Question

[Xuan Xuannajimex]

Cho a,b > 1 và a + b \(\ge\) 4 . Tìm GTNN của P = \(\dfrac{a^4}{\left(b-1\right)^3}+\dfrac{b^4}{\left(a-1\right)^3}\)

Answer 1

[Nguyễn Việt Lâm]

\(\dfrac{a^4}{\left(b-1\right)^3}+\dfrac{256}{81}\left(b-1\right)+\dfrac{256}{81}\left(b-1\right)+\dfrac{256}{81}\left(b-1\right)\ge4\sqrt[4]{\dfrac{a^4.256^3.\left(b-1\right)^3}{81^3\left(b-1\right)^3}}=\dfrac{256a}{27}\)

\(\dfrac{b^4}{\left(a-1\right)^3}+\dfrac{256}{81}\left(a-1\right)+\dfrac{256}{81}\left(a-1\right)+\dfrac{256}{81}\left(a-1\right)\ge\dfrac{256b}{27}\)

Cộng vế với vế: 

\(P+\dfrac{256}{27}\left(a+b\right)-\dfrac{512}{27}\ge\dfrac{256}{27}\left(a+b\right)\)

\(\Rightarrow P\ge\dfrac{512}{27}\)

Dấu "=" xảy ra khi \(a=b=4\)