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

Question

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.

Answer 1

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

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

Answer 2

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 $$(u+\sqrt{u^2+1})^3>(3+2\sqrt3)u.$$

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Another approach uses $\frac1{u+\sqrt{u^2+1}}=\sqrt{u^2+1}-u.$ So if $w=\sqrt{u^2+1}-u,$ then $0<w<1$ 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 $u=\frac{\sqrt2}4.$

Answer 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.

Answer 4

Another way to prove your inequality :

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

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Addendum:

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.

Answer 5

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 $$E=\frac{2\sinh(t)}{e^{3t}} \Rightarrow E=\frac{e^t-e^{-t}}{e^{3t}}=e^{-2t}-e^{-4t}$$ $$\Rightarrow E=\frac{1}{4}-\left(e^{-2 t}-\frac{1}{2}\right)^2 \leqslant \frac{1}{4}<\frac{1}{3}$$