Algebra - minimum value...

1708

$\hspace{-16}$Let $\bf{z=xy}$ and Given $\bf{x^2+y^2+xy=1}$\\\\\\ $\bf{(x+y)^2=1+xy\Leftrightarrow (x+y)^2=1+z}$\\\\\\ Now $\bf{(x+y)^2\geq 0\Leftrightarrow z\geq -1}$\\\\\\ Similarly $\bf{x^2+y^2+xy=1}$ (Given)\\\\\\ $\bf{(x-y)^2=1-3xy\Leftrightarrow (x-y)^2=1-3xy}$\\\\\\ Now $\bf{(x-y)^2\geq 0\Leftrightarrow z\leq \frac{1}{3}}$\\\\\\ So $\bf{-1 \leq z\leq \frac{1}{3}\Leftrightarrow z^2\leq 1}$\\\\\\ Now Given $\bf{x^2y+xy^3+4=(xy).(x^2+y^2)+4}$\\\\\\ $\bf{=(xy).(1-xy)+4=z.(1-z)+4}$\\\\\\ $\bf{=z-z^2+4\geq -1-1+4=2}$\\\\\\ Which is occur at $\bf{z=-1\Leftrightarrow xy=-1}$\\\\\\ So $\boxed{\boxed{\bf{\left(x^3y+xy^3+4\right)_{Min.} = 2}}}$ at $\bf{xy=-1}$\\\\\\ Means either $\bf{(x,y)=(1,-1)}$ or $\bf{(x,y)=(-1,1)}$$

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Hardik Sheth ·Feb 22 '13 at 0:27

is it 38/9??

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Hardik Sheth oh..this is perhaps the maximum value...min value can neva b greater than 4.
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Akshay Ginodia ·Feb 22 '13 at 3:02

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