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I am working on the following problem:

If $a, b \in \mathbb{R}^+$, then $$a^2 + b^2 + 1 > a \sqrt{b^2 + 1} + b \sqrt{a^2 + 1}$$

My book provides the following solution. The following inequality is clearly true: $$(a - \sqrt{b^2 + 1})^2 + (b - \sqrt{a^2 + 1})^2 \ge 0.$$ and the desired inequality follows directly.

However, I don't understand the steps that were taken to reach the sum of squares above. Was it just guess-and-check?

Or does someone understand the steps on how to come up with the sum of squares? Or can you think of a different solution?

user56202
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  • There is no magical "steps" to solving any mathematical problem. You try it in several ways and keep trying until you reach a solution. The author of the textbook of course isn't going to include every "scratch paper" detail of it into the text.

    Have you actually tried to solve this problem, other than just staring at it for a few seconds and immediately look up the solution?

     – dezdichado Jan 11 '23 at 03:17
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    @dezdichado Yes of course. I tried various algebraic manipulations for about 45 minutes but nothing proved fruitful. – user56202 Jan 11 '23 at 03:19
  • IMO With a passing familiarity of AM-GM, the $ a^2 + b^2 + 1 \geq 2 a \sqrt{b^2+1}$ is almost immediate. That gives you the first square, and likewise with the second. – Calvin Lin Jan 11 '23 at 03:34
  • @CalvinLin Thanks. I like your approach the most. – user56202 Jan 11 '23 at 03:36
  • The answer question https://math.stackexchange.com/questions/2093930/prove-the-following-inequality-x2y21x-sqrty21y-sqrtx21?noredirect=1&lq=1 – Moustapha_M_I Jan 11 '23 at 03:37

3 Answers3

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Multiply the given inequality by $2$ and rearrange it in order to obtain the following relation \begin{align*} a^{2} + b^{2} + 1 > a\sqrt{b^{2} + 1} + b\sqrt{a^{2} + 1} & \Longleftrightarrow 2a^{2} + 2b^{2} + 2 > 2a\sqrt{b^{2} + 1} + 2b\sqrt{a^{2} + 1}\\\\ & \Longleftrightarrow (a^{2} - 2a\sqrt{b^{2} + 1} + b^{2} + 1) + (b^{2} - 2b\sqrt{a^{2} + 1} + a^{2} + 1) > 0\\\\ & \Longleftrightarrow (a - \sqrt{b^{2} + 1})^{2} + (b - \sqrt{a^{2} + 1})^{2} > 0 \end{align*} which is clearly true.

Hopefully this helps!

Átila Correia
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  • Not quite true right? The end is just $ \geq 0$, till you show that equality can't hold. – Calvin Lin Jan 11 '23 at 03:27
  • @CalvinLin Indeed, you are right. But I believe the OP is able to take it from there. Should I include the proof of such result in the body of my answer though? – Átila Correia Jan 11 '23 at 03:29
  • @ÁtilaCorreia Thank you. Is there any strong intuition for the first and second line? Or is it just experience and/or do semi-random manipulations until something works? Indeed, showing strict inequality is easy. – user56202 Jan 11 '23 at 03:31
  • I think at least a mention of that would be sufficient. It's not immediately clear to me if OP understands it, or they just repeated the solution from the book. – Calvin Lin Jan 11 '23 at 03:31
  • @CalvinLin Thanks for the comment. Based on such discussion, I believe the OP is already aware of such need. – Átila Correia Jan 11 '23 at 03:42
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Let's try prove this inequality via different method.

Consider vectors $x:=(a,b)$ and $y:=(\sqrt{a^2+1},\sqrt{b^2+1})$. Then Cauchy-Schwarz inequality tells us that $\langle x,y\rangle\leq \lVert x \rVert \lVert y \rVert$, where $\langle\cdot,\cdot \rangle$ is a dot product and $\lVert \cdot \rVert$ is a length of vector.

Hence $$a\sqrt{b^2+1}+b\sqrt{a^2+1}\leq\sqrt{a^2+b^2}\cdot \sqrt{a^2+b^2+2}. \quad \quad \quad (*)$$ But this is not enough yet! If we can show that $$\sqrt{a^2+b^2}\cdot \sqrt{a^2+b^2+2}<a^2+b^2+1, \quad \quad \quad (**)$$ then we are done.

Assume $t:=a^2+b^2,$ then the inequality $(**)$ can be written as $\sqrt{t(t+2)}<t+1$ which is obviously true. Combining inequalities $(*)$ and $(**)$ we obtain the desired one!

Hopefully this helps!

RFZ
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Here is another way, using AM-GM: $$a^2+(b^2+1) \geqslant 2a\sqrt{b^2+1}\tag1$$ $$(a^2+1)+b^2 \geqslant 2b\sqrt{a^2+1}\tag2$$

Now add these and divide by two to get what you want. As equality is not possible simultaneously for $(1)$ and $(2)$, the inequality is strict.

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P.S. $a, b$ can be any real numbers, the positivity condition is obviously unnecessary.

Macavity
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