Prove that $\sin(x) + \cos(x) \geq 1$ $\forall x\in[0,\pi/2]: \sin{x}+\cos{x} \ge 1.$
I am really bad at trigonometric functions, how could I prove it?
 A: A straight line is the shortest path between two points, so the hypotenuse of a right-angled triangle is shorter than the sum of its legs. Hence the required inequality.
A: 
If you look at the trigonometric unit circle, you will see that $\sin \theta$ and $\cos\theta$ are the legs of a right triangle with  hypotenuse=$1$ and using triangle inequality you get that their sum is greater than or equal to the  hypotenuse.
A: Here is a way: $\sin x+\cos x=\sqrt{2}(\sin x\cos\frac{\pi}{4}+\cos x\sin\frac{\pi}{4})=\sqrt{2}\sin(x+\frac{\pi}{4})$ So you need to show that $\sqrt{2}\sin(x+\frac{\pi}{4})$ is greather or equal to $1$ on your given inteval. Divide both sides by $\sqrt{2}$ and see what you get...
A: Hint Since $0 \leq \sin(x) , \cos(x) \leq 1$ you have
$$\sin(x) \geq \sin^2(x) \\
\cos(x) \geq \cos^2(x) \\
\sin(x) +\cos(x) \geq \sin^2(x)+\cos^2(x) =1$$
A: squaring one times and observe that $$\sin(x)^2+\cos(x)^2=1$$ we get $$\sin(2x)<0$$ which is impossible. If we assume that $$\sin(x)+\cos(x)<1 $$
Sonnhard.
A: Re-arrange to get $\cos(x) < 1 - \sin(x)$
We know that $\sin^2 (x) + \cos^2(x) = 1$ and therefor that $|\cos(x)| = \sqrt{1-\sin^2(x)}$
Now we want to know if $\sqrt{1-\sin^2(x)} < 1 - \sin(x)$
Now we square both sides: $1-\sin^2(x) < (1 - \sin(x))^2 = 1 - 2\sin(x) + \sin^2(x)$
Re-arranging again we get $0 < 2\sin^2(x) - 2\sin(x)$
In order for $\sin^2(x) > \sin(x)$ we would need a $\sin(x)$ value greater than 1. This is not possible, and therefor neither is the original form of the inequality $\cos(x) < 1 - \sin(x)$.
A: you can consider the function $f(x) = \sin{x} +\cos{x} -1$ then see that in $[0,\pi /4)$ ,
$f'(x) >0$ 
and in $(\pi/4, \pi/2], f'(x) <0$ 
now see what happens at $0$ and $\pi/2$
A: you can write $$\sin x + \cos x = \sqrt 2 \left(\sin x \cos (\pi/4) + \cos x\sin(\pi/4)\right) = \sqrt 2 \sin(x+\pi/4)$$ now use the fact that 
$\frac{ \sqrt 2} 2 \le \sin(x + \pi/4) \le 1 \text{ for } 0 \le x \le \pi/2$ to conclude $$1 \le \sin x + \cos x  \le \sqrt 2 \text{ for } 0 \le x \le \pi/2.$$
A: Let $E=\sin(x)+\cos(x)$
Multiply and divide with $\sqrt2$. 
$$\begin{align}
E&=\sqrt2\left(\frac{\sin(x)}{\sqrt2} +\frac{\cos(x)}{\sqrt2}\right)\\ 
\implies E&= \sqrt2\sin\left(\frac \pi 4 + x\right)\\
\end{align}$$
Let $(\frac \pi4 + x)$ be $y$ then range of $y\in \left(\frac\pi4,\frac {3\pi}4\right)$ 
So $E= \sqrt2\sin(y)$.
Since max value of $\sin(y)$ is $1$ and min value is $\frac1{\sqrt2}$.
Therefore $\sqrt2 \ge E \ge 1$.
