A normed linear space $X$ is said to be rotund if for all $x,y\in X$ with $\|x\|=1=\|y\|$, $\|x+y\|<2$.
I want to prove that a normed linear space $X$ is rotund iff the function $\varphi:X\to \mathbb{R}$ defined by $\varphi(x)=\frac{1}{2}\|x\|^2$ for all $x\in X$ is strictly convex.
Suppose $\varphi$ is strictly convex. Let $x,y\in X$ such that $\|x\|=1=\|y\|$. Then $\|\frac{x+y}{2}\|^2=2\varphi(\frac{x}{2}+\frac{y}{2})<\varphi(x)+\varphi(y)=\frac{1}{2}+\frac{1}{2}=1$. Thus $\|x+y\|<2$ and so $X$ is rotund. But how to prove the converse. Any hint will be thankfully appreciated.
We can use a different equivalent definition of rotund given in an "An introduction to Banach space theory" by R.E. Megginson.
A normed space is rotund if for all $x,y\in X$ with $\|x\|=\|y\|=1$ and all $t\in (0,1)$ we have $\|tx+(1-t)y\|<1$. This is equivalent to your definition by Proposition 5.1.2 in this book.
Let $x,y\in X$ and let $\lambda\in(0,1)$. We have to show that $\varphi(\lambda x+(1-\lambda)y)<\lambda\varphi(x)+(1-\lambda)\varphi(y)$. Let $x'=\frac{x}{\|x\|}$, let $y'=\frac{y}{\|y\|}$, let $c=\lambda\|x\|+(1-\lambda)\|y\|$ and let $\mu=\frac{\lambda\|x\|}{c}\in(0,1)$. Note that $$\varphi(\lambda x+(1-\lambda)y)=c^{2}\varphi(\mu x'+(1-\mu)y')=\frac{c^{2}}{2}\|\mu x'+(1-\mu)y'\|^{2}<\frac{c^{2}}{2}\\=\frac{\lambda^{2}\|x\|^{2}+2(\lambda-\lambda^{2})\|x\|\|y\|+(1-\lambda)^{2}\|y\|^{2}}{2}\\=\frac{\lambda\|x\|^{2}+(1-\lambda)\|y\|^{2}+(\lambda^{2}-\lambda)\|x\|^{2}+2(\lambda-\lambda^{2})\|x\|\|y\|+(\lambda^{2}-\lambda)\|y\|^{2}}{2}=(*).$$ Since $2\|x\|\|y\|\leq\|x\|^{2}+\|y\|^{2}$ we find $$(*)\leq\frac{\lambda\|x\|^{2}+(1-\lambda)\|y\|^{2}}{2}=\lambda\|x\|^{2}\varphi(x')+(1-\lambda)\|y\|^{2}\varphi(y')=\lambda\varphi(x)+(1-\lambda)\varphi(y).$$ So $\varphi$ is strictly convex.