Prove $4$ is the smallest value for $k$ so this equality holds

Question

Let $a$, $b$ and $k$ be positive reals, such that

$$\frac{1}{3} = k\cdot 2 \frac{ab^2}{(a + \sqrt{a^2 + b^2})^3}.$$

Prove $4$ is the smallest possible value for $k$.

This problem is related to this one I asked in another question, but they are asking different things. This time, I am not really asking for a hint or solution (even though this would also be fine), but for some clarification about the problem. In the other question, user Angelo proved the following inequality:

$$\frac{1}{4} \geq 2 \frac{ab^2}{(a + \sqrt{a^2 + b^2})^3}$$

Then, they showed equality is reached when $b = 2a\sqrt{2}$. But doesn't this contradicts the statement of the problem? If I set $k = \frac{4}{3}$:

$\dfrac{1}{3} = \dfrac{4}{3}\cdot 2 \dfrac{ab^2}{(a + \sqrt{a^2 + b^2})^3}$

$\dfrac{1}{3} \cdot \dfrac{3}{4} = 2 \dfrac{ab^2}{(a + \sqrt{a^2 + b^2})^3}$

$\dfrac{1}{4} = 2 \dfrac{ab^2}{(a + \sqrt{a^2 + b^2})^3}$

Which is true when $b= 2a\sqrt{2}$, so $4$ is not actually the smallest possible value for $k$. Am I missing something? Is the problem simply wrong or am I misinterpreting it? Can anyone shed some light?

Answer 1

This is not true.

For $a=\dfrac{1}{2}\sqrt{\dfrac{3}{2}-\dfrac{5}{2\sqrt{3}}}$ and $b=1$, one has $$\frac{ab^2}{(a + \sqrt{a^2 + b^2})^3}=\dfrac{1}{12}$$

and therefore, $k=2$ works.

(I interpreted the question with $k$ being an integer, but if $k$ can take any positive real value, then $a=b=1$ gives an easier counterexample)

Answer 2

We have that

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

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

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

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

$$\begin{align} \iff &(6k+1)\cos^2\theta-(6k-2)\cos\theta+1=0 \end{align} $$ which leads to

$$\cos\theta=\frac{(3k-1)\pm\sqrt{3k(3k-4)}}{6k+1}$$

which requires $k\ge \frac 43$ and for $k=\frac43$ we obtain $\cos \theta=\frac13$, that is

• $a=\frac{\rho }3$
• $b=\frac{2\sqrt 2\rho }3$

indeed

$$\frac{1}{3} =2\frac43 \frac{\frac{\rho }3\left(\frac{2\sqrt 2\rho}3\right)^2}{\left(\frac{\rho }3 + \sqrt{\left(\frac \rho 3\right)^2 + \left(\frac{2\sqrt 2\rho }3\right)^2}\right)^3}$$

therefore the minimum positive value is $\boxed{k=\frac43}$.

Answer 3

A really nice linear substitution works here.

Let $b = u \cdot a \,, \space u > 0$ for variable $u$.

Then,

$$\dfrac{1}{3} = \dfrac{k \cdot 2a\cdot b^2}{\left(a + \sqrt{a^2 + b^2}\right)^3} \\\iff \dfrac{k \cdot 2a^3\cdot u^2}{\left(a + |a|\sqrt{1 + u^2}\right)^3} = \dfrac{2k \cdot u^2}{\left(1 + \sqrt{1 + u^2}\right)^3} \,, \text{since } \space a > 0$$

Now, if $k$ has a minimum value; then this min value must occur when $f(u) = \dfrac{2\cdot u^2}{\left(1 + \sqrt{1 + u^2}\right)^3}$ reaches a maximum. The little tedious but doable part is to find $f'(u) = 0$, such that we can obtain $\max{(f(u))}$.

$$\left(\dfrac{w}{v}\right)' = 0 \implies \dfrac{v\cdot w' - w\cdot v'}{v^2} = 0 \implies v\cdot w' - w\cdot v' = 0 \,, \text{since } \space v^2 \neq 0$$

Hence,

$$f'(u) = 0 \implies 16u\sqrt{u^2 + 1} - 2u\cdot\left(u^4 - 4u^2 - 8\right) = 0 \\\iff 8\sqrt{u^2 + 1} = u^4 - 4u^2 - 8 \,, \space u \neq 0 \\\iff u^8 - 8u^6 + 64u^2 + 64 = 64u^2 + 64 \\\iff u^8 - 8u^6 = 0 \\\iff u^2 - 8 = 0\,, \space u \ge 0 \\ \therefore u = 2\sqrt{2} \\ \quad \\ \dfrac{1}{3} = k \cdot f(2\sqrt{2}) = k \cdot \dfrac{1}{4} \\ \therefore k_{\text{min}} = \dfrac{4}{3}$$

Answer 4

$\dfrac13=k\cdot2\dfrac{ab^2}{\left(a+\sqrt{a^2+ b^2}\right)^3}$

is equivalent to each of the following equalities :

$\left(a+\sqrt{a^2+b^2}\right)^3=6kab^2\;\;,$

$\left(a+\sqrt{a^2+b^2}\right)^3\!=3kb^2\!\left(a+\sqrt{a^2+b^2}+a-\sqrt{a^2+b^2}\right)\;\;,$

$\left(a+\sqrt{a^2+b^2}\right)^4=3kb^2\left[\left(a+\sqrt{a^2+b^2}\right)^2-b^2\right]\;\;,$

$\left(a+\sqrt{a^2+b^2}\right)^4-3kb^2\left(a+\sqrt{a^2+b^2}\right)^2=-3kb^4\;\;,$

$\left[\left(a+\sqrt{a^2+b^2}\right)^2-\dfrac32kb^2\right]^2-\dfrac94k^2b^4=-3kb^4\;\;,$

$\left[\left(a+\sqrt{a^2+b^2}\right)^2-\dfrac32kb^2\right]^2=\dfrac94k^2b^4-3kb^4\;\;,$

$\left[\left(a+\sqrt{a^2+b^2}\right)^2-\dfrac32kb^2\right]^2=\dfrac94kb^4\left(k-\dfrac43\right)\;,$

$\dfrac4{9kb^4}\left[\left(a+\sqrt{a^2+b^2}\right)^2-\dfrac32kb^2\right]^2\!=k-\dfrac43\;.$

Since the left hand side of the last equality is nonnegative, it follows that

$k-\dfrac43\geqslant0\;\;,\quad$ consequently ,

$k\geqslant\dfrac43\,.$