# Prove $\frac{1}{3} > \frac{2ab^2}{(a + \sqrt{a^2 + b^2})^3}$

Let $$a$$ and $$b$$ be positive reals. Prove $$\frac{1}{3} > \frac{2ab^2}{(a + \sqrt{a^2 + b^2})^3}\;.$$

I am a high school student preparing for math contests, and this question was sent to me by one of my classmates on a contest math training course, but I wasn't able to solve it. I do not know where he got this problem from. I tried using AM-GM, since $$2ab \leq 2a^2 + b^2$$, but I couldn't progress further. Also, I also tried to manipulate the expression to use the HM-AM-GM-QM inequality, and I got to this:

$$\sqrt{\dfrac{a^2 + b^2}{2}} > \left(\dfrac{b\sqrt{3}}{a + \sqrt{a^2 + b^2}} - 1\right)\left(\dfrac{b\sqrt{3}}{a + \sqrt{a^2 + b^2}} + \dfrac{a\sqrt{2}}{2}\right)$$

Which didn't help much (and I am not even sure if this is correct). After these attempts, I got stuck. If anyone could give a hint, it would be very appreciated. Full solutions are also fine, but I would like to try to work out with the hints first.

• Dividing numerator and denominator by $b^3$ and letting $u=a/b,$ this can be written as: $$\frac13>\frac{2u}{(u+\sqrt{u^2+1})^3}.$$ One variable is easier than two. Jul 5 at 15:27
• @Ingrid, could you see my answer? I have used neither AM-GM inequality nor Cauchy-Schwartz inequality. Tell me if you like my answer. Jul 5 at 16:45
• @Ingrid, in my answer I have only used very basic rules of algebra, for example “square of a binomial” and “conjugate binomial” which is the product of the sum of two terms multiplied by the subtraction of these terms. I hope you will like my solution. Jul 5 at 16:53
• @Ingrid, thank you for your beautiful words. Do not worry, even though you can just accept one answer, you can upvote all the answers you like. So if you want, you can upvote my answer. Jul 6 at 1:26
• @Ingrid, please, look at the addendum I have written in my answer. Jul 6 at 16:49

Dividing numerator and denominator by $$b^3,$$ and letting $$u=\frac ab,$$ then this becomes:

$$\frac13>\frac{2u}{(u+\sqrt{u^2+1})^3}$$ for $$u>0.$$

This can be written as: $$(u+\sqrt{1+u^2})^3\geq 6u.$$

But $$(u+\sqrt{1+u^2})^3=u^3+3u^2\sqrt{1+u^2}+3u(1+u^2)+(1+u^2)^{3/2}$$

The third term contributes $$3u,$$ and via the second and fourth term, we get, via AM/GM, $$3u^2(1+u^2)^{1/2}+(1+u^2)^{3/2}\geq 2\sqrt 3 u(1+u^2)>3u.$$

So, instead of $$6,$$ we actually have the stronger $$(u+\sqrt{u^2+1})^3>(3+2\sqrt3)u.$$

Another approach uses $$\frac1{u+\sqrt{u^2+1}}=\sqrt{u^2+1}-u.$$ So if $$w=\sqrt{u^2+1}-u,$$ then $$0 and $$2u=\frac1w-w,$$ so we want:

$$\frac13>(w^{-1}-w)w^3=w^2-w^4$$

The maximum of the right side is when $$2w=4w^3,$$ or $$w=\frac1{\sqrt 2}.$$ (we can obviously exclude $$w=0.$$) So the maximum of $$w^2-w^4$$ is $$\frac14.$$

Since $$2u=w^{-1}-w,$$ the maximum occurs when $$u=\frac{\sqrt2}4.$$

Let $$a=\frac{xb}{2\sqrt2}.$$

Thus, by C-S and AM-GM we obtain: $$\frac{\left(a+\sqrt{a^2+b^2}\right)^3}{ab^2}=\frac{\left(\frac{x}{2\sqrt2}+\sqrt{\frac{x^2}{8}+1}\right)^3}{\frac{x}{2\sqrt2}}=\frac{\left(x+\sqrt{x^2+8}\right)^3}{8x}=$$ $$=\frac{\left(x+\frac{1}{3}\sqrt{(x^2+8)(1+8)}\right)^3}{8x}\geq\frac{\left(x+\frac{1}{3}(x+8)\right)^3}{8x}=\frac{8(x+2)^3}{27x}\geq\frac{8\left(3\sqrt[3]{x\cdot1^2}\right)^3}{27x}=8>6.$$

• You've been around long enough to know how to use \begin{align}...\end{align} Jul 5 at 16:09

Let $$x=\frac{b}{a}>0$$. We get $$E=\frac{2 a b^2}{\left(a+\sqrt{a^2+b^2}\right)^3}=\frac{2 x}{\left(x+\sqrt{1+x^2}\right)^3}$$ Now let $$x=\sinh(t) \Rightarrow t>0$$. We get $$x+\sqrt{1+x^2}=\cosh(t)+\sinh(t)=e^t$$ So we have $$E=\frac{2\sinh(t)}{e^{3t}} \Rightarrow E=\frac{e^t-e^{-t}}{e^{3t}}=e^{-2t}-e^{-4t}$$ $$\Rightarrow E=\frac{1}{4}-\left(e^{-2 t}-\frac{1}{2}\right)^2 \leqslant \frac{1}{4}<\frac{1}{3}$$

We have that

$$\frac{1}{3} > \frac{2ab^2}{(a + \sqrt{a^2 + b^2})^3} \iff (a + \sqrt{a^2 + b^2})^3-6ab^2>0$$

then, by setting $$a=\rho\cos\theta$$ and $$b=\rho\sin\theta$$, from the latter we obtain

\begin{align} \iff &(\rho\cos\theta+\rho)^3-6\rho^3\cos\theta\sin^2\theta>0\\\\ \iff &\rho^3(\cos\theta+1)^3-6\rho^3\cos\theta(1+\cos\theta)(1-\cos\theta)>0 \end{align}

and dividing by $$\rho^3(\cos\theta+1)>0$$ we reduce to

\begin{align} \iff &7\cos^2\theta-4\cos\theta+1>0 \end{align} which is true.

