# Prove $( \frac{\sin ^{2}\theta}{2}+\frac{2}{\cos ^{2}\theta} )^{1/4} + ( \frac{\cos ^{2}\theta}{2}+\frac{2}{\sin ^{2}\theta} )^{1/4}\ge 68^{1/4}$

Let $$0<\theta<\pi/2$$. Prove that $$\left ( \frac{\sin ^{2}\theta}{2}+\frac{2}{\cos ^{2}\theta} \right )^{1/4}+\left ( \frac{\cos ^{2}\theta}{2}+\frac{2}{\sin ^{2}\theta} \right )^{1/4}\geqslant (68)^{1/4}$$ and find when the equality case holds.

This is a competition math problem. The material used should only cover up to pre-calculus.

So I quickly found out that equality holds when both of the $$\sin^2(\theta)$$ and $$\cos^2(\theta)$$ equals to 1/2, but I am not sure how to prove that this equality is true. I tried to substitute for variables and also use trig identities but just can't find out a way to do this. Thank you guys for helping me.

• It clearly blows up close to $0$ and $\pi/2$. So, have you done any calculus to find the minimum? Commented Nov 10, 2017 at 4:02
• the solution is a competition math problem so doing a calculus will be kind of strange-- the material used should only cover up to pre-cal. But yeah calculus will be really helpful here. Commented Nov 10, 2017 at 4:13
• @Randall is there any problem if that blows up? I think it makes the result very fine Commented Nov 10, 2017 at 15:19
• @GuyFsone no, it's a good thing since the inequality is a $\geq$. The question, then, is where does it min out. My point was the fact that it blows up makes it believeable. Commented Nov 10, 2017 at 15:42
• it min out a $\pi/4$ Commented Nov 10, 2017 at 15:53

By Minkowski (see here: https://en.wikipedia.org/wiki/Minkowski_inequality) we obtain $$\sqrt[4]{ \frac{\sin ^{2}\theta}{2}+\frac{2}{\cos ^{2}\theta}}+\sqrt[4]{ \frac{\cos ^{2}\theta}{2}+\frac{2}{\sin ^{2}\theta}}\geq\sqrt[4]{\frac{1}{2}(\sqrt{\sin\theta}+\sqrt{\cos\theta})^4+2\left(\frac{1}{\sqrt{\sin\theta}}+\frac{1}{\sqrt{\cos\theta}}\right)^4}.$$ Thus, it remains to prove that $$\frac{1}{2}(\sqrt{\sin\theta}+\sqrt{\cos\theta})^4+2\left(\frac{1}{\sqrt{\sin\theta}}+\frac{1}{\sqrt{\cos\theta}}\right)^4\geq68.$$

Now, let $\sin\theta+\cos\theta=2k\sqrt{\sin\theta\cos\theta}$.

Thus, $k\geq1$ and we need to prove that $$\frac{1}{2}(\sqrt{\sin\theta}+\sqrt{\cos\theta})^4+2(\sin^2\theta+\cos^2\theta)^2\left(\frac{1}{\sqrt{\sin\theta}}+\frac{1}{\sqrt{\cos\theta}}\right)^4\geq68(\sin^2\theta+\cos^2\theta)$$ or $$(k+1)^2+16(2k^2-1)^2(k+1)^2\geq68(2k^2-1)$$ or $$(k-1)(64k^5+192k^4+192k^3+64k^2-119k-85)\geq0,$$ which is obvious.

Done!

Since you proved that the equality holds for $\sin^2(\theta)=\cos^2(\theta)=\frac 12$, in the same spirit as PTDS's answer, consider the function $$f=\sqrt[4]{\frac{1-t}{2}+\frac{2}{t}}+\sqrt[4]{\frac{t}{2}+\frac{2}{1-t}}$$ and develop it as a Taylor series around $t=\frac 12$.

You should get a polynomial in $(t-\frac 12)^{2k}$ in which all coefficients are positive. Limited to very first orders $$f= \sqrt[4]{68}+\frac{77 }{4 \sqrt{2}\, 17^{3/4}}\left(t-\frac{1}{2}\right)^2+\frac{54803}{1088 \sqrt{2}\, 17^{3/4}} \left(t-\frac{1}{2}\right)^4+O\left(\left(t-\frac{1}{2}\right)^6\right)$$ Then $\sqrt[4]{68}$ is the minimum value.

An alternative formulation:

Let $\frac{\sin^2 \theta}{2} = x$ and $\frac{\cos^2 \theta}{2} = y$

Minimize $f(x,y) = (x+\frac{1}{y})^{\frac{1}{4}}+(y+\frac{1}{x})^{\frac{1}{4}}$ subject to $x+y = \frac{1}{2}, x > 0, y > 0$

$f(x,y)$ has a local minimum at $x = y = \frac{1}{4}$ (easy to show using Lagrange multipliers)

Let $\mathcal{L} = (x+\frac{1}{y})^{\frac{1}{4}}+(y+\frac{1}{x})^{\frac{1}{4}} + \lambda (x+y- \frac{1}{2})$

Set $\frac{\partial \mathcal{L}}{\partial x} = \frac{\partial \mathcal{L}}{\partial y} = \frac{\partial \mathcal{L}}{\partial \lambda} = 0$ and solve.

The first two equations imply $x = y$ and from the third equation, we finally have $x = y = \frac{1}{4}$

• Explain please how you are solving the system. Commented Nov 10, 2017 at 6:21

Clearly, we only need to prove the case when $$\theta \in (0, \pi/4]$$.

Let \begin{align*} A &= \frac{4}{17}\left(\frac{\sin^2 \theta}{2} + \frac{2}{\cos^2\theta}\right), \\ B &= \frac{4}{17}\left(\frac{\cos^2 \theta}{2} + \frac{2}{\sin^2\theta}\right). \end{align*} It suffices to prove that $$A^{1/4} + B^{1/4} \ge 2$$ or $$B^{1/4} - 1 \ge 1 - A^{1/4}$$ or $$\frac{B - 1}{(B^{1/4} + 1)(B^{1/2} + 1)} \ge \frac{1 - A}{(1 + A^{1/4})(1 + A^{1/2})}$$ or $$\frac{B - 1}{B^{3/4} + B^{1/2} + B^{1/4} + 1} \ge \frac{1 - A}{1 + A^{1/4} + A^{1/2} + A^{3/4}}.$$

It is easy to prove that $$B \ge 1 \ge A$$.

It suffices to prove that $$\frac{B - 1}{B + B + B + 1} \ge \frac{1 - A}{1 + A + A + A}$$ or $$3AB - A - B - 1 \ge 0$$ or (clearing the denominators) $$(3\sin^4\theta - 3\sin^2\theta + 56)(1 - 2\sin^2\theta)^2 \ge 0$$ which is true, with equality if $$\sin^2\theta = 1/2$$.

We are done.