Denote $A_{\lambda}=(x_1,\dots,x_n)$ and suppose that $i=n$. Then we have to prove $(n-1)\sigma_k(x_1,\dots,x_{n-1})\leqslant \sigma_1(x_1,\dots,x_{n-1})\sigma_{k-1}(x_1,\dots,x_{n-1})$. Denote $f(t)=(t-x_1)(t-x_2)\dots (t-x_{n-1})$, $g(t)=f^{(k-1)}(t)$, so $g$ is polynomial of degree $k$ with $k$ real roots $u_1,\dots,u_k$. We have to prove $$k^2u_1\dots u_k\leqslant (u_1+\dots+u_k)\sigma_{k-1}(u_1,\dots,u_k).\,\,\,\,(1)$$ (1) is obvious if all $u_i$ are non-negative, so suppose that $u_k<0$.
We are given that $f(t)(t-x_n)=t^{n}-c_1t^{n-1}+c_2t^{n-2}+\dots+(-1)^k c_k t^{n-k}+\dots$ for positive $c_1,\dots,c_k$, and $f(t)$ has only real roots. Taking $k$-th derivative we see by Rolle theorem that $(f(t)(t-x_n))^{(k)}$ has only real roots and they must be positive. So, $(t-x_n)g'(t)+kg(t)$ has $k$ positive roots. Thus $(g(t)\cdot (t-x_n)^k)'$ has a root $x_n$ of multiplicity $k-1$ and $k$ positive roots.
If $x_n\leqslant 0$, then by Rolle theorem polynomial $(g(t)\cdot (t-x_n)^k)'$ has a root between $x_n$ and $u_k$ (or has roots $x_n$ of multiplicity at least $k$ if $x_n=u_k$.) It contradicts to above observations. So, in this case $g$ has $k$ non-negative roots. Analogously, if $x_n\geqslant 0$ but $u_i\leqslant 0$ for some $i\ne k$, we get a negative root of $(g(t)\cdot (t-x_n)^k)'$.
So, $x_n>0$ and $u_1,\dots,u_{k-1}>0$. Since $W(t):=kg(t)+(t-x_n)g'(t)$ has $k$ positive roots and positive leading coefficient, we have $W(0)(-1)^{k}>0$. On the other hand, $g(0)(-1)^k=u_1\dots u_k<0$. Thus $(-1)^{k-1}x_ng'(0)>0$, i.e. $\sigma_{k-1}(u_1,\dots,u_k)>0$. It easily implies that $u_1+\dots+u_k>0$, thus $(u_1+\dots+u_k)\sigma_{k-1}(u_1,\dots,u_k)>0>k^2u_1\dots u_k$ as desired.