Question

Cho a;b;c>0.chứng minh rằng \(\frac{a^3+b^3+c^3}{abc}+\frac{54abc}{\left(a+b+c\right)^3}\ge5\)

Answer 1

Theo Holder , ta có : \(\left(a^3+b^3+c^3\right)\left(1+1+1\right)\left(1+1+1\right)\ge\left(a+b+c\right)^3\)

\(\Rightarrow\left(a+b+c\right)^3\le9\left(a^3+b^3+c^3\right)\)

Ta có :  \(\frac{a^3+b^3+c^3}{abc}+\frac{54abc}{\left(a+b+c\right)^3}\ge\frac{\left(a+b+c\right)^3}{9abc}+\frac{54abc}{\left(a+b+c\right)^3}\)

Đặt \(t=\frac{\left(a+b+c\right)^3}{27abc}\) thì \(t\ge1\) , khi đó :  \(\frac{\left(a+b+c\right)^3}{9abc}+\frac{54abc}{\left(a+b+c\right)^3}=3t+\frac{2}{t}=t+\left(2t+\frac{2}{t}\right)\ge1+2\sqrt{2t.\frac{2}{t}}=5\)

Dấu "=" xảy ra khi a = b = c = 1

Answer 2

a=b=c=1

Answer 3

tôi ko bt kiến thức này chưa hok

Answer 4

kết quả =1 nha bạn Hoàng Lê Bảo Ngọc viết lời giải rồi nha ai k minh minh thank you

Answer 5

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