Question

Cho a,b,c>0. Chứng minh rằng: \(\dfrac{a}{b}+\dfrac{b}{c}+\dfrac{c}{a}+\dfrac{\sqrt[3]{abc}}{a+b+c}\ge\dfrac{10}{3}\)

Answer 1

Từ bài toán này (mà bạn đã hỏi cách đây vài bữa):

cho a,b,c>0. Chứng minh rằng: \(\dfrac{a}{b}+\dfrac{b}{c}+\dfrac{c}{a}\ge\dfrac{a+b+c}{\sqrt[3]{abc}}\) - Hoc24

Ta có: \(\dfrac{a}{b}+\dfrac{b}{c}+\dfrac{c}{a}\ge\dfrac{a+b+c}{\sqrt[3]{abc}}\)

Do đó: \(VT\ge\dfrac{a+b+c}{\sqrt[3]{abc}}+\dfrac{\sqrt[3]{abc}}{a+b+c}\)

Lại có: \(\dfrac{a+b+c}{\sqrt[3]{abc}}\ge\dfrac{3\sqrt[3]{abc}}{\sqrt[3]{abc}}=3\)

Đặt \(\dfrac{a+b+c}{\sqrt[3]{abc}}=x\ge3\Rightarrow VT\ge x+\dfrac{1}{x}=\dfrac{x}{9}+\dfrac{1}{x}+\dfrac{8x}{9}\ge2\sqrt{\dfrac{x}{9x}}+\dfrac{8}{9}.3=\dfrac{10}{3}\) (đpcm)