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.

Here's a trick that doesn't involve calculus:
You have seen that
$$(a - b)^2 + 4ab + b^2 = 7$$
Now, adding this equation to S times the original equation, you get:
$$(a - b)^2 + Sa^2 + (4 + 2S)ab + (1 + 2S)b^2 = 7 + 7S$$
You would like to turn the last three terms into a perfect square, ie:
$$Sa^2  + (4 + 2S)ab + (1 + 2S)b^2 = (\sqrt{S}a + \sqrt{1 + 2S}b)^2$$
So, matching the middle terms, we get:
$$4 + 2S = 2\sqrt{S}\sqrt{1 + 2S}$$
$$(2 + S)^2 = S(1 + 2S)$$
$$4 + 3S- S^2 = 0$$
Now, $S = 4$ and $S = -1$ are solutions, but we want the positive one. Plugging in $S = 4$, we go back to the second line:
$$(a - b)^2 + 4a^2 + 12ab + 9b^2 = 7 + 7 \cdot 4$$
$$(a - b)^2 + (2a + 3b)^2 = 35$$
Since squares are positive, the largest possible value of $(a - b)^2$ is clearly $35$, and it is achieved exactly when $2a + 3b = 0$. 
You might think this is a convoluted and needlessly tricky way of solving this problem. That's probably true, but I think the technique of creating terms which are perfect squares is handy in proving inequalities, because of the fact that perfect squares are nonnegative.
