Question

[Nguyễn Phan Ngọc Tú]

chứng minh bất đẳng thức \(\frac{1}{x^3+y^3}+\frac{1}{xy}\ge4+2\sqrt{3}\)(x+y=1)

Answer 1

[Thắng Nguyễn]

\(P=\frac{1}{x^3+y^3}+\frac{1}{xy}\)

Ta có:

\(x+y=1\Rightarrow\left(x+y\right)^3=1\)

\(\Rightarrow x^3+y^3+3xy\left(x+y\right)=1\)

\(\Rightarrow x^3+y^3+3xy=1\)

\(\Rightarrow P=\frac{x^3+y^3+3xy}{x^3+y^3}+\frac{x^3+y^3+3xy}{xy}\)\(=4+\frac{3xy}{x^3+y^3}+\frac{x^3+y^3}{xy}\left(1\right)\)

Áp dụng Bđt Cô si ta có:

\(\frac{3xy}{x^3+y^3}+\frac{x^3+y^3}{xy}\ge2\sqrt{\frac{3xy}{x^3+y^3}\cdot\frac{x^3+y^3}{xy}}=2\sqrt{3}\)

\(\Rightarrow P\ge4+2\sqrt{3}\)(Đpcm)

Dấu = khi \(\hept{\begin{cases}x+y=1\\x^3+y^3=\sqrt{3xy}\end{cases}}\Leftrightarrow\hept{\begin{cases}x+y=1\\1-3xy=\sqrt{3xy}\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}x+y=1\\3\sqrt{xy}=\frac{-1+\sqrt{5}}{2}\end{cases}}\)\(\Leftrightarrow\hept{\begin{cases}x+y=1\\xy=\frac{6-2\sqrt{5}}{12}\end{cases}}\)

\(\Leftrightarrow x^2-x+\frac{6-2\sqrt{5}}{12}=0\)\(\Leftrightarrow x,y=\frac{1\pm\sqrt{\frac{2\sqrt{5}-3}{3}}}{2}\)

Answer 2

[hoang phuc]

chiu

tk nhe

xin do

bye