Hi Divya,

I assume you're going ahead in the following manner;

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

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

Now, I don't understand what you meant by 'root of 2 gives..', but I'll tell you how to go ahead from here. 

We know that the squareroot function in the denominator must be defined over all real $$x$$, i.e

$$1+x^4>0 \Rightarrow x^4>-1$$ but since $$x^4$$ is always positive, this will be true for all $$x$$.

We know, to maximise $$y$$, we must maximise $$y^2$$ and hence we must minimise the denominator in $$\text{(1)}$$. Or, we must minimise $$\frac{1}{x}+x$$.

We see that $$x$$ should be taken to be always positive if we want to maximise $$y$$, hence the minimum value of $$\frac{1}{x}{x}$$ can be calculated by AM-GM and it comes out to be $$2$$. (Sidenote: this happens when $$x=1$$).

And hence,

$$y^2 = \frac{1}{2^2-2} \Rightarrow y^2 = \frac{1}{2}$$

Again, since we assumed $$x$$ to be positive, we get $$y=\frac{1}{\sqrt{2}}$$

Note that we have put $$x=1$$ so I don't understand what you were objecting at.

Hope this helps

Tiny correction: I had mistakenly typed $$\frac{1}{x}x$$, that is supposed to be, of course, $$\frac{1}{x}+x$$

Yes. Thank you. I have basically missed to take the square root of y which I squared in the starting. Hence, my answer was 1/2. Another silly mistake. Thank you for the help.

You're welcome, always feel free to post your doubts!