,b,c > 0 CM a^3/bc + b^3/ac + c^3/ab = a+b+C 05/11/2021 Bởi Aubrey cho a,b,c > 0 CM a^3/bc + b^3/ac + c^3/ab = a+b+C
$\frac{a^{3}}{bc}$ + b + c $\geq$ 3$\sqrt[3]{\frac{a^{3}}{bc}bc}$ = 3a $\frac{b^{3}}{ac}$ + a + c $\geq$ 3$\sqrt[3]{\frac{b^{3}}{ac}ac}$ = 3b $\frac{c^{3}}{ab}$ + a + b $\geq$ 3$\sqrt[3]{\frac{c^{3}}{ab}ab}$ = 3c Cộng lại được $\frac{a^{3}}{bc}$ + $\frac{b^{3}}{ac}$ + $\frac{c^{3}}{ab}$ + 2(a + b + c) $\geq$ 3(a + b + c) ⇒ $\frac{a^{3}}{bc}$ + $\frac{b^{3}}{ac}$ + $\frac{c^{3}}{ab}$ $\geq$ a + b + c Bình luận
$\frac{a^{3}}{bc}$ + b + c $\geq$ 3$\sqrt[3]{\frac{a^{3}}{bc}bc}$ = 3a
$\frac{b^{3}}{ac}$ + a + c $\geq$ 3$\sqrt[3]{\frac{b^{3}}{ac}ac}$ = 3b
$\frac{c^{3}}{ab}$ + a + b $\geq$ 3$\sqrt[3]{\frac{c^{3}}{ab}ab}$ = 3c
Cộng lại được
$\frac{a^{3}}{bc}$ + $\frac{b^{3}}{ac}$ + $\frac{c^{3}}{ab}$ + 2(a + b + c) $\geq$ 3(a + b + c)
⇒ $\frac{a^{3}}{bc}$ + $\frac{b^{3}}{ac}$ + $\frac{c^{3}}{ab}$ $\geq$ a + b + c