Cho a,b,c ≥0 sử dụng bắt đẳng thức cauchy CMR (1+a)(1+b)(1+c) ≥ (1+ ∛abc) ³ 11/08/2021 Bởi Reese Cho a,b,c ≥0 sử dụng bắt đẳng thức cauchy CMR (1+a)(1+b)(1+c) ≥ (1+ ∛abc) ³
Giải thích các bước giải: $\dfrac{1}{a+1}+\dfrac{1}{b+1}+\dfrac{1}{c+1}\ge 3\sqrt[3]{\dfrac{1}{a+1}.\dfrac{1}{b+1}.\dfrac{1}{c+1}}=\dfrac{3}{\sqrt[3]{(a+1)(b+1)(c+1)}}$ $\dfrac{a}{a+1}+\dfrac{b}{b+1}+\dfrac{c}{c+1}\ge 3\sqrt[3]{\dfrac{a}{a+1}.\dfrac{b}{b+1}.\dfrac{c}{c+1}}=\dfrac{3\sqrt[3]{abc}}{\sqrt[3]{(a+1)(b+1)(c+1)}}$ $\rightarrow\dfrac{1}{a+1}+\dfrac{1}{b+1}+\dfrac{1}{c+1}+\dfrac{a}{a+1}+\dfrac{b}{b+1}+\dfrac{c}{c+1}\ge \dfrac{3(1+\sqrt[3]{abc})}{\sqrt[3]{(a+1)(b+1)(c+1)}} $ $\rightarrow 3\ge \dfrac{3(1+\sqrt[3]{abc})}{\sqrt[3]{(a+1)(b+1)(c+1)}}$ $\rightarrow (1+a)(1+b)(1+c)\ge (1+\sqrt[3]{abc})^3\rightarrow dpcm$ Bình luận
Giải thích các bước giải:
$\dfrac{1}{a+1}+\dfrac{1}{b+1}+\dfrac{1}{c+1}\ge 3\sqrt[3]{\dfrac{1}{a+1}.\dfrac{1}{b+1}.\dfrac{1}{c+1}}=\dfrac{3}{\sqrt[3]{(a+1)(b+1)(c+1)}}$
$\dfrac{a}{a+1}+\dfrac{b}{b+1}+\dfrac{c}{c+1}\ge 3\sqrt[3]{\dfrac{a}{a+1}.\dfrac{b}{b+1}.\dfrac{c}{c+1}}=\dfrac{3\sqrt[3]{abc}}{\sqrt[3]{(a+1)(b+1)(c+1)}}$
$\rightarrow\dfrac{1}{a+1}+\dfrac{1}{b+1}+\dfrac{1}{c+1}+\dfrac{a}{a+1}+\dfrac{b}{b+1}+\dfrac{c}{c+1}\ge \dfrac{3(1+\sqrt[3]{abc})}{\sqrt[3]{(a+1)(b+1)(c+1)}} $
$\rightarrow 3\ge \dfrac{3(1+\sqrt[3]{abc})}{\sqrt[3]{(a+1)(b+1)(c+1)}}$
$\rightarrow (1+a)(1+b)(1+c)\ge (1+\sqrt[3]{abc})^3\rightarrow dpcm$