Prove every real positive a, b satisfy 1/sqrt[a^2+b] + 1/sqrt[a+b^2] < 1/a + 1/b

[Jimmy44]

1. Proof that every couple of real and positive numbers (a, b) satisfy this inequality:

. . .\(\displaystyle \dfrac{1}{\sqrt{\strut a^2\, +\, b\,}}\, +\, \dfrac{1}{\sqrt{\strut a\, +\, b^2\,}}\, <\, \dfrac{1}{a}\, +\, \dfrac{1}{b}\)

I have spent a lot of time to solve this, but I couldn't proof this. Can you help me please?

 

[ksdhart2]

I have spent a lot of time to solve this, but I couldn't proof this.

Great! That means you ought to have loads of work to share with us, as per the rules laid out in the Read Before Postinghttps://www.freemathhelp.com/forum/threads/46555-Read-Before-Posting! thread that's stickied at the top of each sub-forum (you did read it, right? ). Please share with us all of the work you've done on this problem, even the parts you know for sure are wrong. Thank you.

 

[Jimmy44]

Ok, so let's get started :cool: Here are my conversions:

\(\displaystyle \frac{1}{\sqrt{a^ {2}+b}} + \frac{1}{\sqrt{a+b^ {2}} } < \frac{1}{a} + \frac{1}{b}\)

multiply both sides of inequality by \(\displaystyle \sqrt{(a+ b^{2} )(a ^{2}+b)}\)

\(\displaystyle \sqrt{a+b^{2}} + \sqrt{a^2+b}< \frac{\sqrt{(a+ b^{2} )(a ^{2}+b)}}{a} + \frac{\sqrt{(a+ b^{2} )(a ^{2}+b)}}{b}\)

multiply both sides by \(\displaystyle ab\)

\(\displaystyle ab(\sqrt{a+b^{2}} + \sqrt{a^2+b})<b\sqrt{(a+ b^{2} )(a ^{2}+b)}+ a\sqrt{(a+ b^{2} )(a ^{2}+b)}\)

\(\displaystyle ab(\sqrt{a+b^{2}} + \sqrt{a^2+b})< (a+b)(\sqrt{(a+ b^{2} )(a ^{2}+b)})\)

intensify both sides ^2

\(\displaystyle a^{2}b^{2}(a+b^{2}+ 2\sqrt{(a+ b^{2} )(a ^{2}+b)} +a^{2}+ b)\)

. . . . .\(\displaystyle <(a^{2}+ 2ab+b^{2})((a+ b^{2} )(a ^{2}+b))\)

\(\displaystyle a^{3}b^{2}+a^{2}b^{4}+2a^{2}b^{2}\sqrt{(a+ b^2)(a^{2}+b)}+a^{4}b^{2}+a^{2}b^{3}\)

. . . . .\(\displaystyle <(a^{2}+ 2ab+b^{2})(a^{3}+ab+a^{2}b^{2}+b^{3})\)

\(\displaystyle a^{3}b^{2}+a^{2}b^{4}+2a^{2}b^{2}\sqrt{(a+ b^2)(a^{2}+b)}+a^{4}b^{2}+a^{2}b^{3}\)

. . . . .\(\displaystyle <a^{5}+a^{3}b+a^{4}b^{2}+a^{2}b^{3}+2a^{4}b+2a^{2}b^{2}+2a^{3}b^{3}\)

. . . . . . . .\(\displaystyle +2ab^{4}+a^{3}b^{2}+ab^{3}+a^{2}b^{4}+b^{5}\)

\(\displaystyle 2a^{2}b^{2}\sqrt{(a+ b^2)(a^{2}+b)}\)

. . . . .\(\displaystyle <a^{5}+a^{3}b+2a^{4}b+2a^{2}b^{2}+2a^{3}b^{3}+2ab^{4}+ab^{3}+b^{5}\)

and what to do next? This is moment when I got stuck.

 

[MarkFL]

Given \(\displaystyle a,b>0\in\mathbb{R}\), we may state:

\(\displaystyle a^2+b>a^2\)

Hence:

\(\displaystyle \sqrt{a^2+b}>a\)

And thus:

\(\displaystyle \displaystyle \frac{1}{\sqrt{a^2+b}}<\frac{1}{a}\tag{1}\)

Likewise:

\(\displaystyle \displaystyle \frac{1}{\sqrt{a+b^2}}<\frac{1}{b}\tag{2}\)

Adding (1) and (2), we obtain:

\(\displaystyle \displaystyle \frac{1}{\sqrt{a^2+b}}+\frac{1}{\sqrt{a+b^2}}< \frac{1}{a}+\frac{1}{b}\)