Proving that $(x^TAx)^2 - x^TA^2x \le 0$ for $x^Tx = 1$. Given arbitrary $x \in \mathbb{R}^n$ and $A = A^T \in \mathbb{R}^{n \times n}$ such that $x^Tx = 1$ prove that:
$$(x^TAx)^2 - x^TA^2x \le 0 $$
My attempt
Diagonalize $A$ as $A = L\Lambda L^T$. Thus $A^2 = L\Lambda^2 L^T$. Let $w = L^Tx$.
Thus we need to prove
$$(w^T\Lambda w)^2 - w^T\Lambda^2w \le 0 \quad \text{for} \quad w^Tw=1 $$
I'm stuck here. 
 A: Use the standard inner product space structure $(\cdot|\cdot)$ on $\mathbb R^n$, the LHS of your inequality regarding $w$'s is just 
$$
|(w | \varLambda w)|^2 - |\varLambda w|^2.
$$
The assumption is $|w|^2 = 1$, so the inequality is just
$$
|(w | \varLambda w)|^2 - |\varLambda w|^2 |w|^2 \leqslant 0,  
$$
which is true by the Cauchy--Schwartz inequality of inner products. 
A: $w^{T}\Lambda w=\sum d_iw_i^{2}$ where $d_i$'s are the diagonal elements. What you need is $(\sum d_iw_i^{2})^{2} \leq \sum d_i^{2}w_i^{2}$. Apply Cauchy -Scawrtaz inequlaity $(\sum a_ib_i)^{2} \leq  \sum a_i^{2}\sum b_i^{2}$ with $a_i=d_iw_i$ and $b_i=w_i$. 
A: Another approach.  
The claim 
$$(x^TAx)^2 - x^TA^2x \le 0 $$
would be proved if starting from a little transformed inequality
$$ x^TA^2x-(x^TAx)(x^TAx)= x^T(A^2- Ax x^TA)x \ge 0  $$
we could prove that $(A^2- Ax x^TA)$ is positive semi-definite.    
Indeed $V= x x^T$ is projection matrix ( $x^Tx=1$ , it is   projection onto a line) hence we have $V=V^T$ and $V=V^2$.
$A^2- Ax x^TA= AIA-AVA=A(I-V)A^T$ 
But $P=I-V$ is also a projection matrix with $P=P^2=PP^T$. 
Then $APP^TA^T=(AP)(AP)^T=BB^T  $ what means that 
the matrix   $(A^2- Ax x^TA)$ is PSD  .