• Perhaps you mean $\cos\theta\neq -1?$ Jul 5 at 17:04
• @ThomasAndrews Exactly, I mean the case $a<0$ with $b=0$ for the original expression! Thanks, I fix that.
– user
Jul 5 at 17:07
• (+1) I hope that my editing is acceptable. If not, you can of course revert to the original. Jul 5 at 17:28
• It’s a little bit heavy but I let it in this way for the moment. Thanks
– user
Jul 5 at 17:32
• The fourth line,$$(\rho\cos\theta+\rho)^3-6\rho^3\cos\theta\sin^2\theta>0,$$ is an unproven assertion. Jul 6 at 6:46

Another way to prove your inequality :

\begin{align}\dfrac13&=\dfrac{2ab^2}{6ab^2}\geqslant\dfrac{2ab^2}{b(2a-b)^2+6ab^2}=\dfrac{2ab^2}{b\left(4a^2+b^2\right)+2ab^2}>\\[3pt]&\!\!\!\underset{\overbrace{\text{because }b}\dfrac{2ab^2}{\left(a+\sqrt{a^2+b^2}\right)\left(4a^2+b^2\right)+2ab^2}=\\[3pt]&\!\!\!\!\!\!\!=\!\dfrac{2ab^2}{\left(a\!+\!\sqrt{a^2\!+\!b^2}\right)\!\left[\left(a\!+\!\sqrt{a^2\!+\!b^2}\right)^2\!\!+\!2a\!\left(a\!-\!\sqrt{a^2\!+\!b^2}\right)\right]\!+\!2ab^2}\!=\\[3pt]&=\dfrac{2ab^2}{\left(a+\sqrt{a^2+b^2}\right)^3+2a\left[a^2-(a^2+b^2)\right]+2ab^2}=\\[3pt]&=\dfrac{2ab^2}{\left(a+\sqrt{a^2+b^2}\right)^3}\;.\\\\\end{align}

Actually it is possible to prove that $$\dfrac14\geqslant\dfrac{2ab^2}{\left(a+\sqrt{a^2+b^2}\right)^3}\;\;.$$

Proof:

\begin{align}\dfrac14&=\dfrac{2ab^2}{8ab^2}=\dfrac{2ab^2}{4b^2\left[\left(a+\sqrt{a^2+b^2}\right)+\left(a-\sqrt{a^2+b^2}\right)\right]}=\\[3pt]&=\dfrac{2ab^2\left(a+\sqrt{a^2+b^2}\right)}{4b^2\left[\left(a+\sqrt{a^2+b^2}\right)^2+\left(a^2-(a^2+b^2)\right)\right]}=\\[3pt]&=\dfrac{2ab^2\left(a+\sqrt{a^2+b^2}\right)}{4b^2\left(a+\sqrt{a^2+b^2}\right)^2\!-4b^4}\geqslant\\[3pt]&\geqslant\dfrac{2ab^2\left(a+\sqrt{a^2+b^2}\right)}{\left[\left(a\!+\!\sqrt{a^2\!+\!b^2}\right)^2\!\!-\!2b^2\right]^2 \!\!+\!4b^2\left(a\!+\!\sqrt{a^2\!+\!b^2}\right)^2\!\!-\!4b^4}=\\[3pt]&=\dfrac{2ab^2\left(a+\sqrt{a^2+b^2}\right)}{\left(a+\sqrt{a^2+b^2}\right)^4}=\dfrac{2ab^2}{\left(a+\sqrt{a^2+b^2}\right)^3}\;.\end{align}

Moreover ,

\begin{align}\dfrac14&=\dfrac{2ab^2}{\left(a\!+\!\sqrt{a^2\!+\!b^2}\right)^3}\iff\left(a\!+\!\sqrt{a^2\!+\!b^2}\right)^2\!=2b^2\iff\\[3pt]&\iff a\!+\!\sqrt{a^2\!+\!b^2}=b\sqrt2\iff\sqrt{a^2\!+\!b^2}=b\sqrt2\!-\!a\\[3pt]&\iff a^2+b^2=2b^2+a^2-2ab\sqrt2\iff\\[3pt]&\iff b\left(b\!-\!2a\sqrt2\right)=0\iff b=2a\sqrt2\;.\end{align}

In my answer (addendum included) I have only used very basic rules of algebra, for example “square of a binomial” and “conjugate binomial” which is the product of the sum of two terms multiplied by the subtraction of these terms.

• Thank you a lot for the addendum, I find it really interesting, as it helps solving the second part of the problem (I didn't post it here). But it is making me a little bit confused. In the second part of the problem, we need to prove that if the RHS of the inequality is multiplied by a number k, then equality is only reached if k is greater than or equal to 4, but your solution shows that this is not true, since k = 4/3 works. Is the second part of the problem simply wrong? I can post another question with the full problem, if needed. Jul 6 at 23:04
• @Ingrid, you can modify your post in order to complete it by adding the second part of the problem. If you do it, I will you help to solve the second part too. Jul 7 at 5:25
• I am going to post another question, since adding the second part here will make the other answers incomplete. Jul 7 at 11:37