Let $\alpha \in \left( 0, \dfrac{\pi}{2}\right)$.Find the minimum value of the expression $$ P = (1+\cos\alpha)\left(1+\frac{1}{\sin \alpha} \right)+(1+\sin \alpha)\left(1+\frac{1}{\cos \alpha} \right) .$$
Problem
Source: MDA TST 2016, 9
Tags: algebra
NikoIsLife
26.03.2019 13:26
The expression is also equal to
$$P=\left(1+\frac1{\sin\alpha}\right)\left(1+\frac1{\cos\alpha}\right)$$And, using Cauchy-Schwarz,
\begin{align*}
P&\ge\left(1+\frac1{\sqrt{\sin\alpha\cos\alpha}}\right)^2\\
&=\left(1+\frac1{\sqrt{\frac12\sin2\alpha}}\right)^2\\
&\ge\left(1+\frac1{\sqrt{\frac12}}\right)^2\\
&=\boxed{3+2\sqrt2}
\end{align*}and equality occurs when $\alpha=\frac\pi4$.
Vrangr
26.03.2019 15:42
You wouldn't believe it was almost a 21st Century Putnam problem if I didn't tell you. This proof is the same idea as the Putnam proof, Replace $\alpha$ with $x$ for convenience. \[P = \sin(x) + \cos(x) + \tan(x) + \cot(x) + \sec(x) + \csc(x) + 2\]Let $t + 1 = \sin(x) + \cos(x) = \sqrt{2} \sin\left(x + \frac{\pi}{4}\right)$, $t \in (0, \sqrt{2} - 1]$. \[P(t) = t + \frac{2}{t} + 3\]Now, $P'(t) = 1 - 2t^{-2} \le0$ for $t^2 \le 2$, so $P(t)$ is decreasing in the interval $(0, \sqrt{2} - 1]$. Hence, minimum is achieved when $t = \sqrt{2} - 1$, ie, when $x = \frac{\pi}{4}$. \[P_{\min} = 4 + 3\sqrt{2}.\] @above's proof is incorrect because the first step itself is wrong.