In a Real inner product space $S$, $a \perp b \iff ||a + kb|| \ge ||a||$ for all $a, b \in S$ and all $k \in \mathbb{R}$.

Question

In a Real inner product space $S$, $a \perp b \iff ||a + kb|| \ge ||a||$ for all $a, b \in S$ and all $k \in \mathbb{R}$.

My attempt :

Let, $a\perp b$. Then $<a,b>=0$.

Now for any $k\in \mathbb{R}$,

$$|| a+kb || \\ = \sqrt{<a,a>+2k<a,b>+k^2<b,b>} \\ =\sqrt{<a,a>+k^2<b,b>}\ \ \ [\because<a,b>=0] \\ \ge \sqrt{<a,a>} \ \ \ [\because k^2 \ge 0 \ \text{and} <b,b> \ \ge 0] \\ =||a||$$

But I can't prove the converse part. Is the converse part wrong?

Can anyone help me?

Answer 1

We have : $$\forall k \in \mathbb{R}, \|a + k b\| \geq \|a\|$$ then : $$\forall k \in \mathbb{R}, \|a + k b\|^2 \geq \|a\|^2$$ then : $$\forall k \in \mathbb{R}, \|a\|^2 + 2 k a \cdot b + k^2 \|b\|^2 \geq \|a\|^2$$ we deduce that : $$\forall k \in \mathbb{R}, 2 k a \cdot b + k^2 \|b\|^2 \geq 0$$

1. Method 1 : Consider the polynom : $$p(k) = 2 k a \cdot b + k^2 \|b\|^2 = k^2 \|b\|^2 + 2 k a \cdot b$$ We have : $$\forall k \in \mathbb{R}, p(k) \geq 0$$ then it's discriminant : $$\Delta \leq 0$$ then : $$(a \cdot b)^2 \leq 0$$ and because : $$(a \cdot b)^2 \geq 0$$ we deduce that ; $$a \cdot b = 0$$ then $a \perp b$.
2. Method 2 : Assume that $a \cdot b \neq 0$ then for : $$k = -\dfrac{a \cdot b}{\|b\|^2}$$ $$2 k a \cdot b + k^2 \|b\|^2 = - 2 \dfrac{(a \cdot b)^2}{\|b\|^2} + \dfrac{(a \cdot b)^2}{\|b\|^4} \|b\|^2 = - 2 \dfrac{(a \cdot b)^2}{\|b\|^2} + \dfrac{(a \cdot b)^2}{\|b\|^2} = \dfrac{(a \cdot b)^2}{\|b\|^2} < 0$$ Absurd then $a \cdot b = 0$. We conclude that $a \perp b$.