Validity of $\sum\limits_{i=1}^n(a_i^2+b_i^2+c_i^2+d_i^2)\lambda_i\geq\lambda_1+\lambda_2+\lambda_3+\lambda_4$? Suppose that $\lambda_1\leq\lambda_2\leq\dots\leq\lambda_n$ is a sequence of real numbers. Clearly, if $a=(a_1,\dots, a_n)$ is a unit vector, then $\sum\limits_{i=1}^na_i^2\lambda_i\geq \lambda_1$. I want to see if the following generalization is true or not: 
If  $a=(a_1,\dots, a_n)$,  $b=(b_1,\dots, b_n)$, $c=(c_1,\dots, c_n)$, and $d=(d_1,\dots, d_n)$ ($n\geq 4$) form an orthonormal set, I wonder if we have
 $\sum\limits_{i=1}^n(a_i^2+b_i^2+c_i^2+d_i^2)\lambda_i\geq\lambda_1+\lambda_2+\lambda_3+\lambda_4$.

It doesnt hold: if $\lambda_1=x<0$ and $\lambda_i=0, i=2..n$, your inequality becomes $(a_1^2+b_1^2+c_1^2+d_1^2)x\geq x$ which becomes false if we find and orthogonal system $(a,b,c,d)$ such as $ a_1^2+b_1^2+c_1^2+d_1^2>1$. For example
$a=(\frac{\sqrt{2}}{2},  -\frac{\sqrt{2}}{2},0,...,0)$, 
$b=(\frac{\sqrt{3}}{3},\frac{\sqrt{3}}{3},\frac{\sqrt{6}}{6},\frac{\sqrt{6}}{6},0,...,0)$,
$c=(\frac{\sqrt{6}}{6},\frac{\sqrt{6}}{6},-\frac{\sqrt{3}}{3},-\frac{\sqrt{3}}{3},0...,0)$,
$d=(0,0,0,0,1,...,0)$.
