I want to show that for $x,y \in \mathbb{R}^d$ and $\lambda \in [0,1]$:
$$ \lambda(1-\lambda) \|x-y\|_2^2 + \|\lambda x + (1-\lambda)y\|_2^2 \le (\lambda \|x\|_2^2 + (1-\lambda)\|y\|_2^2) $$
My idea is to try to show that the left hand side has a maximum at $\lambda=1/2$ and use convexity. At least in the the case $d=1$ we have equality, i.e.
$$ \frac{1}{4}(x-y)^2 + (\frac{1}{2}x - \frac{1}{2}y)^2 = \frac{1}{2}x^2 + \frac{1}{2}y^2 $$
which follows by factoring out. But I don't know how to generalize this for arbitrary $d$. Do I have to calculate the derivative of the left-hand side?
Notice that for $d=1$, it is equality: $\lambda(1-\lambda)(x-y)^2 + (\lambda x + (1-\lambda)y)^2 = \lambda x^2 + (1-\lambda)y^2$.
Now take $x = (x_1, ..., x_d)$ and $y = (y_1, ..., y_d)$, then $||x||_{2}^{2} = \sum x_i^2$.
Writing inequality like this, we have $$\lambda(1-\lambda)||x-y||_2^2 + ||\lambda x + (1-\lambda)y||_2^2 = \lambda(1-\lambda)\sum(x_i - y_i)^2 + \sum(\lambda x_i + (1-\lambda)y_i)^2 = $$ $$= \lambda\sum x_i^2 + (1-\lambda)\sum y_i^2= \lambda ||x||_2^2 + (1-\lambda)||y||_2^2$$