Question

If $x,y>0$, then the minimum value of $2x^2+\frac{2}{x}-2x+2y^2+\frac{2}{y}-2y+2$ is equal to

• A. $4$
• B. $5$
• C. $6$
• D. $7$

Answer 1

The correct option is C $6$

Take $2$ common,

$2(x^2+\frac{1}{x}-x+y^2+\frac{1}{y}-y+1)$
$=2(x^2+\frac{1}{x}-2x+x+y^2+\frac{1}{y}-2y+y+1)$

$=2(x^2-2x+1+y^2-2y+1+x+\frac{1}{x}-2+y+\frac{1}{y}-2+3)$

$=2\left(\right(x-1{)}^2+(y-1{)}^2+{(\sqrt{x}-\frac{1}{\sqrt{x}})}^2+{(\sqrt{y}-\frac{1}{\sqrt{y}})}^2+3)$

Now when $x=1,y=1$ then all perfect squares will be zero and that can be the minimum values of the perfect squares,

which means it will be the minimum value of the given expression.

$=2(0+0+0+0+3)=6$

$\therefore$ min value is $6$.

Hence, $\left(C\right)$