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Suppose that $x,y\in [0,1]$. Prove that $\frac{1}{\sqrt{1+x^2}}+\frac{1}{\sqrt{1+y^2}}\leq \frac{2}{\sqrt{1+xy}}.$

I suppose that this problem can be solved by some application of AM-GM inequality. I was trying to do the following: since $xy\leq \frac{x^2+y^2}{2}$ then $\frac{2}{\sqrt{1+xy}}\geq \frac{2}{\sqrt{1+x^2/2+y^2/2}}$. But the inequality $\frac{2}{\sqrt{1+x^2/2+y^2/2}}\geq \frac{1}{\sqrt{1+x^2}}+\frac{1}{\sqrt{1+y^2}}$ is obviously false. So I guess we have to use something which is non-trivial.

Would be grateful if someone can show the solution.

I have spent probably 2-3 hours and did not get it.

RFZ
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    See solution 2 - https://www.cut-the-knot.org/arithmetic/algebra/PowerOfSubstitution3.shtml – Math Lover Sep 13 '20 at 12:03
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    The solution @MathLover mentioned contains an unfortunate typo: $a+a^2$ should read $1+a^2$. It's otherwise correct. – J.G. Sep 13 '20 at 12:10

2 Answers2

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We have $$\left(\frac{1}{\sqrt{1+x^2}}+\frac{1}{\sqrt{1+y^2}}\right)^2 = \frac{1}{1+x^2}+\frac{1}{1+y^2}+\frac{2}{\sqrt{(1+x^2)(1+y^2)}}$$ Using the AM-GM we have $$\frac{2}{\sqrt{(1+x^2)(1+y^2)}} \leqslant \frac{1}{1+x^2}+\frac{1}{1+y^2}.$$ Therefore, we need to prove $$\frac{1}{1+x^2}+\frac{1}{1+y^2}\leqslant \frac{2}{1+xy},$$ equivalent to $$\frac{(xy-1)(x-y)^2}{(1+x^2)(1+y^2+1)(1+xy)} \leqslant 0.$$ Which is true.

nguyenhuyenag
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Apply Jensen's inequality to $$f(t)=\frac{1}{\sqrt{1+e^{-t}}}.$$ One has $$f'(t)=\frac{e^{-t}}{2(1+e^{-t})^{3/2}}$$ and $$f''(t)=\frac{3e^{-2t}}{4(1+e^{-t})^{5/2}}-\frac{e^{-t}}{2(1+e^{-t})^{3/2}} =\frac{e^{-2t}-2e^{-t}}{4(1+e^{-t})^{5/2}}<0$$ for $t>0$.

So $f$ is concave, and $f(2t)+f(2u)\le f(t+u)$ for $t$, $u\ge0$. Take $t=-\ln x$ and $u=-\ln y$.

Angina Seng
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    This proof is correct but I'd like to see something with AM-GM inequality. – RFZ Sep 13 '20 at 11:53
  • BTW,if this looks like it's coming out of nowhere, the restriction to [0,1] suggests that we might be using a transformation that relies on that restriction, and indeed -ln takes (0,1] to the nonnegative reals. Technically, the case that x and/or y is zero needs to be taken separately, but that's rather trivial. – Acccumulation Sep 13 '20 at 19:59