If $\|x_1\| \leq \|y_1\|$ and $\|x_2\| \leq \|y_2\|$ then $\|\lambda x_1 + (1 - \lambda) x_2\| \leq \|\lambda y_1 + (1 - \lambda) y_2\|$ I have being running in circles in this problem. Let $x_1,x_2,y_1,y_2 \in \mathbb{R}^n_+$ such that $\|x_1\| \leq \|y_1\|$,$\|x_2\| \leq \|y_2\|$ and $y_1$ and $y_2$ are l.d.. Is it true that for any $\lambda \in (0,1)$ we have that
$$\|\lambda x_1 + (1- \lambda)x_2\| \leq \|\lambda y_1 + (1 - \lambda)y_2\|?$$
I attempt to square the inequality and use the convexity of $f(x) = x^2$ but it was fruitless.
 A: Take $x_1=x_2=y_1=e_1$ and $y_2=e_2$, where $(e_n)_n$ is the standard orthonormal basis in $\mathbb{R}^n$ (for $n\geq 2$). You have
$$
\lVert x_1\rVert_2 = \lVert x_2\rVert_2 = \lVert y_1\rVert_2 =\lVert y_2\rVert_2 =1
$$
and, for all $\lambda \in (0,1)$,
$$
\lVert \lambda x_1+(1-\lambda)x_2\rVert_2 = \lVert x_1\rVert_2 =1
$$
but
$$
\lVert \lambda y_1+(1-\lambda)y_2\rVert_2 = \sqrt{\lambda^2+(1-\lambda)^2} < 1
$$

Under the extra assumption that $y_1,y_2$ are linearly dependent but not that they have non-negative coordinates: take $e_1=x_1=x_2=y_1$ and $y_1=-2y_2$. Then, for $\lambda = 1/3$,
$$
\lVert \lambda y_1 + (1-\lambda) y_2\rVert_2
= |(1-3\lambda)|\lVert y_2\rVert_2 = 0
$$
but
$$
\lVert \lambda x_1 + (1-\lambda) x_2\rVert_2 = \frac{\sqrt{5}}{3} > 0.
$$

With all the assumptions: it's true. Since $y_1,y_2$ are linearly dependent, there exist $(\alpha,\beta)\neq (0,0)$ such that $\alpha y_1+\beta y_2=0_n$.

*

*Suppose either $\alpha=0$ or $\beta=0$; say, wlog, $\alpha =0$. Then$y_2=0_n$, and therefore $x_2=0$, and therefore the statement holds since $\lambda\lVert x_1\rVert_2 \leq \lambda\lVert y_1\rVert_2$.


*Otherwise, rewrite for convenience $y_2 = \gamma y_1$ for $\gamma> 0$ (the sign of $\gamma$ follows from the fact that $y_1,y_\in\mathbb{R}_+^n$).
$$
\lVert \lambda y_1+(1-\lambda)y_2\rVert_2 = \lVert \lambda y_1+\gamma(1-\lambda)y_1\rVert_2 = (\lambda+\gamma(1-\lambda)) \lVert y_1\rVert_2
$$
(note that $\lambda+\gamma(1-\lambda) \geq 0$); while
$$\begin{align}
\lVert \lambda x_1+(1-\lambda)x_2\rVert_2
&\leq \lambda\lVert x_1\rVert + (1-\lambda)\rVert x_2\rVert_2
\leq \lambda\lVert y_1\rVert + (1-\lambda)\rVert y_2\rVert_2\\
&=  \lambda\lVert y_1\rVert + \gamma (1-\lambda)\rVert y_1\rVert_2\\
&=  (\lambda+\gamma(1-\lambda)) \lVert y_1\rVert_2
\end{align}$$
proving the claim.
