Proving a Trigonometric inequality For this question $x$ satisfies $0 \leq x < \pi/2$
Prove that:  $1 \leq \sec x \leq 1 + \tan x$
I'm not sure how to start this problem.
I tried changing $\sec x$ to $1/\cos x$ and $\tan x$ to $\sin x/\cos x$ to get:
$1 \leq 1/\cos x \leq 1 + \sin x/\cos x$
And then multiplying by $\cos x$ to get:
$\cos x \leq 1 \leq \cos x + \sin x$
I'm not sure if that is the correct way to solve this and if it is, I don't know where to go from here.
 A: For $0\le x<\dfrac\pi2,1\ge\cos x>0\implies1\le\sec x<\infty$
Now for $\sec x\le1+\tan x\iff1\le\cos x+\sin x$ as $\cos x>0$
Now $(\cos x+\sin x)^2=1+\sin2x\ge1+0$ as $0\le2x\le\pi$
$\implies\cos x+\sin x\ge1$
A: $1>\cos(x)>0$, on this interval, therefore $1 \le\sec(x)<+\infty$ on this interval. 
$\sec(x)=\displaystyle \frac{1}{\cos(x)}<1+\tan(x)=\frac{\cos+\sin(x)}{\cos(x)}$.
This is because $\cos(x)+\sin(x)>1$, since $\cos(x)+\sin(x)=\sqrt{2}\sin(x+\frac{π}{4})$, which $>1$ on the interval.
Therefore, both sides of the inequality are true, and it holds in the interval $0 \le x<\displaystyle \frac{π}{2}$
A: The left inequality it's $1\geq\cos{x}$, which is obvious.
The right inequality it's
$$\sin{x}+\cos{x}\geq1$$ or
$$2\sin{\frac{x}{2}}\cos\frac{x}{2}\geq2\sin^2\frac{x}{2}$$ or
$$\tan\frac{x}{2}\leq1,$$
which is true for given value of $x$.
A: The first part is obvious:
$$\cos x \leq 1$$
The second part is also trivial:
$$\cos x + \sin x  = \sqrt{2} (\sin{\pi \over {4}} \sin x + \cos{\pi \over {4}} \cos x) = \sqrt{2}\cos(x - {\pi \over {4}})$$
For angle $x$ between 0 and $\pi/2$:
$$- {\pi \over 4} \le x - {\pi \over 4} \leq {\pi \over 4}$$
which means that:
$$\cos{(x - {\pi \over {4}})} \ge {\sqrt{2} \over 2}$$
Therefore:
$$\cos{x} + \sin{x} \ge 1$$
A: $0 \leq x < \pi/2 $ (1)
(1) $\implies\cos x$ is decreasing and $\cos(0)=1 \geq0$, $\cos(\pi/2) = 0\geq 0$ (2)
(1) $\implies\sin x$ is increasing and $\sin(0)=0 \geq0$, $\sin(\pi/2) = 1\geq 0$ (3)
(2) & (3) $\implies 2\cos(x)\sin(x)\geq 0 \implies (\cos(x)+\sin(x))^2\geq \cos^2(x)+\sin^2(x) = 1 $  (4)
(2) & (4) $\implies \cos x \leq 1 \leq \cos x + \sin x$ and now by doing the reverse manipulations you did to the inequality you're supposed to prove you prove it
