I am getting too much confused with this stuff. If \((a,b)\) are positive reals and \(a-b=2\)

Then find the smallest value of the constant \(L\) for which

\[\sqrt{x^{2}+ax}-\sqrt{x^{2}+bx}<L\]

for all \(x>0\)

Note* a solution that doesn't require calc is preferred, but feel free to post a calc solution if u like.

Also, this should be doable by hand.

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TopNewestHint:What happens when you multiply by the conjugate?It then because a 1-2 line problem. – Calvin Lin Staff · 2 years, 5 months ago

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Right! In case it helps anybody, the steps would be as follows:

\((\sqrt{x^{2} + ax} - \sqrt{x^{2} + bx})*\dfrac{(\sqrt{x^{2} + ax} + \sqrt{x^{2} + bx})}{\sqrt{x^{2} + ax} + \sqrt{x^{2} + bx}} =\)

\(\dfrac{x^{2} + ax - x^{2} - bx}{\sqrt{x^{2} + ax} + \sqrt{x^{2} + bx}} = \dfrac{(a - b)x}{x*\sqrt{1 + \frac{a}{x}} + x*\sqrt{1 + \frac{b}{x}}} = \dfrac{2}{\sqrt{1 + \frac{a}{x}} + \sqrt{1 + \frac{b}{x}}}\),

where we were able to take the \(x\) outside the square roots without absolute value signs since it is specified that \(x \gt 0\). The fraction is then maximized when the denominator is minimized, which approaches a minimum of \(\sqrt{1} + \sqrt{1} = 2\) as \(x \rightarrow \infty\). Thus the expression has a strict upper limit of \(\frac{2}{2} = 1\).

(P.S.. Calvin Lin I can delete this if it spoils the fun for others, but I just thought it might help to flesh out the details of your hint. Have there been any thoughts about having the option of "hiding" portions of text, so that they can be "revealed" by the viewer once they've given the problem (or hint) a try themselves? I've seen it on other sites but I don't know how onerous a task it is to create such an option.) – Brian Charlesworth · 2 years, 5 months ago

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In the meantime, let me add some spoiler space to the start of your solution. – Calvin Lin Staff · 2 years, 5 months ago

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@Calvin Lin for the hint – Rajat Bisht · 2 years, 5 months ago

I think i must improve my observational skills. Well thanksLog in to reply

I think you mean greatest value of the expression? as you have shown \( expression~<~L\)

For greatest -

\( D = \sqrt{x}(\sqrt{( \sqrt{x} - 0)^2 + ( 0 - \sqrt{a})^2} - \sqrt{( \sqrt{x} - 0)^2 + ( 0 - \sqrt{b})^2})\)

\( A(0 , \sqrt{a})~ ~B(0,\sqrt{b})~~C(x,0)\)

\( D = CA - CB\)

\( |CA - CB| \leq |AB|\)

\( |AB| = \sqrt{a} - \sqrt{b}\)

\( Thus~\sqrt{x}(\sqrt{a} - \sqrt{b})\) – Megh Choksi · 2 years, 5 months ago

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– Rajat Bisht · 2 years, 5 months ago

Dont think so We need the least numerical value of LLog in to reply

– Megh Choksi · 2 years, 5 months ago

What have you tried till now?Log in to reply

– Trevor Arashiro · 2 years, 5 months ago

There was a messy step, but it becomes pretty after a bash, or wolfram lol. And at the end I used stupidity cuz I was too lazy to maximize the function \(\sqrt{x+x\sqrt{1+x^2}}-\sqrt{x\sqrt{1+x^2}-x}\) by "proper" methods.Log in to reply

– Brian Charlesworth · 2 years, 5 months ago

Unrelated note: I just accidently un-followed you and then re-followed you. Just thought you might have been wondering why you received a notification to this effect. :)Log in to reply

– Trevor Arashiro · 2 years, 5 months ago

Haha, no worries, didn't even see it. Email must be buried in my 3000 unread mail XDLog in to reply

@brian charlesworth sir can u please refer me some text related to higher algebra. I always seek for it...but never found the right thing. Well also thanks for that solution. – Rajat Bisht · 2 years, 5 months ago

Lol that also happens with me. WellLog in to reply

@megh choksi or @Trevor Arashiro might have some ideas. – Brian Charlesworth · 2 years, 5 months ago

I'm sorry, but I'm not up to date on what the best texts are. Mine are all out of print and weren't that good to start with. If I have to check on anything these days I rely mostly on the internet. PerhapsLog in to reply

@Trevor Arashiro tried it with calculus using wolfram and got the ans. But i wish an easy solution for this – Rajat Bisht · 2 years, 5 months ago

Ohh js tried with the derivative. And the equation became such messy.Log in to reply

My approach is somehow different but obviously the result is the same for L > 1.

If we look separately the two expressions under the radicals y can assume there are the equations of two hyperbola y^2 - x^2 - ax = 0 and y^2 -x^2 -bx =0.

Just at sight we can see the asymptotes have the same slope m=1 and identify that distance between centers of two conics is (a-b )/2, what means centers are 1 unit apart, since a-b= 2 by hypothesis .

As the slope is 1 the difference of ordinates at two asymptotes is 1 unit too at any value of x. So at infinity where the two hyperbolas meet there asymptotes and the distance between them are widening from x=0. – Mariano PerezdelaCruz · 2 years, 5 months ago

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My first instinct is to get rid of the square root. So far I've got \[x^2 +ax < L^2 + (x^2 +bx) + 2L\sqrt{x^2 +bx}\] \(\Rightarrow\) using a-b = 2 \[2x < L^2 +2L\sqrt{x^2 +bx} \] However, once I continue the method there doesn't seem to be a clear path to the solution – Curtis Clement · 2 years, 5 months ago

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