If $a,b \in \mathbb R$ satisfy $a^2+2ab+2b^2=7,$ then find the largest possible value of $|a-b|$ I came across the following problem that says:  

If $a,b \in \mathbb R$ satisfy $a^2+2ab+2b^2=7,$ then the largest possible value of $|a-b|$ is which of the following?
  $(1)\sqrt 7$,  $(2)\sqrt{7/2}$ , $(3)\sqrt {35}$  $(4)7$  

My Attempt:  We notice, $(a-b)^2=7-(4ab+b^2)$ and hence $|a-b|=\sqrt {7-(4ab+b^2)}$ and so $|a-b|$ will be maximum whenever $(4ab+b^2)$ will be minimum. But now I am not sure how to progress further hereon.Can someone point me in the right direction? Thanks in advance for your time.

You can solve the given equation as a quadratic to get $a=-b \pm \sqrt {b^2-2b^2+7}=-b \pm \sqrt {7-b^2}$.  To minimize $4ab+b^2$ you want $a$ as negative as possible, so take the minus sign.  Now you have $|a-b|=\sqrt{7-4b(-b-\sqrt{7-b^2})-b^2}=\sqrt{7+3b^2+4b\sqrt{7-b^2}}$.  Now you can take the derivative, set to zero, etc.
Added:  Wolfram Alpha shows the maximum at $b=2\sqrt{\frac 75}, a=-3\sqrt {\frac 75}$ with $|a-b|=\sqrt {35}$
