Show $\|x\|^2$ is strictly convex Show $f(x)=\|x\|^2 = x^Tx$ is strictly convex by proving:
For distinct $x,y \in \mathbb{R}$ and $0<\lambda <1$, $f(\lambda x+(1-\lambda) y) < \lambda f(x)+(1-\lambda)f(y)$

I have tried to expand and simplify both sides (taking up a full page each try) but I am having trouble exactly relating the two. It would be too much to show all my work here but I will show the point that I have gotten to.
\begin{align}
\text{LHS:} \quad & \lambda^2\sum_{i=1}^n x_i^2+(1-\lambda)^2\sum_{i=1}^n y_i^2 +2\lambda(1-\lambda)\sum_{i=1}^n x_iy_i \\[10pt]
\text{RHS:} \quad & \lambda\sum_{i=1}^n x_i^2+(1-\lambda)\sum_{i=1}^n y_i^2
\end{align}
I think I have to use the fact that $\lambda^2<\lambda$ so I have that 
$$\lambda^2\sum_{i=1}^n x_i^2+(1-\lambda)^2\sum_{i=1}^n y_i^2 < \lambda\sum_{i=1}^n x_i^2+(1-\lambda)\sum_{i=1}^n y_i^2$$
But how can I show that the difference between the two sides of the inequality is greater than $2\lambda(1-\lambda)\sum_{i=1}^n x_iy_i$
 A: Suppose $\|x\|\neq \|y\|$
$$\lambda (1-\lambda)(\|x\|-\|y\|)^2>0\\\Rightarrow \lambda (1-\lambda)\|x\|^2+\lambda (1-\lambda)\|y\|^2-2\lambda (1-\lambda)\|x\|\|y\|>0\\\Rightarrow [\lambda-\lambda^2]\|x\|^2+[(1-\lambda)-(1-\lambda)^2]\|y\|^2-2\lambda (1-\lambda)\|x\|\|y\|>0\\\Rightarrow \lambda^2\|x\|^2+(1-\lambda)^2\|y\|^2+2\lambda (1-\lambda)\|x\|\|y\|<\lambda\|x\|^2+(1-\lambda)\|y\|^2\\\Rightarrow [\lambda\|x\|+(1-\lambda)\|y\|]^2<\lambda\|x\|^2+(1-\lambda)\|y\|^2$$
By Triangle inequality $$\|\lambda x+(1-\lambda) y\|\leq\lambda\|x\|+(1-\lambda)\|y\|$$
Hence $$\|\lambda x+(1-\lambda) y\|^2<\lambda\|x\|^2+(1-\lambda)\|y\|^2$$
Then $\|x\|^2$ is strictly convex. $\blacksquare$
A: $f(x) =\|x\|^2$  then $f''(x)=2I$ which is positive definite.
If you want use definition , you first compute 
$$\text{RHS} - \text{LHS}$$
And don't use that fact you mentioned.
A: You can use your working to show that
\begin{align}
& \lambda f(x) + (1-\lambda) f(y) - f(\lambda x + (1-\lambda)y) \\[10pt]
= {} & (\lambda-\lambda^2) x^Tx + ((1-\lambda)-(1-\lambda)^2) y^T y - 2\lambda(1-\lambda) x^T y \\[10pt]
= {} & \lambda(1-\lambda) (x-y)^T(x-y) = \lambda(1-\lambda) f(x-y) > 0
\end{align}
 if $x \neq y$ and $0<\lambda<1$. (This identity actually holds for any quadratic form.)
