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Let we have $(X,\langle,\rangle)$ inner product space , we know that we can define a norm on $X$ by $\|x\|=\sqrt{\langle x,x\rangle}$ how can I prove that $(X,\|\cdot\|)$ is strictly convex normed space?

By using the following definition of strictly convex normed space : If $x$ does not equal to $0$ and $y$ does not equal to $0$ and $\|x+y\|=\|x\|+\|y\|$ then $y=ax$ where $a>0$

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    Note that proper notation is $\langle x,x\rangle$, not $$. – 2017-02-07
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    The parallelogram identity shows that an inner product space is even uniformly convex. – 2017-02-07

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Let us say that we have two distinct points $x,y$ on the unit sphere. Then

$$\|\tfrac{1}{2}(x+y)\|^2=\tfrac{1}{2}\|x\|^2+\tfrac{1}{2}\|y\|^2-\tfrac{1}{4}\underbrace{\|x-y\|^2}_{>0}< 1.$$

Consequently, $\|x-y\|<2$.

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    Can you see my question please ? Thank you sir – 2017-02-08
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    Is the last line supposed to say $\|x+y\|<2$ rather than $\|x-y\|<2$? (I guess it is just a typo, but I've decided to ask in a comment rather than just edit the answer - just in case I missed something.) – 2017-09-14