Is |$\vec{a}+\lambda\vec{b}$|>$|\vec{a}|$ true or false? 
|$\vec{a}+\lambda\vec{b}|\geq|\vec{a}|$ for all real $\lambda$ if
  $\vec{a}$ and $\vec{b}$ are perpendicular.

I need to verify the truth or falsity of this statement.What would be the best method?
P.S:Now I realize it was a really silly question which I asked.I guess my book's answer key is wrong as it claims the statement is false.
Thanks a lot to all those who answered!
 A: $|\vec{a}+\lambda \vec{b}|^2=(\vec{a}+\lambda \vec{b})\cdot (\vec{a}+\lambda \vec{b})=|\vec{a}|^2+\lambda^2|\vec{b}|^2+2\lambda\vec{a}\cdot\vec{b}=|\vec{a}|^2+\lambda^2|\vec{b}|^2\geq |\vec{a}|^2$
Thus the statement is true.
Another approach may be a visual one seeing that $\vec{a}$ is a side in a right triangle with hypotenuse $\vec{a}+\lambda\vec{b}$
A: Here is a picture displaying (a small part) of vectors having the form $u+\lambda \vec{v}$. 
One can observe values of $\lambda$ for which $\|u+\lambda \vec{v}\|<\|u\|$.

A: $$|\vec a+\lambda\vec b|=\sqrt{|\vec a|^2+\lambda^2|\vec b|^2+2|\vec a|\cdot|\vec b|\cos\frac{\pi}2}\ge|\vec a|$$
since $\vec a$ and $\vec b$ are perpendicular.
So, the inequality is true.
A: Let $\lambda\in\mathbb{R}$, and $a,b\in V$ be vectors in a real inner product space $V$. As the quantities $|a+\lambda b|$ and $|a|$ are positive, the inequality in question holds exactly when it holds for the squares:
$$ |a+\lambda b|^2 \geq |a|^2 $$
Apply the fact that $|v|^2 = v\cdot v$ for all $v\in V$:
$$ |a+\lambda b|^2 = (a+\lambda b)\cdot (a+\lambda b) = a\cdot a + 2\lambda a\cdot b + b\cdot b $$
As $a,b$ are perpendicular, $a\cdot b = 0$, and also $b\cdot b \geq 0$. This establishes the desired result.
